33 representations for Catalan's constant

by Victor Adamchik
Carnegie Mellon University
Computer Science Department
adamchik@cs.cmu.edu


Abstract

This page contains 33 series and integral representations for Catalan's constant. Most of these formulas are proved just by evaluating them directly in Mathematica. Others are proved by a few steps of Mathematica computation

Integral Representations

Entry 1.

Proof.

In[1]:= Integrate[ArcTan[x]/x, {x, 0, 1}]

Out[1]= Catalan

Entry 2.

Proof.

In[2]:= -Integrate[Log[x]/(x^2 + 1), {x, 0, 1}]

Out[2]= Catalan

Entry 3.

Proof.

In[3]:= 1/2*Integrate[x*Sech[x], {x, 0, Infinity}]

Out[3]= Catalan

Entry 4.

Proof.

In[4]:= -2*Integrate[Log[2*Sin[x]], {x, 0, Pi/4}]

            -Catalan   I    2   Pi Log[1 - I]   Pi Log[2]
Out[4]= -2 (-------- - -- Pi  - ------------- + ---------)
               2       16             4             8

In[5]:= FullSimplify[%]

Out[5]= Catalan

Entry 5.

Proof.

In[6]:= 2*Integrate[Log[2*Cos[x]], {x, 0, Pi/4}]

                        2
        8 Catalan + I Pi  - 4 Pi Log[1 + I] + 2 Pi Log[2]
Out[6]= -------------------------------------------------
                                8

In[7]:= FullSimplify[%]

Out[7]= Catalan

Remark 1.

From 4 and 5 it immediately follows

Remark 2.

Actually formulas 4 and 5 are particular cases of the following more general representations

where p is a positive integer.

Entry 6.

Proof.

In[8]:= -(1/4)*Pi*Log[2] + Integrate[(x*Csc[x])/(Cos[x] + Sin[x]), {x, 0, Pi/2}]

Out[8]= Catalan

Remark .

Here are related representations:

Entry 7.

Proof.

In[9]:= 1/4*Pi*Log[2] - Integrate[(x*Csc[x])/(Cos[x] - Sin[x]), {x, 0, Pi/2}, PrincipalValue -> True]

Out[9]= Catalan

Entry 8.

Proof.

In[10]:= (7*Zeta[3])/(4*Pi) + 2/Pi*Integrate[ArcTan[x]^2/x, {x, 0, 1}]

         4 Catalan Pi - 7 Zeta[3]   7 Zeta[3]
Out[10]= ------------------------ + ---------
                   4 Pi               4 Pi

In[11]:= Expand[%]

Out[11]= Catalan

Entry 9.

Proof.

In[12]:= 1/2*Integrate[x/Sin[x], {x, 0, Pi/2}]

Out[12]= Catalan

Entry 10.

Proof.

In[13]:= (7*Zeta[3])/(4*Pi) + 1/(2*Pi)*Integrate[x^2/Sin[x], {x, 0, Pi/2}]

                           2                2
Out[13]= (8 Catalan Pi + Pi  Log[1 - I] - Pi  Log[1 + I] - 8 PolyLog[3, -I] + 
 
                                                 7 Zeta[3]
>       8 PolyLog[3, I] - 14 Zeta[3]) / (8 Pi) + ---------
                                                   4 Pi

It is known that the polylogarithm of roots of unity can be expressed in terms of derivatives of the gamma function

where parameters p and q are integer. From here by setting p=3, q=4 and p=1, q=4 we obtain

Applying them to the Out[13], we have

In[14]:= %/. {
	PolyLog[n_, -I] :> -(I*(-(1/4))^n*
		(PolyGamma[n - 1, 1/4] - PolyGamma[n - 1, 3/4]))/Gamma[n] - 
		((-2 + 2^n)*Zeta[n])/2^(2*n), 
	PolyLog[n_, I] :> (I*(-1/4))^n*
		(PolyGamma[n - 1, 1/4] - PolyGamma[n - 1, 3/4]))/Gamma[n] - 
		((-2 + 2^n)*Zeta[n])/2^(2*n)
	}

                           2                2
Out[15]= (8 Catalan Pi + Pi  Log[1 - I] - Pi  Log[1 + I] + 
 
           -I                1                 3     3 Zeta[3]
>       8 (--- (PolyGamma[2, -] - PolyGamma[2, -]) - ---------) - 
           128               4                 4        32
 
            I                1                 3     3 Zeta[3]
>       8 (--- (PolyGamma[2, -] - PolyGamma[2, -]) - ---------) - 14 Zeta[3])\
           128               4                 4        32
 
                  7 Zeta[3]
>      / (8 Pi) + ---------
                    4 Pi

In[16]:= FullSimplify[%]

Out[16]= Catalan

Remark .

Formulas 9 and 10 are particular cases of the more general integral

where p is a positive integer.

Entry 11.

Proof.

