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This notebook \ presents thirty-two integral and series representations of Catalan's \ constant, some of them new. For each representation, a proof is given, \ accessible by pressing the proof button. Often the proof consists just in \ evaluating the representation directly in ", StyleBox["Mathematica", FontSlant->"Italic"], ". 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Applying them to the result of integration we \ obtain\ \>", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(%%% /. %%\) /. %\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/4\), " ", "\[ImaginaryI]", " ", RowBox[{"(", RowBox[{ RowBox[{\(-2\), " ", RowBox[{"(", RowBox[{ RowBox[{"\[ImaginaryI]", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\^2\/48\), "-", \(1\/2\ \(\(log\^2\)(1\/2 - \[ImaginaryI]\/2)\)\)}], ")"}]}], "+", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{\(-\[ImaginaryI]\), " ", TagBox["C", Catalan]}], "+", \(\[Pi]\^2\/48\), "-", \(1\/2\ \(\(log\^2\)(1\/2 + \[ImaginaryI]\/2)\)\)}], ")"}]}], "+", \(\(log(1\/2 - \[ImaginaryI]\/2)\)\ \(log(2)\)\), "-", \(\(log(1\/2 + \[ImaginaryI]\/2)\)\ \(log(2)\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(-\(\[Integral]\_0\%1 \(\(log(\ \(1 - x\^2\)\/2)\)\/\(x\^2 + 1\)\) \[DifferentialD]x\)\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[BoxData[ \(\(-\(\[Integral]\_0\%1\( Log[1\/2\ \((1 - x\^2)\)]\/\(x\^2 + 1\)\) \[DifferentialD]x\)\)\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/4\ \[ImaginaryI]\ \(( \[Pi]\^2 + \(log\^2\)(\(-1\) - \[ImaginaryI]) - \(log\^2\)(\(-1\) + \[ImaginaryI]) - 2\ \[ImaginaryI]\ \[Pi]\ \(log(1\/2 - \[ImaginaryI]\/2)\) + 2\ \[ImaginaryI]\ \[Pi]\ \(log(1\/2 + \[ImaginaryI]\/2)\) + \[ImaginaryI]\ \[Pi]\ \(log(2)\) + 2\ \(\(Li\_2\)(1\/2 - \[ImaginaryI]\/2)\) - 2\ \(\(Li\_2\)(1\/2 + \[ImaginaryI]\/2)\))\)\)], "Output"] }, Open ]], Cell["In the same manner as in the previous section ", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(% /. {\n\t\t PolyLog[2, 1\/2 - \[ImaginaryI]\/2] \[Rule] \[Pi]\^2\/48 - 1\/2\ Log[1\/2 + \[ImaginaryI]\/2]\^2 - \ \[ImaginaryI]\ Catalan\ , PolyLog[2, 1\/2 + \[ImaginaryI]\/2] \[Rule] \[Pi]\^2\/48 - 1\/2\ Log[1\/2 - \[ImaginaryI]\/2]\^2 + \[ImaginaryI]\ Catalan\ \n\t\t}\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/4\), " ", "\[ImaginaryI]", " ", RowBox[{"(", RowBox[{ \(\[Pi]\^2\), "+", \(\(log\^2\)(\(-1\) - \[ImaginaryI])\), "-", \(\(log\^2\)(\(-1\) + \[ImaginaryI])\), "-", \(2\ \[ImaginaryI]\ \[Pi]\ \(log(1\/2 - \[ImaginaryI]\/2)\)\), "-", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{"\[ImaginaryI]", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\^2\/48\), "-", \(1\/2\ \(\(log\^2\)(1\/2 - \[ImaginaryI]\/2)\)\)}], ")"}]}], "+", \(2\ \[ImaginaryI]\ \[Pi]\ \(log(1\/2 + \[ImaginaryI]\/2)\)\), "+", RowBox[{"2", " ", RowBox[{"(", RowBox[{ RowBox[{\(-\[ImaginaryI]\), " ", TagBox["C", Catalan]}], "+", \(\[Pi]\^2\/48\), "-", \(1\/2\ \(\(log\^2\)(1\/2 + \[ImaginaryI]\/2)\)\)}], ")"}]}], "+", \(\[ImaginaryI]\ \[Pi]\ \(log(2)\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(3\/4\ \(\[Integral]\_0\%\(\[Pi]\/6\)\(x\/\(sin(x)\)\) \[DifferentialD]x\) + 1\/8\ \[Pi]\ log \((\@3 + 2)\)\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[BoxData[ \(3\/4\ \(\[Integral]\_0\%\(\[Pi]\/6\)\(x\/Sin[x]\) \[DifferentialD]x\) + 1\/8\ \[Pi]\ Log[\@3 + 2]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/8\ \[Pi]\ \(log(2 + \@3)\) + 1\/16\ \[ImaginaryI]\ \(( 3\ \[Pi]\^2 - 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 - \@\(-1\)\%6)\) + 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 + \@\(-1\)\%6)\) + 12\ \(\(Li\_2\)(\(-\@\(-1\)\%6\))\) - 12\ \(\(Li\_2\)(\@\(-1\)\%6)\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "From the reflection formula for the dilogarithm ", Cell[BoxData[ \(TraditionalForm \`\(Li\_2\)(z) = \(-\(1\/2\)\)\ \(\(log\^2\)(\(-z\))\) - \[Pi]\^2\/6 - \(Li\_2\)(1\/z)\)]], "\n", "with ", Cell[BoxData[ \(TraditionalForm\`z = \(-\@\(-1\)\%6\)\)]], " it follows that" }], "Text"], Cell[BoxData[ \(\(PolyLog[2, \(-\((\(-1\))\)\^\(1/6\)\)] \[Rule] \(-\(\(11\ \[Pi]\^2\)\/72\)\) - PolyLog[2, \((\(-1\))\)\^\(5/6\)]; \)\)], "Input"], Cell["Applying this to the result of the integration, we have", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%% /. %]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/96\ \[ImaginaryI]\ \(( \[Pi]\^2 - 6\ \[ImaginaryI]\ \[Pi]\ \(log(2 - \@3)\) - 6\ \[ImaginaryI]\ \[Pi]\ \(log(2 + \@3)\) - 72\ \(\(Li\_2\)(\@\(-1\)\%6)\) - 72\ \(\(Li\_2\)(\((\(-1\))\)\^\(5/6\))\))\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(% /. \[Rho]_\ Log[\[Alpha]_]\ + \ \[Rho]_\ Log[\[Beta]_]\ \[RuleDelayed] \[Rho]\ Log[\[Alpha]\ \[Beta] // Expand]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/96\ \[ImaginaryI]\ \(( \[Pi]\^2 - 72\ \(\(Li\_2\)(\@\(-1\)\%6)\) - 72\ \(\(Li\_2\)(\((\(-1\))\)\^\(5/6\))\))\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Collect[%, \ Pi, \ Simplify]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(\[ImaginaryI]\ \[Pi]\^2\)\/96 - 3\/4\ \[ImaginaryI]\ \(( \(Li\_2\)(\@\(-1\)\%6) + \(Li\_2\)(\((\(-1\))\)\^\(5/6\)))\)\)], "Output"] }, Open ]], Cell[TextData[{ "Using the following well-known property of the dilogarithm ", Cell[BoxData[ \(TraditionalForm \`\(Li\_2\)(t\^q) = q\ \(\[Sum]\+\(j = 0\)\%\(q - 1\)\(Li\_2\)( \[ExponentialE]\^\(\(2\ \[Pi]\ \[ImaginaryI]\ j\)\/q\)\ t) \)\)]], "\nwith ", Cell[BoxData[ \(TraditionalForm\`t = \(-\[ImaginaryI]\)\)]], " and ", Cell[BoxData[ \(TraditionalForm\`q = 3\)]] }], "Text"], Cell["\<\ PolyLog[2, t^q] = q*Sum[PolyLog[2, Exp[(2*Pi*I*j)/q]*t], {j, 0, q - \ 1}]\ \>", "Input", FormatType->StandardForm], Cell[BoxData[ \(\(PolyLog[2, \((\(-1\))\)\^\(1/6\)] + PolyLog[2, \((\(-1\))\)\^\(5/6\)] \[Rule] \(4\ \[ImaginaryI]\)\/3\ \ Catalan + \[Pi]\^2\/72; \)\)], "Input"], Cell["we finally obtain", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[%% /. %]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { RowBox[{ RowBox[{"-", StyleBox[\(\[Pi]\^2\/4\), ScriptLevel->0]}], StyleBox[" ", ScriptLevel->0], StyleBox[ \(\[Integral]\_0\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x\), ScriptLevel->0]}], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[TextData[{ "We divide the interval of integration into two parts (0,", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], ") and (", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], ",1) so that ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_0\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x = \[Integral]\_0\%\(1\/2\)\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x + \[Integral]\_\(1\/2\)\%1\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\) \[DifferentialD]x\)]], "\n", "Changing the variable x\[Rule]1-z in the first integral, we obtain\n ", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_0\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x = \(\[Integral]\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x + \[Integral]\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x = \ 2\ \(\[Integral]\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x\)\)\)]], ".", "\n", "This leads to the following identity ", Cell[BoxData[ \(TraditionalForm \`\(-\(\[Pi]\^2\/4\)\)\ \(\[Integral]\_0\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x\) = \(-\(\[Pi]\^2\/2\)\)\ \(\[Integral]\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(\[Pi]\ x)\)\) \[DifferentialD]x\)\)]], "\n", "the right side of which can be easily evaluated:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(-\(\[Pi]\^2\/2\)\)\ \ \(\[Integral]\_\(1\/2\)\%1\(\((x - 1\/2)\)\ Sec[\[Pi]\ x]\) \[DifferentialD]x\)\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(1\/2\ \(\[Integral]\_0\%1\(K(x\^2)\) \[DifferentialD]x\)\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[TextData[{ "Using the integral representation\n ", Cell[BoxData[ \(TraditionalForm\`\(\n K(x)\ = \ \[Integral]\_0\%1 1\/\@\(1 - x\ \(\(sin\^2\)(t)\)\)\ \[DifferentialD]t\)\)]] }], "Text"], Cell[TextData[{ "and changing the order of integration, we get\n\n", Cell[BoxData[ \(TraditionalForm \`1\/2\ \(\[Integral]\_0\%1\( K(x\^2)\) \[DifferentialD]x\)\ = \ 1\/2\ \(\[Integral]\_0 \%\(\[Pi]\/2\)\(\[Integral]\_0\%1 \( 1\/\@\(1 - x\^2\ \(\(sin\^2\)(t)\)\)\) \[DifferentialD]x\ \[DifferentialD]t\)\)\)]], ".\n" }], "Text"], Cell[TextData[{ "Now we use ", StyleBox["Mathematica", FontSlant->"Italic"], " to evaluate integrals from the right side:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%1\( 1\/\@\(1 - x\^2\ Sin[t]\^2\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(\(sin\^\(-1\)\)(sin(t))\)\ \(csc(t)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(1\/2\ \(\[Integral]\_0\%\(\[Pi]\/2\)PowerExpand[%] \[DifferentialD]t\)\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(\[Integral]\_0\%1\(E(x\^2)\) \[DifferentialD]x - 1\/2\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[TextData[{ "In the same manner as in the previous section using the integral \ representations for the elliptic function\n ", Cell[BoxData[ \(TraditionalForm \`E(x) = \[Integral]\_0\%\(\[Pi]\/2\)\(\@\(1 - x\ \(\(sin\^2\)(t)\)\)\) \[DifferentialD]t\)]], ":\n \nwe have\n\n", Cell[BoxData[ \(TraditionalForm \`\(-\(1\/2\)\) + \[Integral]\_0\%1\( E(x\^2)\) \[DifferentialD]x\ = \(-\(1\/2\)\) + \[Integral]\_0 \%\(\[Pi]\/2\)\(\[Integral]\_0\%1 \(\@\( 1 - x\^2\ \(\(sin\^2\)(t)\)\)\) \[DifferentialD]x\ \[DifferentialD]t\)\)]], ".