Thirty-Two Representations of Catalan's Constant

Victor Adamchik

Catalan's constant C is a numerical constant (called Catalan in Mathematica) that appears in many combinatorial and other settings. 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 Mathematica. In other cases, explicit symbolic manipulation in Mathematica is required to develop a proof.

({{(∫_0^1)(((tan^(-1))(x))/x) 
                  dx, proof}})

Proof

(∫_0^1) ArcTan[x]/x \[DifferentialD]x

■

({{(1/2) (∫_0^((π/2)))(x/(sin(x))) 
                    dx, proof}})

Proof

(1/2) 
      (∫_0^((π/2)))(x/Sin)[x] \[DifferentialD]x

■

({{-(∫_0^1)((log(x))/((x^2) + 1)) 
                    dx, proof}})

Proof

-(∫_0^1) Log[x]/((x^2) + 1) 
          \[DifferentialD]x

■

({{(1/2) (∫_0^∞)x sech(x) 
                    dx, proof}})

Proof

(1/2) 
      (∫_0^∞) x Sech[x] \[DifferentialD]x

■

({{-2 
                (∫_0^((π/4)))log(2 sin(x)) 
                    dx, proof}})

Proof

-2 
      (∫_0^((π/4)))Log[2 Sin[x]] \[DifferentialD]x

FullSimplify[%]

■

({{2 (∫_0^((π/4)))log(2 cos(x)) 
                    dx, proof}})

Proof

2 (∫_0^((π/4)))Log[2  Cos[x]] \[DifferentialD]x

FullSimplify[%]

■

({{ 
              (∫_0^((π/4)))log(cot(x)) dx, proof}})

Proof

(∫_0^((π/4)))Log[Cot[x]] \[DifferentialD]x

■

({{(∫_0
                    ^((π/2)))((x csc(x))/(cos(x) + sin(x))) 
                    dx - (1/4) π log (2), proof}})

Proof

(∫_0^((π/2)))((x Csc[x])/(Cos[x] + Sin[x])) 
          \[DifferentialD]x - (1/4) π Log[2]

■

({{-(∫_0
                      ^((π/2)
                          ))((x csc(x))/(cos(x) - sin(x))) 
                      dx+(1/4) π log(2), proof}})

Proof

(1/4) π Log[2] - 
      Integrate[((x Csc[x])/(Cos[x] - Sin[x])), {x, 0, (π/2)}, 
        PrincipalValue -> True]

■

({{(2/π) 
                  (∫_0^1)((((tan^(-1))(x))^2)/x) 
                      dx+((7 ζ(3))/(4 π)), proof}})

Proof

\(2\/π\ \(∫\_0\%1\( ArcTan[x]\^2\/x\) \[DifferentialD]x\) + 
      \(7\ Zeta[3]\)\/\(4\ π\)\)

Expand[%]

■

({{(1/(2 π)) (∫_0^1)((x^2)/(sin(x))) 
                      dx+((7 ζ(3))/(4 π)), proof}})

Proof

((7 Zeta[3])/(4 π)) + 
      (1/(2 π)) 
        (∫_0^((π/2)))((x^2)/Sin)[x] \[DifferentialD]x

\(% /. {
		
        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\^\(2\ n\)
}\)

FullSimplify[%]

■

({{-(∫_0^1)((log(((1 - x)/(Sqrt[2]))))/((x^2) + 1)) 
                    dx, proof}})

Proof

-(∫_0^1) Log[((1 - x)/(Sqrt[2]))]/((x^2) + 1) 
          \[DifferentialD]x

From the Landen identity for the dilogarithm \(TraditionalForm
      \`\(Li\_2\)(z) -> 
        \(-\(1\/2\)\)\ \(\(log\^2\)(1 - z)\) - \(Li\_2\)(z\/\(z - 1\))\)
with z = (1/2) - (i/2) and z = (1/2) + (i/2) we have

PolyLog[2, (1/2) - (I/2)] -> 
      -I Catalan + ((π^2)/48) - 
        (1/2) Log[(1/2) + (I/2)]^2;

and

PolyLog[2, (1/2) + (I/2)] -> 
      I Catalan + ((π^2)/48) - 
        (1/2) Log[(1/2) - (I/2)]^2;

respectively. Applying them to the result of integration we obtain

%%% /. %% /. %

FullSimplify[%]

■

({{-(∫_0^1)
                    ((log( ((1 - (x^2))/2)))/((x^2) + 1)) 
                    dx, proof}})