In[17]:= -Integrate[Log[(1 - x)/Sqrt[2]]/(x^2 + 1), {x, 0, 1}]

         I      1   I               1   I                        1   I
Out[17]= - (Log[- - -] Log[2] - Log[- + -] Log[2] + 2 PolyLog[2, - - -] - 
         4      2   2               2   2                        2   2
 
                    1   I
>      2 PolyLog[2, - + -])
                    2   2

From the Landen identity for the dilogarithm

with z=1/2-i/2 and z = 1/2+i/2 follows

respectively. Applying them to the Out[17], we obtain

In[18]:= %/.{
         PolyLog[2, 1/2 - I/2] -> -I*Catalan + Pi^2/48 - 1/2*Log[1/2 + I/2]^2,
         PolyLog[2, 1/2 + I/2] ->  I*Catalan + Pi^2/48 - 1/2*Log[1/2 - I/2]^2
         }

                                      1   I 2
                              2   Log[- - -]
         I                  Pi        2   2
Out[18]= - (-2 (I Catalan + --- - -----------) + 
         4                  48         2
 
                                 1   I 2
                         2   Log[- + -]
                       Pi        2   2          1   I
>      2 (-I Catalan + --- - -----------) + Log[- - -] Log[2] - 
                       48         2             2   2
 
           1   I
>      Log[- + -] Log[2])
           2   2

In[19]:= FullSimplify[%]

Out[19]= Catalan

Entry 12.

Proof.

In[20]:= -Integrate[Log[1/2*(1 - x^2)]/(x^2 + 1), {x, 0, 1}]

         I    2              2              2              1   I
Out[20]= - (Pi  + Log[-1 - I]  - Log[-1 + I]  - 2 I Pi Log[- - -] + 
         4                                                 2   2
 
                  1   I                               1   I
>      2 I Pi Log[- + -] + I Pi Log[2] + 2 PolyLog[2, - - -] - 
                  2   2                               2   2
 
                    1   I
>      2 PolyLog[2, - + -])
                    2   2

We will proceed in the same manner as in the previous section

In[21]:= %/.{
	PolyLog[2, 1/2 - I/2] -> -I*Catalan + Pi^2/48 - 1/2*Log[1/2 + I/2]^2,
	PolyLog[2, 1/2 + I/2] ->  I*Catalan + Pi^2/48 - 1/2*Log[1/2 - I/2]^2
         }
         
         I    2              2              2              1   I
Out[21]= - (Pi  + Log[-1 - I]  - Log[-1 + I]  - 2 I Pi Log[- - -] - 
         4                                                 2   2
 
                                1   I 2
                        2   Log[- - -]
                      Pi        2   2                 1   I
>      2 (I Catalan + --- - -----------) + 2 I Pi Log[- + -] + 
                      48         2                    2   2
 
                                 1   I 2
                         2   Log[- + -]
                       Pi        2   2
>      2 (-I Catalan + --- - -----------) + I Pi Log[2])
                       48         2

In[22]:= FullSimplify[%]

Out[22]= Catalan

Entry 13.

Proof.

In[23]:= 1/8*Pi*Log[Sqrt[3] + 2] + 3/4*Integrate[x/Sin[x], {x, 0, Pi/6}]

         Pi Log[2 + Sqrt[3]]   I
Out[23]= ------------------- + -- 
                  8            16
 
           2                      1/6                       1/6
>     (3 Pi  - 2 I Pi Log[1 - (-1)   ] + 2 I Pi Log[1 + (-1)   ] + 
 
                           1/6                      1/6
>       12 PolyLog[2, -(-1)   ] - 12 PolyLog[2, (-1)   ])

From the reflection formula for the dilogarithm

with it follows

Applying it to the result of integration Out[23], we have

In[24]:= FullSimplify[%/.PolyLog[2, -(-1)^(1/6)] -> (-11*Pi^2)/72 - PolyLog[2, (-1)^(5/6)]]

         I     2
Out[24]= -- (Pi  - 6 I Pi Log[2 - Sqrt[3]] - 6 I Pi Log[2 + Sqrt[3]] - 
         96
 
                         1/6                      5/6
>      72 PolyLog[2, (-1)   ] - 72 PolyLog[2, (-1)   ])

In[25]:= %/.r_ Log[w_] + r_ Log[e_] :> r Log[w e//Expand]

         I     2                     1/6                      5/6
Out[25]= -- (Pi  - 72 PolyLog[2, (-1)   ] - 72 PolyLog[2, (-1)   ])
         96

In[26]:= Collect[%, Pi, Simplify]

         I    2   3 I                 1/6                   5/6
Out[26]= -- Pi  - --- (PolyLog[2, (-1)   ] + PolyLog[2, (-1)   ])
         96        4

Using the following well-known property of the dilogarithm

with t=-i and q=3, we derive

Applying this formula to Out[26], we finally obtain

In[27]:= %/.PolyLog[2, (-1)^(1/6)] + PolyLog[2, (-1)^(5/6)] -> (4*I)/3*Catalan + Pi^2/72         

                                       2
         I    2   3 I  4 I           Pi
Out[27]= -- Pi  - --- (--- Catalan + ---)
         96        4    3            72

In[28]:= Expand[%]

Out[28]= Catalan

Entry 14.