\n\nSo we have:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%1\(\@\( 1 - x\^2\ Sin[t]\^2\)\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/2\ \((\(\(sin\^\(-1\)\)(sin(t))\)\ \(csc(t)\) + \@\(\(cos\^2\)(t)\)) \)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\(-\(1\/2\)\) + \[Integral]\_0\%\(\[Pi]\/2\)PowerExpand[%] \[DifferentialD]t\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(\(sinh\^\(-1\)\)(sin(x))\) \[DifferentialD]x\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[TextData[{ "The integral can be evaluated to Catalan`s constant by using the integral \ representation for ", Cell[BoxData[ \(TraditionalForm\`\(sinh\^\(-1\)\)(z)\)]], "\n", Cell[BoxData[ \(TraditionalForm \`\(sinh\^\(-1\)\)(z) = \[Integral]\_0\%1\( z\/\@\(t\^2\ z\^2 + 1\)\) \[DifferentialD]t\)]], "\n", "Substituting it into the given integral and changing the order of \ integration, we have\n", Cell[BoxData[ \(TraditionalForm \`\[Integral]\_0\%\(\[Pi]\/2\)\(\(sinh\^\(-1\)\)(sin(x))\) \[DifferentialD]x = \[Integral]\_0\%1 \((\[Integral]\_0 \%\(\[Pi]\/2 \)\(\(sin(x)\)\/\@\(z\^2\ \(\(sin\^2\)(x)\) + 1\)\) \[DifferentialD]x)\) \[DifferentialD]z\)]], "\n", "Then with the use of ", StyleBox["Mathematica", FontSlant->"Italic"], " we obtain the desirable result." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%1 \((\[Integral]\_0\%\(\[Pi]\/2\)\(Sin[x]\/\@\(z\^2\ Sin[x]\^2 + 1\)\) \[DifferentialD]x)\) \[DifferentialD]z\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "From ", Cell[BoxData[ FormBox[ StyleBox[ \(\[Integral]\_0\%\(\[Pi]\/2\)\(\(sinh\^\(-1\)\)(sin(x))\) \[DifferentialD]x = C\), ScriptLevel->0], TraditionalForm]]], " it immediately follows \n", Cell[BoxData[ \(TraditionalForm \`C = \[Integral]\_0\%\(\[Pi]\/2\)\(\(sinh\^\(-1\)\)(cos(x))\) \[DifferentialD]x\)]], "\n", Cell[BoxData[ \(TraditionalForm \`C = \[Integral]\_0\%\(\[Pi]\/2\)\(\(csch\^\(-1\)\)(csc(x))\) \[DifferentialD]x\)]], "\n", Cell[BoxData[ \(TraditionalForm \`C = \[Integral]\_0\%\(\[Pi]\/2\)\(\(csch\^\(-1\)\)(sec(x))\) \[DifferentialD]x\)]] }], "Text"], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["1", ScriptLevel->0]], RowBox[{ SubsuperscriptBox["\[Integral]", "0", StyleBox["1", ScriptLevel->0]], \(\(1\/\(\@\(1 - x\)\ \@\(1 - y\)\ \((x + y)\)\)\) \[DifferentialD]x \[DifferentialD]y\)}]}], ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[TextData[{ "To prove this we consider the more general integral\n\n", Cell[BoxData[ FormBox[ RowBox[{\(1\/4\), StyleBox[ RowBox[{\(\[Integral]\_0\%1\), RowBox[{\(\[Integral]\_0\%1\), RowBox[{ FractionBox[ StyleBox[ SuperscriptBox[ StyleBox["x", ScriptLevel->0], "\[Lambda]"], ScriptLevel->0], \(\@\(1 - x\)\ \@\(1 - y\)\ \((x + y)\)\)], \(\[DifferentialD]x\), \(\[DifferentialD]y\)}]}]}], ScriptLevel->0]}], TraditionalForm]]], "\n\nwhere \[Lambda] is an arbitrary parameter. 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At first we find the asymptotic expansion of ", Cell[BoxData[ \(TraditionalForm\`\(\(\[ThinSpace]\_3\)F\_2\)\)]], ". For that we convert the hypergeometric function into the series and \ consider the summand:" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(HypergeometricPFQ[{1, 1, 1\/2 - \[Lambda]}, {3\/2, 1 - \[Lambda]}, \(-1\)] /. 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CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell["Comparing", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(1\/2\ \[Pi]\ Log[2] - 1\/32\ \[Pi]\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\((2\ k + 1)\)!\)\^2\/\(16\^k\ \(k!