Proof

-(∫_0^1) Log[(1/2) (1 - (x^2))]/((x^2) + 1) 
          \[DifferentialD]x

In the same manner as  in the previous section

\(% /. {
		
        PolyLog[2, 1\/2 - I\/2] -> 
          π\^2\/48 - 1\/2\ Log[1\/2 + I\/2]\^2 - \ 
            I\ Catalan\ , 
        PolyLog[2, 1\/2 + I\/2] -> 
          π\^2\/48 - 1\/2\ Log[1\/2 - I\/2]\^2 + 
            I\ Catalan\ 
		}\)

FullSimplify[%]

■

({{(3/4) (∫_0^((π/6)))(x/(sin(x))) 
                      dx + (1/8) π log ((Sqrt[3]) + 2), proof}})

Proof

(3/4) (∫_0^((π/6)))(x/Sin)[x] \[DifferentialD]x + 
      (1/8) π Log[(Sqrt[3]) + 2]

From the reflection formula for the dilogarithm \(TraditionalForm
      \`\(Li\_2\)(z) = 
        \(-\(1\/2\)\)\ \(\(log\^2\)(\(-z\))\) - π\^2\/6 - 
          \(Li\_2\)(1\/z)\)
with z = -((-1)^(1/6)) it follows that

PolyLog[2, -(((-1))^(1/6))] -> 
      -((11 (π^2))/72) - PolyLog[2, (((-1))^(5/6))];

Applying this to the result of the integration, we have

FullSimplify[%% /. %]

% /. 
      ρ_ Log[α_]  +  ρ_ Log[β_]  :> 
        ρ Log[α β // Expand]

Collect[%,  Pi,  Simplify]

Using the following well-known property of the dilogarithm \(TraditionalForm
      \`\(Li\_2\)(t\^q) = 
        q\ \(∑\+\(j = 0\)\%\(q - 1\)\(Li\_2\)(
              e\^\(\(2\ π\ i\ j\)\/q\)\ t)
              \)\)
with t = -i and q = 3

PolyLog[2, t^q] = q*Sum[PolyLog[2, Exp[(2*Pi*I*j)/q]*t], {j, 0, q - 1}]

PolyLog[2, (((-1))^(1/6))] + PolyLog[2, (((-1))^(5/6))] 
      -> ((4 I)/3)  Catalan + ((π^2)/72);

we finally obtain

Expand[%% /. %]

■

({{-((π^2)/4) (∫_0^1)(x - (1/2)) sec(π x) 
                    dx, proof}})

Proof

We divide the interval of integration into two parts (0,(1/2)) and ((1/2),1) so that \(TraditionalForm
      \`∫\_0\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
            dx = 
        ∫\_0\%\(1\/2\)\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
              dx + 
          ∫\_\(1\/2\)\%1\((x - 1\/2)\)\ \(sec(π\ x)\) 
              dx\)
Changing the variable x->1-z in the first integral, we obtain
 \(TraditionalForm
      \`∫\_0\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
            dx = 
        \(∫\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
                dx + 
            ∫\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
                dx = \ 
          2\ \(∫\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
                dx\)\)\).
This leads to the following identity \(TraditionalForm
      \`\(-\(π\^2\/4\)\)\ 
          \(∫\_0\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
              dx\) = 
        \(-\(π\^2\/2\)\)\ 
          \(∫\_\(1\/2\)\%1\(\((x - 1\/2)\)\ \(sec(π\ x)\)\) 
              dx\)\)
the right side of which can be easily evaluated:

-((π^2)/2)  
      (∫_((1/2))^1)(x - (1/2)) Sec[π x] 
          \[DifferentialD]x

■

({{(1/2) (∫_0^1)K((x^2)) dx, proof}})

Proof

Using the integral representation
 \(TraditionalForm\`\(

      K(x)\  = \ 
        ∫\_0\%1 1\/\@\(1 - x\ \(\(sin\^2\)(t)\)\)\ 
            dt\)\)

and changing the order of integration, we get

\(TraditionalForm
      \`1\/2\ \(∫\_0\%1\( K(x\^2)\) dx\)\  = \ 
        1\/2\ \(∫\_0
              \%\(π\/2\)\(∫\_0\%1
                \( 1\/\@\(1 - x\^2\ \(\(sin\^2\)(t)\)\)\) dx\ 
                dt\)\)\).