Proof.

In[29]:= -(Pi^2/4)*Integrate[(x - 1/2)*Sec[Pi*x],  {x, 0, 1}]

                                                   3  3
Out[29]= (32 Catalan - Pi HypergeometricPFQ[{1, 1, -, -}, {2, 2, 2}, 1] + 
                                                   2  2
 
>      8 Pi Log[4]) / 64

To force Mathematica to do this quite simple integral, we need to divide the interval of integration into two parts (0,1/2) and (1/2,1) and then change the variable x -> 1-z in the first integral. This leads to the following identity

the right side of which can be easily evaluated:

In[30]:= -(Pi^2/2)*Integrate[(x - 1/2)*Sec[Pi*x], {x, 1/2, 1}]

Out[30]= Catalan

Entry 15.

Proof.


In[31]:= 1/2*Integrate[EllipticK[x^2], {x, 0, 1}]

                               1  1  1       3
         Pi HypergeometricPFQ[{-, -, -}, {1, -}, 1]
                               2  2  2       2
Out[31]= ------------------------------------------
                             4

To see how this integral can be evaluated to Catalan`s constant we have to use the integral representation for the elliptic function K(x):

Substituting it into the given integral and changing the order of integration, we have

Now we can use Mathematica to evaluate integrals from the right side:

In[32]:= Integrate[1/Sqrt[1 - x^2*Sin[t]^2], {x, 0, 1}]

Out[32]= ArcSin[Sin[t]] Csc[t]

In[33]:= 1/2*Integrate[PowerExpand[%], {t, 0, Pi/2}]

Out[33]= Catalan

Entry 16.

Proof.

In[34]:= -(1/2) + Integrate[EllipticE[x^2], {x, 0, 1}]

                                        1   1  1       3
                Pi HypergeometricPFQ[{-(-), -, -}, {1, -}, 1]
           1                            2   2  2       2
Out[34]= -(-) + ---------------------------------------------
           2                          2

In the same manner as in the previous section using the integral representations for the elliptic function E(x):

we have

The integrals in the right side of this equality can be successfully computed by Mathematica:

In[35]:= Integrate[Sqrt[1 - x^2*Sin[t]^2], {x, 0, 1}]

                    2
         Sqrt[Cos[t] ] + ArcSin[Sin[t]] Csc[t]
Out[35]= -------------------------------------
                           2

In[36]:= -(1/2) + Integrate[PowerExpand[%], {t, 0, Pi/2}]

Out[36]= Catalan

Entry 17.

Proof.

In[37]:= Integrate[ArcSinh[Sin[x]], {x, 0, Pi/2}]

                                           Pi
Out[37]= Integrate[ArcSinh[Sin[x]], {x, 0, --}]
                                           2

The integral can be evaluated to Catalan`s constant by using the integral representation for

Substituting it into the given integral and changing the order of integration, we have

Then with use of Mathematica we obtain the desirable result

In[38]:= Integrate[Integrate[Sin[x]/Sqrt[t^2*Sin[x]^2 + 1], {x, 0, Pi/2}], {t, 0, 1}]

Out[38]= Catalan

Remark.

It follows immediately from 17 that

Entry 18.

Proof.

To prove this we consider the more general integral

where p is an arbitrary parameter. Evaluating this integral with Mathematica, we have

In[39]:= 1/4 Integrate[x^p/(Sqrt[1-x] Sqrt[1-y] (x+y)), {y, 0, 1}, {x, 0, 1}, 
                            GenerateConditions -> False] //Expand

             3/2                 1   p
         -(Pi    Csc[p Pi] Gamma[- + -])
                                 2   2
Out[39]= ------------------------------- + 
                            p
                  4 p Gamma[-]
                            2
 
                                                1        3
     Sqrt[Pi] Gamma[p] HypergeometricPFQ[{1, 1, - - p}, {-, 1 - p}, -1]
                                                2        2
>    ------------------------------------------------------------------
                                       1
                               2 Gamma[- + p]
                                       2

Now we need to find the limit of this expression when p tends to infinity. At first we find the asymptotic expansion of HypergeometricPFQ. For that we convert the hypergeometric function into the series and consider the summand:

In[40]:= HypergeometricPFQ[{1,1,1/2-p},{3/2,1-p},-1]/.
                HypergeometricPFQ[a_,b_,z_] :>
                      (Times@@Pochhammer[a,k])/(Times@@Pochhammer[b,k]) z^k/k!

             k                 2            1
         (-1)  Pochhammer[1, k]  Pochhammer[- - p, k]
                                            2
Out[40]= --------------------------------------------
                         3
           k! Pochhammer[-, k] Pochhammer[1 - p, k]
                         2