\)\^4\ \((k + 1)\)\^3\)\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(-\(1\/32\)\), " ", "\[Pi]", " ", TagBox[ RowBox[{\(\(\[ThinSpace]\_4\)F\_3\), "(", RowBox[{ TagBox[ TagBox[ RowBox[{ TagBox["1", (Editable -> True)], ",", TagBox["1", (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[ TagBox[ RowBox[{ TagBox["2", (Editable -> True)], ",", TagBox["2", (Editable -> True)], ",", TagBox["2", (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox["1", (Editable -> True)]}], ")"}], InterpretTemplate[ HypergeometricPFQ[ #, #2, #3]&]]}], "+", \(1\/2\ \[Pi]\ \(log(2)\)\)}], TraditionalForm]], "Output"] }, Open ]], Cell["and", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%1\(\((x - 1\/2)\)\ Sec[\[Pi]\ x]\) \[DifferentialD]x\)], "Input"], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ RowBox[{\(-32\), " ", TagBox["C", Catalan]}], "+", RowBox[{"\[Pi]", " ", TagBox[ RowBox[{\(\(\[ThinSpace]\_4\)F\_3\), "(", RowBox[{ TagBox[ TagBox[ RowBox[{ TagBox["1", (Editable -> True)], ",", TagBox["1", (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[ TagBox[ RowBox[{ TagBox["2", (Editable -> True)], ",", TagBox["2", (Editable -> True)], ",", TagBox["2", (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox["1", (Editable -> True)]}], ")"}], InterpretTemplate[ HypergeometricPFQ[ #, #2, #3]&]]}], "-", \(8\ \[Pi]\ \(log(4)\)\)}], \(16\ \[Pi]\^2\)], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "and taking into account the result of the section ", StyleBox["15 ", FontWeight->"Bold"], "Catalan`s constant pops up immediately ." }], "Text"], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(\@2\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\^2\)\) - \(\ \[Pi]\)\/4\ \(log(2)\)\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[BoxData[ \(\(-\(1\/4\)\)\ \[Pi]\ Log[2] + \@2\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\^2\)\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\@2\), " ", TagBox[ RowBox[{\(\(\[ThinSpace]\_3\)F\_2\), "(", RowBox[{ TagBox[ TagBox[ RowBox[{ TagBox[\(1\/2\), (Editable -> True)], ",", TagBox[\(1\/2\), (Editable -> True)], ",", TagBox[\(1\/2\), (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[ TagBox[ RowBox[{ TagBox[\(3\/2\), (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[\(1\/2\), (Editable -> True)]}], ")"}], InterpretTemplate[ HypergeometricPFQ[ #, #2, #3]&]]}], "-", \(1\/4\ \[Pi]\ \(log(2)\)\)}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "We observe that ", Cell[BoxData[ \(TraditionalForm \`1\/\(2\ k + 1\) = \[Integral]\_0\%1\( t\^\(2\ k\)\) \[DifferentialD]t\)]], ".\nSubstituting it into the above series and changing the order of \ summation and integration, we obtain\n\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(k = 0\)\%\[Infinity]\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\^2\) = \(\[Sum]\+\(k = 0 \)\%\[Infinity]\(\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\)\) \((\[Integral]\_0\%1\( t\^\(2 k\)\) \[DifferentialD]t)\) = \[Integral]\_0\%1 \((\[Sum]\+\(k = 0\)\%\[Infinity]\(\(\((2\ k)\)!\) t\^\(2 k\)\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\))\) \[DifferentialD]\)\)]], "\n\nwhich leads to the following equality \n\n", Cell[BoxData[ \(TraditionalForm \`\(-\(1\/4\)\)\ \[Pi]\ \(log(2)\) + \@2\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\^2\)\) = \(-\(1\/4\)\)\ \[Pi]\ \(log(2)\) + \@2\ \(\[Integral]\_0\%1 \((\[Sum]\+\(k = 0\)\%\[Infinity]\(\(\((2\ k)\)!\)\ t\^\(2\ k\)\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\))\) \[DifferentialD]t\)\)]], "\n\nSeries and integral in the right side of this equality can be easily \ evaluated by ", StyleBox["Mathematica", FontSlant->"Italic"], " ." }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\(\((2\ k)\)!\)\ t\^\(2\ k\)\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\)\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(\@2\ \(\(sin\^\(-1\)\)(\@t\^2\/\@2)\)\)\/\@t\^2\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(PowerExpand[%, \ t]\)], "Input"], Cell[BoxData[ \(TraditionalForm\`\(\@2\ \(\(sin\^\(-1\)\)(t\/\@2)\)\)\/t\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Integrate[%, \ {t, \ 0, \ 1}]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/16\), " ", RowBox[{"(", RowBox[{ RowBox[{"8", " ", \(\@2\), " ", TagBox["C", Catalan]}], "+", \(\[ImaginaryI]\ \@2\ \[Pi]\^2\), "+", \(4\ \@2\ \[Pi]\ \(log(1 - \[ImaginaryI])\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell["Finally, we have", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\(-\(1\/4\)\)\ \[Pi]\ Log[2] + \@2\ %\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ FractionBox[ RowBox[{ RowBox[{"8", " ", \(\@2\), " ", TagBox["C", Catalan]}], "+", \(\[ImaginaryI]\ \@2\ \[Pi]\^2\), "+", \(4\ \@2\ \[Pi]\ \(log(1 - \[ImaginaryI])\)\)}], \(8\ \@2\)], "-", \(1\/4\ \[Pi]\ \(log(2)\)\)}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { StyleBox[ \(\[Pi]\/8\ \(log(\@3 + 2)\) + 3\/8\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(k!