# 33 representations for Catalan's constant

## 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
```

#### 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
```