In[41]:= FunctionExpand[%]

             k                                          1
         (-1)  Sqrt[Pi] Gamma[1 + k] Gamma[1 - p] Gamma[- + k - p]
                                                        2
Out[41]= ---------------------------------------------------------
                       3            1
               2 Gamma[- + k] Gamma[- - p] Gamma[1 + k - p]
                       2            2

Then we expand this with respect to p at the neighborhood of zero:

In[42]:= Series[%, {p, 0, 1}]//Normal//FullSimplify

Out[42]= -(
 
           k                                 1
       (-1)  (-1 + p Log[4] + p PolyGamma[0, - + k] - p PolyGamma[0, 1 + k])
                                             2
>      ---------------------------------------------------------------------)
                                      1 + 2 k

In[43]:= %/.PolyGamma[0,1/2+k] -> PolyGamma[0,3/2+k]-2/(1+2k)

                k
Out[43]= -(((-1)  (-1 + p Log[4] - p PolyGamma[0, 1 + k] + 
 
                -2                   3
>          p (------- + PolyGamma[0, - + k]))) / (1 + 2 k))
              1 + 2 k                2

and sum it up

In[44]:= Sum[%, {k,0,Infinity}]//Simplify

         8 Catalan p + Pi - 3 p Pi Log[2]
Out[44]= --------------------------------
                        4

We obtain the folowing asymptotic formula

Substituting it into the result of the integratyion Out[39], we finally prove the given identity

In[45]:= %39/.HypergeometricPFQ[{1,1,1/2-p},{3/2,1-p},-1]->%

             3/2                 1   p
         -(Pi    Csc[p Pi] Gamma[- + -])
                                 2   2
Out[45]= ------------------------------- + 
                            p
                  4 p Gamma[-]
                            2
 
     Sqrt[Pi] Gamma[p] (8 Catalan p + Pi - 3 p Pi Log[2])
>    ----------------------------------------------------
                                1
                        8 Gamma[- + p]
                                2

In[46]:= Series[%, {p, 0, 0}]//Normal//FullSimplify

Out[46]= Catalan

Remark.

The above integral can be generalized to this one

which generates Catalan's number when a, b, c and d are positive integers.

Series Representations

Entry 19.

Proof.

In[47]:= Sum[(-1)^k/(2*k + 1)^2, {k, 0, Infinity}]

Out[47]= Catalan

Entry 20.

Proof.

In[48]:= -(Pi^2/8) + 2*Sum[1/(4*k + 1)^2, {k, 0, Infinity}]

            2                 2
         -Pi    8 Catalan + Pi
Out[48]= ---- + ---------------
          8            8

In[49]:= Expand[%]

Out[49]= Catalan

Entry 21.

Proof.

In[50]:= Pi^2/8 - 2*Sum[1/(4*k + 3)^2, {k, 0, Infinity}]

           2                 2
         Pi    8 Catalan - Pi
Out[50]= --- + ---------------
          8           8

In[51]:= Expand[%]

Out[51]= Catalan

Entry 22.

Proof.

In[52]:= 1/2*Sum[(4^k*k!^2)/((2*k)!*(2*k + 1)^2), {k, 0, Infinity}]

Out[52]= Catalan

Entry 23.

Proof.

In[53]:= 1/2*Log[2]*Pi + Sum[((-1)^k*(PolyGamma[k + 1] + EulerGamma))/(2*k + 1), {k, 0, Infinity}]

         EulerGamma Pi   Pi Log[2]   4 Catalan - EulerGamma Pi - 2 Pi Log[2]
Out[53]= ------------- + --------- + ---------------------------------------
               4             2                          4

In[54]:= Expand[%]

Out[54]= Catalan

Entry 24.

Proof.

In[55]:= 1/4*Pi*Log[2] + Sum[((-1)^k*(PolyGamma[k + 3/2] + EulerGamma))/(2*k + 1), {k, 0, Infinity}]

         EulerGamma Pi   Pi Log[2]   4 Catalan - EulerGamma Pi - Pi Log[2]
Out[55]= ------------- + --------- + -------------------------------------
               4             4                         4

In[56]:= Expand[%]

Out[56]= Catalan

Entry 25.

Proof.

In[57]:= 1 - Sum[(k*Zeta[2*k + 1])/4^(2*k), {k, 1, Infinity}]

Out[57]= Catalan

Entry 26.

Proof.

In[58]:= 1/16*Sum[((3^k - 1)*(k + 1)*Zeta[k + 2])/4^k, {k, 1, Infinity}]

Out[58]= Catalan

Entry 27.

Proof.

In[59]:= 1/8*Sum[k/2^k*Zeta[k + 1, 3/4], {k, 2, Infinity}]

Out[59]= Catalan

Entry 28.

Proof.

In[60]:= 1 - 1/8*Sum[k/2^k*Zeta[k + 1, 5/4], {k, 2, Infinity}]

Out[60]= Catalan

Entry 29.

Proof.