\)\^2 \/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\^2\)\)\), ScriptLevel->0], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[BoxData[ \(1\/8\ \[Pi]\ Log[\@3 + 2] + 3\/8\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\( k!\)\^2\/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\^2\)\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(3\/8\), " ", TagBox[ RowBox[{\(\(\[ThinSpace]\_3\)F\_2\), "(", RowBox[{ TagBox[ TagBox[ RowBox[{ TagBox[\(1\/2\), (Editable -> True)], ",", TagBox["1", (Editable -> True)], ",", TagBox["1", (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[ TagBox[ RowBox[{ TagBox[\(3\/2\), (Editable -> True)], ",", TagBox[\(3\/2\), (Editable -> True)]}], InterpretTemplate[ { SlotSequence[ 1]}&]], (Editable -> False)], ";", TagBox[\(1\/4\), (Editable -> True)]}], ")"}], InterpretTemplate[ HypergeometricPFQ[ #, #2, #3]&]]}], "+", \(1\/8\ \[Pi]\ \(log(2 + \@3)\)\)}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ "In the the same manner as in the previous section replacing ", Cell[BoxData[ \(TraditionalForm \`\(1\/\(2\ k + 1\)\) by\ \ \(\[Integral]\_0\%1\( t\^\(2\ k\)\) \[DifferentialD]t\)\)]], " and changing the order of summation and integration, we obtain\n\n", Cell[BoxData[ \(TraditionalForm \`1\/8\ \[Pi]\ \(log(\@3 + 2)\) + 3\/8\ \(\[Sum]\+\(k = 0\)\%\[Infinity]\( k!\)\^2 \/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\^2\)\)\ = \ 1\/8\ \[Pi]\ \(log(\@3 + 2)\) + \(3\/8\) \(\[Integral]\_0\%1 \(\[Sum]\+\(k = 0 \)\%\[Infinity]\(\(\(\( k!\)\^2\) t\^\(2\ k\)\)\/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\)\) \[DifferentialD]t\)\)\)]], "." }], "Text"], Cell[TextData[{ "Now using ", StyleBox["Mathematica", FontSlant->"Italic"], " we evaluate the sum and the integral in the right side" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sum]\+\(k = 0\)\%\[Infinity]\(\( k!\)\^2\ t\^\(2\ k\)\)\/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\)\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(2\ \(\(sin\^\(-1\)\)(\@t\^2\/2)\)\)\/\(\@t\^2\ \@\(1 - t\^2\/4\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(PowerExpand[%, \ t]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(2\ \(\(sin\^\(-1\)\)(t\/2)\)\)\/\(t\ \@\(1 - t\^2\/4\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Integrate[%, \ {t, \ 0, \ 1}]\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/6\ \[ImaginaryI]\ \(( 3\ \[Pi]\^2 - 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 - \@\(-1\)\%6)\) + 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 + \@\(-1\)\%6)\) + 12\ \(\(Li\_2\)(\(-\@\(-1\)\%6\))\) - 12\ \(\(Li\_2\)(\@\(-1\)\%6)\)) \)\)], "Output"] }, Open ]], Cell["Thus, the given sum is", "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(1\/8\ \[Pi]\ Log[\@3 + 2] + \(3\ %\)\/8\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/8\ \[Pi]\ \(log(2 + \@3)\) + 1\/16\ \[ImaginaryI]\ \(( 3\ \[Pi]\^2 - 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 - \@\(-1\)\%6)\) + 2\ \[ImaginaryI]\ \[Pi]\ \(log(1 + \@\(-1\)\%6)\) + 12\ \(\(Li\_2\)(\(-\@\(-1\)\%6\))\) - 12\ \(\(Li\_2\)(\@\(-1\)\%6)\))\)\)], "Output"] }, Open ]], Cell[TextData[{ "which could be further simplified by using the transformation rules from \ the section ", StyleBox["14", FontWeight->"Bold"], ":" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(% /. { PolyLog[2, \(-\((\(-1\))\)\^\(1/6\)\)] \[Rule] \(-\(\(11\ \[Pi]\^2\)\/72\)\) - PolyLog[2, \((\(-1\))\)\^\(5/6\)], PolyLog[2, \((\(-1\))\)\^\(1/6\)] \[Rule] \(4\ \[ImaginaryI]\)\/3\ \ Catalan + \[Pi]\^2\/72 - PolyLog[2, \((\(-1\))\)\^\(5/6\)]}\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/8\ \[Pi]\ \(log(2 + \@3)\)\), "+", RowBox[{\(1\/16\), " ", "\[ImaginaryI]", " ", RowBox[{"(", RowBox[{ \(3\ \[Pi]\^2\), "-", \(2\ \[ImaginaryI]\ \[Pi]\ \(log(1 - \@\(-1\)\%6)\)\), "+", \(2\ \[ImaginaryI]\ \[Pi]\ \(log(1 + \@\(-1\)\%6)\)\), "+", \(12\ \(( \(-\(\(11\ \[Pi]\^2\)\/72\)\) - \(Li\_2\)(\((\(-1\))\)\^\(5/6\)))\)\), "-", RowBox[{"12", " ", RowBox[{"(", RowBox[{ FractionBox[ RowBox[{"4", " ", "\[ImaginaryI]", " ", TagBox["C", Catalan]}], "3"], "+", \(\[Pi]\^2\/72\), "-", \(\(Li\_2\)(\((\(-1\))\)\^\(5/6\))\)}], ")"}]}]}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/16\), " ", RowBox[{"(", RowBox[{ RowBox[{"16", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\ \(log(2 - \@3)\)\), "+", \(\[Pi]\ \(log(2 + \@3)\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(% /. \[Rho]_\ Log[\[Alpha]_]\ + \ \[Rho]_\ Log[\[Beta]_]\ :> \ \[Rho]\ Log[\[Alpha]\ \[Beta] // Expand]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell[BoxData[ FormBox[GridBox[{ { RowBox[{ StyleBox[ RowBox[{\(1\/4\), " ", RowBox[{ UnderoverscriptBox["\[Sum]", RowBox[{ StyleBox["k", ScriptLevel->0], "=", "0"}], "\[Infinity]"], RowBox[{ FractionBox[ RowBox[{ SuperscriptBox["2", StyleBox["k", ScriptLevel->0]], " ", SuperscriptBox[ RowBox[{ StyleBox["k", ScriptLevel->0], "!"