In[61]:= 1/2*Pi*Log[2] - 1/32*Pi*Sum[(2*k + 1)!^2/(16^k*k!^4*(k + 1)^3), {k, 0, Infinity}]

                                       3  3
         -(Pi HypergeometricPFQ[{1, 1, -, -}, {2, 2, 2}, 1])
                                       2  2                    Pi Log[2]
Out[61]= --------------------------------------------------- + ---------
                                 32                                2

Catalan`s constant pops up immediately taking into account the result of the section 14.

Entry 30.

Proof.

In[62]:= -(1/4)*Pi*Log[2] + Sqrt[2]*Sum[(2*k)!/(8^k*k!^2*(2*k + 1)^2), {k, 0, Infinity}]

                                    1  1  1    3  3   1    Pi Log[2]
Out[62]= Sqrt[2] HypergeometricPFQ[{-, -, -}, {-, -}, -] - ---------
                                    2  2  2    2  2   2        4

We observe that

Substituting it into the above series and changing the order of summation and integration, we obtain

Series and integral in the right side of this equality can be easily evaluated by Mathematica

In[63]:= Sum[((2*k)!*t^(2*k))/(8^k*k!^2*(2*k + 1)), {k, 0, Infinity}]

                              2
                        Sqrt[t ]
         Sqrt[2] ArcSin[--------]
                        Sqrt[2]
Out[63]= ------------------------
                       2
                 Sqrt[t ]

In[64]:= PowerExpand[%, t]

                           t
         Sqrt[2] ArcSin[-------]
                        Sqrt[2]
Out[64]= -----------------------
                    t

In[65]:= Integrate[%, {t, 0, 1}]

                                         2
         8 Sqrt[2] Catalan + I Sqrt[2] Pi  + 4 Sqrt[2] Pi Log[1 - I]
Out[65]= -----------------------------------------------------------
                                     16

Finally, we have

In[66]:= -(1/4)*Pi*Log[2] + Sqrt[2]*%

                                         2
         8 Sqrt[2] Catalan + I Sqrt[2] Pi  + 4 Sqrt[2] Pi Log[1 - I]
Out[66]= ----------------------------------------------------------- - 
                                  8 Sqrt[2]
 
     Pi Log[2]
>    ---------
         4

In[67]:= FullSimplify[%]

Out[67]= Catalan

Entry 31.

Proof.

In[68]:= 1/8*Pi*Log[Sqrt[3] + 2] + 3/8*Sum[k!^2/((2*k)!*(2*k + 1)^2), {k, 0, Infinity}]

                              1          3  3   1
         3 HypergeometricPFQ[{-, 1, 1}, {-, -}, -]
                              2          2  2   4    Pi Log[2 + Sqrt[3]]
Out[68]= ----------------------------------------- + -------------------
                             8                                8

In the the same manner as in the previous section we obtain

Using Mathematica we evaluate the sum and the integral in the right side

In[69]:= Sum[(k!^2*t^(2*k))/((2*k)!*(2*k + 1)), {k, 0, Infinity}]

                         2
                   Sqrt[t ]
          2 ArcSin[--------]
                      2
Out[69]= ---------------------
                            2
               2           t
         Sqrt[t ] Sqrt[1 - --]
                           4

In[70]:= PowerExpand[%, t]

                   t
          2 ArcSin[-]
                   2
Out[70]= --------------
                     2
                    t
         t Sqrt[1 - --]
                    4

In[71]:= Integrate[%, {t, 0, 1}]

         I      2                      1/6                       1/6
Out[71]= - (3 Pi  - 2 I Pi Log[1 - (-1)   ] + 2 I Pi Log[1 + (-1)   ] + 
         6
 
                          1/6                      1/6
>      12 PolyLog[2, -(-1)   ] - 12 PolyLog[2, (-1)   ])

Thus, the given sum is

In[72]:= 1/8*Pi*Log[Sqrt[3] + 2] + (3*%)/8

         Pi Log[2 + Sqrt[3]]   I
Out[72]= ------------------- + -- 
                  8            16
 
           2                      1/6                       1/6
>     (3 Pi  - 2 I Pi Log[1 - (-1)   ] + 2 I Pi Log[1 + (-1)   ] + 
 
                           1/6                      1/6
>       12 PolyLog[2, -(-1)   ] - 12 PolyLog[2, (-1)   ])

which could be further simplified by using the transformation rules from the section 14:

In[73]:= % /. {
	PolyLog[2, -(-1)^(1/6)] -> -((11*Pi^2)/72) - PolyLog[2, (-1)^(5/6)], 
	PolyLog[2, (-1)^(1/6)] -> (4*I)/3*Catalan + Pi^2/72 - PolyLog[2, (-1)^(5/6)]
	}

         Pi Log[2 + Sqrt[3]]   I
Out[73]= ------------------- + -- 
                  8            16
 
           2                      1/6                       1/6
>     (3 Pi  - 2 I Pi Log[1 - (-1)   ] + 2 I Pi Log[1 + (-1)   ] + 
 
                  2
            -11 Pi                   5/6
>       12 (------- - PolyLog[2, (-1)   ]) - 
              72
 
                            2
            4 I           Pi                   5/6
>       12 (--- Catalan + --- - PolyLog[2, (-1)   ]))
             3            72

In[74]:= FullSimplify[%]

         16 Catalan + Pi Log[2 - Sqrt[3]] + Pi Log[2 + Sqrt[3]]
Out[74]= ------------------------------------------------------
                                   16

In[75]:= %/.r_ Log[w_] + r_ Log[e_] :> r Log[w e//Expand]

Out[75]= Catalan

Entry 32.