}], "2"]}], RowBox[{ RowBox[{"(", RowBox[{ RowBox[{"2", " ", StyleBox["k", ScriptLevel->0]}], "+", "1"}], ")"}], "!"}]], " ", TagBox["\[Psi]", PolyGamma], \((k + 3\/2)\)}]}]}], ScriptLevel->0], StyleBox["+", ScriptLevel->0], \(\[Pi]\/4\ \ \(log(2)\)\), "+", \(\(\[Pi]\/8\) \[Gamma]\)}], ButtonBox[ StyleBox["proof", FontSlant->"Italic"], ButtonFunction:>(FrontEndExecute[ { FrontEnd`SelectionMove[ FrontEnd`SelectedNotebook[ ], All, ButtonCell], FrontEndToken[ "SelectionOpenAllGroups"]}]&), Active->True, ButtonExpandable->False, ButtonFrame->"DialogBox"]} }, ColumnWidths->{0.7, 0.3}, ColumnAlignments->{Left, Right}], TraditionalForm]], "Subsection", CellDingbat->None, CellMargins->{{Inherited, 65}, {Inherited, Inherited}}, CellFrameLabels->{{Cell[ TextData[ { StyleBox[ "(", FontSlant -> "Italic"], StyleBox[ CounterBox[ "Subsection"], FontSlant -> "Italic"], StyleBox[ ") ", FontSlant -> "Italic"]}]], Inherited}, { Inherited, Inherited}}, FontWeight->"Plain"], Cell[CellGroupData[{ Cell["Proof", "Subsubsection", CellDingbat->"", FontSlant->"Italic"], Cell[CellGroupData[{ Cell[BoxData[ \(\(\[Pi]\ EulerGamma\)\/8 + 1\/4\ \[Pi]\ Log[2] + 1\/4\ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\ PolyGamma[3\/2 + n]\)\/\(\((1 + 2\ n)\)!\)\)\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(1\/4\), " ", RowBox[{\(\[Sum]\+\(n = 0\)\%\[Infinity]\), FractionBox[ RowBox[{\(2\^n\), " ", \(\(n!\)\^2\), " ", RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 3\/2\), ")"}]}], \(\((2\ n + 1)\)!\)]}]}], "+", \(1\/4\ \[Pi]\ \(log(2)\)\), "+", FractionBox[ RowBox[{ TagBox["\[Gamma]", EulerGamma], " ", "\[Pi]"}], "8"]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[{ StyleBox["Mathematica", FontSlant->"Italic"], " proof is possible but quite complicated. First of all by using the \ following property of the polygamma function:\n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 3\/2\), ")"}], "=", RowBox[{ RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 1\/2\), ")"}], "+", \(2\/\(2\ n + 1\)\)}]}], TraditionalForm]]], "\n\nwe divide the given sum in two sums:\n\n", Cell[BoxData[ FormBox[ RowBox[{ RowBox[{\(\[Sum]\+\(n = 0\)\%\[Infinity]\), FractionBox[ RowBox[{\(2\^n\), " ", \(\(n!\)\^2\), " ", RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 3\/2\), ")"}]}], \(\((2\ n + 1)\)!\)]}], "=", RowBox[{ \(2 \(\[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\)\/\(\(\((2\ n + 1)\)!\)\ \((2\ n + 1)\)\)\)\), "+", RowBox[{\(\[Sum]\+\(n = 0\)\%\[Infinity]\), FractionBox[ RowBox[{\(2\^n\), " ", \(\(n!\)\^2\), " ", RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 1\/2\), ")"}]}], \(\((2\ n + 1)\)!\)]}]}]}], TraditionalForm]]], "\n\nand evaluate each of them separately. The first sum can be rewritten \ as\n\n", Cell[BoxData[ \(TraditionalForm \`\[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\)\/\(\(\((2\ n + 1)\)!\)\ \((2\ n + 1)\)\) = \[Integral]\_0\%1 \((\[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\ t\^\(2\ n\)\)\/\(\((2\ n + 1)\)!\))\) \[DifferentialD]t\)]], "\n\nand then evaluated" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\ t\^\(2\ n\)\)\/\(\((2\ n + 1)\)!\)\)], "Input"], Cell[BoxData[ \(TraditionalForm \`\(\@2\ \(\(sin\^\(-1\)\)(\@t\^2\/\@2)\)\)\/\(\@t\^2\ \@\(1 - t\^2\/2\)\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(\[Integral]\_0\%1 PowerExpand[%, t] \[DifferentialD]t\)], "Input"], Cell[BoxData[ \(TraditionalForm \`1\/4\ \(( \[ImaginaryI]\ \@2\ \[Pi]\^2 + \@2\ \[Pi]\ \(log(1 - \@\(-1\)\%4)\) - \@2\ \[Pi]\ \(log(1 + \@\(-1\)\%4)\) + 4\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\(-\@\(-1\)\%4\))\) - 4\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\@\(-1\)\%4)\))\)\)], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ FractionBox[ RowBox[{ RowBox[{\(-48\), " ", TagBox["C", Catalan]}], "+", \(23\ \[ImaginaryI]\ \[Pi]\^2\), "+", \(24\ \[Pi]\ \(log(1 - \@\(-1\)\%4)\)\), "-", \(24\ \[Pi]\ \(log(1 + \@\(-1\)\%4)\)\), "-", \(192\ \[ImaginaryI]\ \(\(Li\_2\)(\@\(-1\)\%4)\)\)}], \(48\ \@2\)], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["Let us name the first sum as \[Alpha]:"], "Text"], Cell[BoxData[ \(\(\[Alpha] = 2\ %; \)\)], "Input"], Cell[TextData[{ "Now we consider the second sum:\n", Cell[BoxData[ FormBox[ RowBox[{\(\[Sum]\+\(n = 0\)\%\[Infinity]\), FractionBox[ RowBox[{\(2\^n\), " ", \(\(n!