Proof.

In[76]:= (Pi*EulerGamma)/8 + 1/4*Pi*Log[2] + 1/4*Sum[(2^k*k!^2*PolyGamma[0, 3/2 + k])/(1 + 2*k)!, {k, 0, Infinity}]         

         EulerGamma Pi   Pi Log[2]
Out[76]= ------------- + --------- + 
               8             4
 
          k   2              3
         2  k!  PolyGamma[0, - + k]
                             2
     Sum[--------------------------, {k, 0, Infinity}]
                 (1 + 2 k)!
>    -------------------------------------------------
                             4

Mathematica proof is possible but quite complicated. First of all by using the following property of the polygamma function:

we divide the given sum in two sums:

and evaluate each of them separately. The first sum can be rewritten as

and then evaluated

In[77]:= Sum[(2^k*k!^2*t^(2*k))/(2*k + 1)!, {k, 0, Infinity}]

                              2
                        Sqrt[t ]
         Sqrt[2] ArcSin[--------]
                        Sqrt[2]
Out[77]= ------------------------
                             2
                2           t
          Sqrt[t ] Sqrt[1 - --]
                            2

In[78]:= Integrate[PowerExpand[%, t], {t, 0, 1}]

                      2                          1/4
Out[78]= (I Sqrt[2] Pi  + Sqrt[2] Pi Log[1 - (-1)   ] - 
 
                              1/4                                1/4
>      Sqrt[2] Pi Log[1 + (-1)   ] + 4 I Sqrt[2] PolyLog[2, -(-1)   ] - 
 
                                  1/4
>      4 I Sqrt[2] PolyLog[2, (-1)   ]) / 4

In[79]:= FullSimplify[%]

                               2                     1/4
Out[79]= (-48 Catalan + 23 I Pi  + 24 Pi Log[1 - (-1)   ] - 
 
                         1/4                         1/4
>      24 Pi Log[1 + (-1)   ] - 192 I PolyLog[2, (-1)   ]) / (48 Sqrt[2])

Let us name the first sum as sum1:

In[80]:= sum1=2 %;

Now we consider the second sum:

At first we need to introduce special values of Tan and Cot at Pi/8:

In[81]:= Unprotect[Tan, Cot];

In[82]:= Tan[Pi/8] = Sqrt[2] - 1; Cot[Pi/8] = Sqrt[2] + 1; 

In[83]:= Protect[Tan, Cot];

Then with the use of Mathematica we calculate that sum :

In[84]:= FullSimplify[Sum[(2^k*k!^2*PolyGamma[0, k + 1/2])/(2*k + 1)!, {k, 0, Infinity}]]

                                2
Out[84]= (-2 I (-3 + Sqrt[2]) Pi  - 8 ArcCot[1 + Sqrt[2]] Log[2] + 
 
>      ArcCot[1 - Sqrt[2]] Log[256] + 
 
>      16 ArcCot[1 + Sqrt[2]] Log[2 - Sqrt[2]] - 
 
>      16 ArcCot[1 - Sqrt[2]] Log[2 + Sqrt[2]] + 
 
>      Pi (-2 EulerGamma - 2 I (-8 + Sqrt[2]) ArcCot[1 - Sqrt[2]] + 
 
>         2 I Sqrt[2] ArcCot[1 + Sqrt[2]] - 8 ArcSinh[1] - 6 Log[2] + 
 
>         5 Log[2 - Sqrt[2]] - Sqrt[2] Log[2 - Sqrt[2]] + 
 
>         3 Log[2 + Sqrt[2]] + Sqrt[2] Log[2 + Sqrt[2]] - Log[99 + 70 Sqrt[2]]
 
                             1   I                   1   I
>         ) + 8 I PolyLog[2, - - -] - 8 I PolyLog[2, - + -] - 
                             2   2                   2   2
 
                              -1 - I                             1 + I
>      8 I Sqrt[2] PolyLog[2, -------] + 8 I Sqrt[2] PolyLog[2, -------] + 
                              Sqrt[2]                           Sqrt[2]
 
                         -1 + I                         1 - I
>      8 I PolyLog[2, ------------] - 8 I PolyLog[2, -----------] - 
                      -2 + Sqrt[2]                   2 + Sqrt[2]
 
                      (1 + I) Sqrt[2]                   (1 + I) Sqrt[2]
>      8 I PolyLog[2, ---------------] + 8 I PolyLog[2, ---------------]) / 4
                      -2 + 2 Sqrt[2]                     2 + 2 Sqrt[2]

Further simplifications can be done by applying the following set of transformation rules :

In[85]:= % /. {
	PolyLog[2, 1/2 - I/2] -> -I*Catalan + Pi^2/48 + 1/32*(Pi + 2*I*Log[2])^2, 
	PolyLog[2, 1/2 + I/2] ->  I*Catalan + Pi^2/48 + 1/32*(Pi - 2*I*Log[2])^2, 
	PolyLog[2, -((1 + I)/Sqrt[2])] -> PolyLog[2, -(-1)^(1/4)], 
	PolyLog[2, (1 + I)/Sqrt[2]] -> PolyLog[2, (-1)^(1/4)], 
	PolyLog[2, ((1 + I)*Sqrt[2])/(-2 + 2*Sqrt[2])] ->  
	   -Pi^2/6 - 1/2*Log[-(-1)^(1/4)*(1 + Sqrt[2])]^2 - PolyLog[2, -(-1)^(3/4)/(1 +Sqrt[2])],
 	PolyLog[2, ((1 + I)*Sqrt[2])/(2 + 2*Sqrt[2])] -> PolyLog[2,(-1)^(1/4)/(1 + Sqrt[2])],
	PolyLog[2, (1 - I)/(2 + Sqrt[2])] -> PolyLog[2, -((-1)^(3/4)/(1 + Sqrt[2]))], 
	PolyLog[2, -((1 - I)/(-2 + Sqrt[2]))] -> 
	   -Pi^2/6 - 1/2*Log[(-1)^(3/4)*(1 + Sqrt[2])]^2 - PolyLog[2, (-1)^(1/4)/(1 + Sqrt[2])]
	}; 

In[86]:= sum2 = FullSimplify[%]

                                            2
Out[86]= (16 Catalan - 2 I (-3 + Sqrt[2]) Pi  - 
 
>      8 ArcCot[1 + Sqrt[2]] Log[2] + ArcCot[1 - Sqrt[2]] Log[256] + 
 
>      16 ArcCot[1 + Sqrt[2]] Log[2 - Sqrt[2]] - 
 
>      16 ArcCot[1 - Sqrt[2]] Log[2 + Sqrt[2]] + 
 
>      Pi (-2 EulerGamma - 2 I (-8 + Sqrt[2]) ArcCot[1 - Sqrt[2]] + 
 
>         2 I Sqrt[2] ArcCot[1 + Sqrt[2]] + 4 ArcSinh[1] - 8 Log[2] + 
 
>         5 Log[2 - Sqrt[2]] - Sqrt[2] Log[2 - Sqrt[2]] + 
 
>         3 Log[2 + Sqrt[2]] + Sqrt[2] Log[2 + Sqrt[2]] - Log[99 + 70 Sqrt[2]]
 
                                          1/4
>         ) - 8 I Sqrt[2] PolyLog[2, -(-1)   ] + 
 
                                  1/4
>      8 I Sqrt[2] PolyLog[2, (-1)   ]) / 4

Finally, combining the values of sum1 and sum2, we get

In[87]:= FullSimplify[(EulerGamma*Pi)/8 + 1/4*Pi*Log[2] +  (sum1 + sum2)/4]

                             2
Out[87]= (16 Catalan + 6 I Pi  + 16 I Pi ArcCot[1 - Sqrt[2]] + 
 
>      4 Pi ArcSinh[1] - 2 ArcCot[1 + Sqrt[2]] Log[16] + 
 
>      ArcCot[1 - Sqrt[2]] Log[256] + 5 Pi Log[2 - Sqrt[2]] + 
 
>      16 ArcCot[1 + Sqrt[2]] Log[2 - Sqrt[2]] + 3 Pi Log[2 + Sqrt[2]] - 
 
>      16 ArcCot[1 - Sqrt[2]] Log[2 + Sqrt[2]] - Pi Log[1584 + 1120 Sqrt[2]])\
 
>     / 16

Then, performing another chain of simplification rules, we obtain the desirable result:

In[88]:= %/.ArcCot[1 - Sqrt[2]]->-ArcCot[1 + Sqrt[2]]-Pi/4;

In[89]:= Collect[%,ArcCot[1 + Sqrt[2]],FullSimplify]

                                      Log[16]   Log[256]
Out[89]= ArcCot[1 + Sqrt[2]] (-I Pi - ------- - -------- + Log[2 - Sqrt[2]] + 
                                         8         16
 
>       Log[2 + Sqrt[2]]) + (64 Catalan + 
 
>       Pi (8 I Pi + 16 ArcSinh[1] - Log[256] + 20 Log[2 - Sqrt[2]] + 
 
>          28 Log[2 + Sqrt[2]] - 4 Log[1584 + 1120 Sqrt[2]])) / 64

In[90]:= %//.r1_. Log[w_] + r2_. Log[e_] :> Log[w^r1 e^r2//Expand]