\)\^2\), " ", RowBox[{ SuperscriptBox[ TagBox["\[Psi]", PolyGamma], \((0)\)], "(", \(n + 1\/2\), ")"}]}], \(\((2\ n + 1)\)!\)]}], TraditionalForm]]], "\n", "At first we need to introduce values of of Tan and Cot at ", Cell[BoxData[ \(TraditionalForm\`\[Pi]\/8\)]], ":" }], "Text"], Cell[BoxData[ \(\(Unprotect[Tan, \ Cot]; \)\)], "Input"], Cell[BoxData[ \(Tan[\[Pi]\/8] = \@2 - 1; Cot[\[Pi]\/8] = \@\(2\ \) + 1; \)], "Input"], Cell[BoxData[ \(\(Protect[Tan, \ Cot]; \)\)], "Input"], Cell[TextData[{ "Then with use of ", StyleBox["Mathematica ", FontSlant->"Italic"], "we calculate that sum :" }], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[ \[Sum]\+\(n = 0\)\%\[Infinity]\( 2\^n\ \(n!\)\^2\ PolyGamma[0, n + 1\/2]\)\/\(\((2\ n + 1)\)!\)]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/4\), " ", RowBox[{"(", RowBox[{ \(\(-2\)\ \[ImaginaryI]\ \((\(-3\) + \@2)\)\ \[Pi]\^2\), "-", \(8\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(2)\)\), "+", \(\(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(256)\)\), "+", \(16\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(2 - \@2)\)\), "-", \(16\ \(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(2 + \@2)\)\), "+", RowBox[{"\[Pi]", " ", RowBox[{"(", RowBox[{ RowBox[{\(-2\), " ", TagBox["\[Gamma]", EulerGamma]}], "-", \(2\ \[ImaginaryI]\ \((\(-8\) + \@2)\)\ \(\(cot\^\(-1\)\)(1 - \@2)\)\), "+", \(2\ \[ImaginaryI]\ \@2\ \(\(cot\^\(-1\)\)(1 + \@2)\)\), "-", \(8\ \(\(sinh\^\(-1\)\)(1)\)\), "-", \(6\ \(log(2)\)\), "+", \(5\ \(log(2 - \@2)\)\), "-", \(\@2\ \(log(2 - \@2)\)\), "+", \(3\ \(log(2 + \@2)\)\), "+", \(\@2\ \(log(2 + \@2)\)\), "-", \(log(99 + 70\ \@2)\)}], ")"}]}], "+", \(8\ \[ImaginaryI]\ \(\(Li\_2\)(1\/2 - \[ImaginaryI]\/2)\)\), "-", \(8\ \[ImaginaryI]\ \(\(Li\_2\)(1\/2 + \[ImaginaryI]\/2)\)\), "-", \(8\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\(-\(\(1 + \[ImaginaryI]\)\/\@2\)\))\)\), "+", \(8\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\(1 + \[ImaginaryI]\)\/\@2)\)\), "+", \(8\ \[ImaginaryI]\ \(\(Li\_2\)(\(-\(\(1 - \[ImaginaryI]\)\/\(\(-2\) + \@2\)\)\)) \)\), "-", \(8\ \[ImaginaryI]\ \(\(Li\_2\)(\(1 - \[ImaginaryI]\)\/\(2 + \@2\))\)\), "-", \(8\ \[ImaginaryI]\ \(\(Li\_2\)( \(\((1 + \[ImaginaryI])\)\ \@2\)\/\(\(-2\) + 2\ \@2\))\)\), "+", \(8\ \[ImaginaryI]\ \(\(Li\_2\)(\(\((1 + \[ImaginaryI])\)\ \@2\)\/\(2 + 2\ \@2\)) \)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ Further simplifications can be done by applying the following set \ of transformation rules:\ \>", "Text"], Cell[BoxData[ \(\(% /. { PolyLog[2, 1\/2 - \[ImaginaryI]\/2] \[Rule] \(-\[ImaginaryI]\)\ Catalan + \[Pi]\^2\/48 + 1\/32\ \((\[Pi] + 2\ \[ImaginaryI]\ Log[2])\)\^2, PolyLog[2, 1\/2 + \[ImaginaryI]\/2] \[Rule] \[ImaginaryI]\ Catalan + \[Pi]\^2\/48 + 1\/32\ \((\[Pi] - 2\ \[ImaginaryI]\ Log[2])\)\^2, PolyLog[2, \(-\(\(1 + \[ImaginaryI]\)\/\@2\)\)] \[Rule] PolyLog[2, \(-\((\(-1\))\)\^\(1/4\)\)], PolyLog[2, \(1 + \[ImaginaryI]\)\/\@2] \[Rule] PolyLog[2, \((\(-1\))\)\^\(1/4\)], PolyLog[2, \(\((1 + \[ImaginaryI])\)\ \@2\)\/\(\(-2\) + 2\ \@2\)] \[Rule] \(-\(\[Pi]\^2\/6\)\) - 1\/2\ Log[\(-\((\((\(-1\))\)\^\(1/4\)\ \((1 + \@2)\))\)\)]\^2 - PolyLog[2, \(-\(\((\(-1\))\)\^\(3/4\)\/\(1 + \@2\)\)\)], PolyLog[2, \(\((1 + \[ImaginaryI])\)\ \@2\)\/\(2 + 2\ \@2\)] \[Rule] PolyLog[2, \((\(-1\))\)\^\(1/4\)\/\(1 + \@2\)], PolyLog[2, \(1 - \[ImaginaryI]\)\/\(2 + \@2\)] \[Rule] PolyLog[2, \(-\(\((\(-1\))\)\^\(3/4\)\/\(1 + \@2\)\)\)], PolyLog[2, \(-\(\(1 - \[ImaginaryI]\)\/\(\(-2\) + \@2\)\)\)] \[Rule] \(-\(\[Pi]\^2\/6\)\) - 1\/2\ Log[\((\(-1\))\)\^\(3/4\)\ \((1 + \@2)\)]\^2 - PolyLog[2, \((\(-1\))\)\^\(1/4\)\/\(1 + \@2\)]}; \)\)], "Input"], Cell[TextData["We name the second sum \[Beta]:"], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(\[Beta] = \ FullSimplify[%]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/4\), " ", RowBox[{"(", RowBox[{ RowBox[{"16", " ", TagBox["C", Catalan]}], "-", \(2\ \[ImaginaryI]\ \((\(-3\) + \@2)\)\ \[Pi]\^2\), "-", \(8\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(2)\)\), "+", \(\(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(256)\)\), "+", \(16\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(2 - \@2)\)\), "-", \(16\ \(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(2 + \@2)\)\), "+", RowBox[{"\[Pi]", " ", RowBox[{"(", RowBox[{ RowBox[{\(-2\), " ", TagBox["\[Gamma]", EulerGamma]}], "-", \(2\ \[ImaginaryI]\ \((\(-8\) + \@2)\)\ \(\(cot\^\(-1\)\)(1 - \@2)\)\), "+", \(2\ \[ImaginaryI]\ \@2\ \(\(cot\^\(-1\)\)(1 + \@2)\)\), "+", \(4\ \(\(sinh\^\(-1\)\)(1)\)\), "-", \(8\ \(log(2)\)\), "+", \(5\ \(log(2 - \@2)\)\), "-", \(\@2\ \(log(2 - \@2)\)\), "+", \(3\ \(log(2 + \@2)\)\), "+", \(\@2\ \(log(2 + \@2)\)\), "-", \(log(99 + 70\ \@2)\)}], ")"}]}], "-", \(8\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\(-\@\(-1\)\%4\))\)\), "+", \(8\ \[ImaginaryI]\ \@2\ \(\(Li\_2\)(\@\(-1\)\%4)\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell[TextData[ "Finally, combining the values of \[Alpha] and \[Beta], we get "], "Text"], Cell[CellGroupData[{ Cell[BoxData[ \(FullSimplify[ \(EulerGamma\ \[Pi]\)\/8 + 1\/4\ \[Pi]\ Log[2] + \(\[Alpha] + \[Beta]\)\/4]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(1\/16\), " ", RowBox[{"(", RowBox[{ RowBox[{"16", " ", TagBox["C", Catalan]}], "+", \(6\ \[ImaginaryI]\ \[Pi]\^2\), "+", \(16\ \[ImaginaryI]\ \[Pi]\ \(\(cot\^\(-1\)\)(1 - \@2)\)\), "+", \(4\ \[Pi]\ \(\(sinh\^\(-1\)\)(1)\)\), "-", \(2\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(16)\)\), "+", \(\(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(256)\)\), "+", \(5\ \[Pi]\ \(log(2 - \@2)\)\), "+", \(16\ \(\(cot\^\(-1\)\)(1 + \@2)\)\ \(log(2 - \@2)\)\), "+", \(3\ \[Pi]\ \(log(2 + \@2)\)\), "-", \(16\ \(\(cot\^\(-1\)\)(1 - \@2)\)\ \(log(2 + \@2)\)\), "-", \(\[Pi]\ \(log(1584 + 1120\ \@2)\)\)}], ")"}]}], TraditionalForm]], "Output"] }, Open ]], Cell["\<\ Now performing the following chain of simplification rules we \ obtain the desirable result:\ \>", "Text"], Cell[BoxData[ \(\(% /. ArcCot[1\ - \ Sqrt[2]] -> \(-ArcCot[1\ + \ Sqrt[2]]\) - Pi/4; \)\)], "Input"], Cell[CellGroupData[{ Cell[BoxData[ \(Collect[%, ArcCot[1\ + \ Sqrt[2]], FullSimplify]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ \(\(\(cot\^\(-1\)\)(1 + \@2)\)\ \((\(-\[ImaginaryI]\)\ \[Pi] - \(log(16)\)\/8 - \(log(256)\)\/16 + log(2 - \@2) + log(2 + \@2))\)\), "+", RowBox[{\(1\/64\), " ", RowBox[{"(", RowBox[{ RowBox[{"64", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\ \(( 8\ \[ImaginaryI]\ \[Pi] + 16\ \(\(sinh\^\(-1\)\)(1)\) - log(256) + 20\ \(log(2 - \@2)\) + 28\ \(log(2 + \@2)\) - 4\ \(log(1584 + 1120\ \@2)\))\)\)}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(% //. r1_.\ Log[w_]\ + \ r2_.\ Log[e_]\ :> \ Log[w^r1\ e^r2 // Expand]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{ \(\(-\[ImaginaryI]\)\ \[Pi]\ \(\(cot\^\(-1\)\)(1 + \@2)\)\), "+", RowBox[{\(1\/64\), " ", RowBox[{"(", RowBox[{ RowBox[{"64", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\ \(( 8\ \[ImaginaryI]\ \[Pi] + 16\ \(\(sinh\^\(-1\)\)(1)\) + log(37814272\/\((1584 + 1120\ \@2)\)\^4 + \(26738688\ \@2\)\/\((1584 + 1120\ \@2)\)\^4))\)\)}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(TrigToExp[%] /. Log[w_]\ :> \ Log[w // FullSimplify]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(-\(\(\[ImaginaryI]\ \[Pi]\^2\)\/8\)\), "+", RowBox[{\(1\/64\), " ", RowBox[{"(", RowBox[{ RowBox[{"64", " ", TagBox["C", Catalan]}], "+", \(\[Pi]\ \(( 8\ \[ImaginaryI]\ \[Pi] + log(665857 - 470832\ \@2) + 16\ \(log(1 + \@2)\))\)\)}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(% //. r1_.\ Log[w_]\ + \ r2_.\ Log[e_]\ :> \ Log[w^r1\ e^r2 // Expand]\)], "Input"], Cell[BoxData[ FormBox[ RowBox[{\(-\(\(\[ImaginaryI]\ \[Pi]\^2\)\/8\)\), "+", RowBox[{\(1\/64\), " ", RowBox[{"(", RowBox[{ RowBox[{"64", " ", TagBox["C", Catalan]}], "+", \(8\ \[ImaginaryI]\ \[Pi]\^2\)}], ")"}]}]}], TraditionalForm]], "Output"] }, Open ]], Cell[CellGroupData[{ Cell[BoxData[ \(Expand[%]\)], "Input"], Cell[BoxData[ FormBox[ TagBox["C", Catalan], TraditionalForm]], "Output"] }, Open ]], Cell[TextData["\[FilledSquare]"], "Text"] }, Open ]] }, Closed]] }, Open ]] }, FrontEndVersion->"NeXT 3.0", ScreenRectangle->{{0, 1053}, {0, 832}}, WindowToolbars->"EditBar", Evaluator->"aconitum", WindowSize->{589, 581}, WindowMargins->{{Automatic, 77}, {Automatic, 12}}, CommonDefaultFormatTypes->{"Output"->TraditionalForm}, StyleDefinitions -> "ArticleModern.nb" ] (*********************************************************************** Cached data follows. 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