Out[90]= -I Pi ArcCot[1 + Sqrt[2]] + 
 
>    (64 Catalan + Pi (8 I Pi + 16 ArcSinh[1] + 
 
                      37814272             26738688 Sqrt[2]
>          Log[---------------------- + ----------------------])) / 64
                                    4                        4
               (1584 + 1120 Sqrt[2])    (1584 + 1120 Sqrt[2])

In[91]:= TrigToExp[%]/.Log[w_] :> Log[w//FullSimplify]

         -I   2
Out[91]= -- Pi  + (64 Catalan + 
         8
 
>       Pi (8 I Pi + Log[665857 - 470832 Sqrt[2]] + 16 Log[1 + Sqrt[2]])) / 64

In[92]:= %//.r1_. Log[w_] + r2_. Log[e_] :> Log[w^r1 e^r2//Expand]

                                     2
         -I   2   64 Catalan + 8 I Pi
Out[92]= -- Pi  + --------------------
         8                 64

In[93]:= Expand[%]

Out[93]= Catalan

Entry 33. (Broadhurst's series.)

Proof.

In[94]:= 
-1/4*Sum[1/4096^k (1/(1 + 8*k)^2 + 1/2 1/(2 + 8*k)^2 + 
         1/8 1/(3 + 8*k)^2 - 1/64 1/(5 + 8*k)^2 - 
         1/128 1/(6 + 8*k)^2 - 1/512 1/(7 + 8*k)^2), {k, 0, Infinity}] + 
 3/2*Sum[1/16^k (1/(1 + 8*k)^2 - 1/(2 + 8*k)^2 + 
         1/2 1/(3 + 8*k)^2 - 1/4 1/(5 + 8*k)^2 + 
         1/4 1/(6 + 8*k)^2 - 1/8 1/(7 + 8*k)^2), {k, 0, Infinity}]//Expand

                    1       1              1       1
         -LerchPhi[----, 2, -]   LerchPhi[----, 2, -]
                   4096     8             4096     4
Out[95]= --------------------- - -------------------- - 
                  256                    512
 
               1       3              1       5              1       3
     LerchPhi[----, 2, -]   LerchPhi[----, 2, -]   LerchPhi[----, 2, -]
              4096     8             4096     8             4096     4
>    -------------------- + -------------------- + -------------------- + 
             2048                  16384                  32768
 
               1       7               1      1               1      1
     LerchPhi[----, 2, -]   3 LerchPhi[--, 2, -]   3 LerchPhi[--, 2, -]
              4096     8               16     8               16     4
>    -------------------- + -------------------- - -------------------- + 
            131072                  128                    128
 
                1      3               1      5               1      3
     3 LerchPhi[--, 2, -]   3 LerchPhi[--, 2, -]   3 LerchPhi[--, 2, -]
                16     8               16     8               16     4
>    -------------------- - -------------------- + -------------------- - 
             256                    512                    512
 
                1      7
     3 LerchPhi[--, 2, -]
                16     8
>    --------------------
             1024

Using the integral representation for the Lerch function:

and changing the variable of integration , we obtain

Evaluating the integral, we found that

In[96]:= Integrate[(2/(8 - 4*x + x^2) - 3/(2 + 2*x + x^2))*Log[x],{x,0,1}]

         I                 1    I                   1    I
Out[96]= - (3 PolyLog[2, -(-) - -] - 3 PolyLog[2, -(-) + -] - 
         2                 2    2                   2    2
 
                  1   I               1   I
>      PolyLog[2, - - -] + PolyLog[2, - + -])
                  4   4               4   4

From the Kummer equation

with x = -I, y = 1 - I and x = -I, y = 1 + I correspondently, we find

From the dublication formula

we find

Applying these formulas along with identities for and to Out[96], we obtain

In[97]:= FullSimplify[%//.{
   PolyLog[2, 1/4 + I/4] -> Pi^2/12 + Log[1 - I]^2/2 - Log[2]^2/2 + 
          PolyLog[2, -1/2 + I/2] + PolyLog[2, -I/2] + PolyLog[2, 1/2 + I/2],
   PolyLog[2, 1/4 - I/4] -> Pi^2/12 + Log[1 + I]^2/2 - Log[2]^2/2 + 
           PolyLog[2, -1/2 - I/2] + PolyLog[2, I/2] + PolyLog[2, 1/2 - I/2],
   PolyLog[2, -1/2 + I/2]->PolyLog[2, -I/2]/2-PolyLog[2, 1/2 - I/2],
   PolyLog[2, -1/2 - I/2]-> PolyLog[2, I/2]/2 - PolyLog[2, 1/2 + I/2],
   PolyLog[2, 1/2 - I/2] -> -I*Catalan + Pi^2/48 - 1/2*Log[1/2 + I/2]^2,
   PolyLog[2, 1/2 + I/2] ->  I*Catalan + Pi^2/48 - 1/2*Log[1/2 - I/2]^2
}]

Out[97]= Catalan