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

(∫_0^1) (1/(Sqrt[1 - (x^2) Sin[t]^2])) \[DifferentialD]x

(1/2) 
      (∫_0^((π/2)))PowerExpand[%] \[DifferentialD]t

■

({{(∫_0^1)E((x^2)) dx - (1/2), proof}})

Proof

In the same manner as in the previous section using the integral representations for the elliptic function
 \(TraditionalForm
      \`E(x) = ∫\_0\%\(π\/2\)\(\@\(1 - x\ \(\(sin\^2\)(t)\)\)\) 
            dt\):
 
we have

\(TraditionalForm
      \`\(-\(1\/2\)\) + ∫\_0\%1\( E(x\^2)\) dx\  = 
        \(-\(1\/2\)\) + 
          ∫\_0
              \%\(π\/2\)\(∫\_0\%1
                \(\@\( 1 - x\^2\ \(\(sin\^2\)(t)\)\)\) dx\ 
                dt\)\).

So we have:

(∫_0^1)(Sqrt[ 1 - (x^2) Sin[t]^2]) \[DifferentialD]x

-(1/2) + 
      (∫_0^((π/2)))PowerExpand[%] \[DifferentialD]t

■

({{(∫_0^((π/2)))(sinh^(-1))(sin(x)) 
                  dx, proof}})

Proof

The integral can be evaluated to Catalan`s constant by using the integral representation for (sinh^(-1))(z)
\(TraditionalForm
      \`\(sinh\^\(-1\)\)(z) = 
        ∫\_0\%1\( z\/\@\(t\^2\ z\^2 + 1\)\) dt\)
Substituting it into the given integral and changing the order of integration, we have
\(TraditionalForm
      \`∫\_0\%\(π\/2\)\(\(sinh\^\(-1\)\)(sin(x))\) 
            dx = 
        ∫\_0\%1
            \((∫\_0
                  \%\(π\/2
                      \)\(\(sin(x)\)\/\@\(z\^2\ \(\(sin\^2\)(x)\) + 1\)\) 
                  dx)\) dz\)
Then with the use of Mathematica we obtain the desirable result.

(∫_0^1)
        ((∫_0^((π/2)))Sin[x]/(Sqrt[(z^2) Sin[x]^2 + 1]) 
              \[DifferentialD]x) \[DifferentialD]z

From (∫_0^((π/2)))(sinh^(-1))(sin(x)) 
                dx = C it immediately follows 
\(TraditionalForm
      \`C = ∫\_0\%\(π\/2\)\(\(sinh\^\(-1\)\)(cos(x))\) 
            dx\)
\(TraditionalForm
      \`C = ∫\_0\%\(π\/2\)\(\(csch\^\(-1\)\)(csc(x))\) 
            dx\)
\(TraditionalForm
      \`C = ∫\_0\%\(π\/2\)\(\(csch\^\(-1\)\)(sec(x))\) 
            dx\)

■

({{(∫_0^1)(∫_0^1)(1/((Sqrt[1 - x]) (Sqrt[1 - y]) (x + y))) 
                    dx dy, proof}})

Proof

To prove this we consider the more general integral

(1/4)(∫_0^1)(∫_0^1)((x^λ)/((Sqrt[1 - x]) (Sqrt[1 - y]) (x + y)))dxdy

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

(1/4) 
      Integrate[
        ((x^λ)/((Sqrt[1 - x]) (Sqrt[1 - y]) (x + y))), {y, 0, 1}, {
          x, 0, 1}, GenerateConditions -> False]

Now we need to find the limit of this expression when λ tends to ∞. At first we find the asymptotic expansion of (Subscript[ , 3])(Subscript[F, 2]). For that we convert the hypergeometric function into the series and consider the summand:

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

FunctionExpand[%]

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

\(\(\(Series[%, \ {λ, \ 0, \ 1}] // Normal\) // FullSimplify

    \)\)

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

and sum it up

Sum[%,  {k, 0, Infinity}] // Simplify

Finally replacing (Subscript[ , 3])(Subscript[F, 2]) in the result of integration by this asymptotic expansion, we obtain

(1/4) 
      (((2 (Sqrt[π]) Gamma[λ] %)/Gamma)[λ + (1/2)] - 
          (((π^(3/2)) Csc[π λ] 
              Gamma[(λ/2) + (1/2)])/(λ 
              Gamma[(λ/2)])))

Series[%,  {λ,  0,  0}] // Normal // FullSimplify

■

({{(∑_(k = 0
                    )^∞)((((-1))^k)/(((2 k + 1))^2)) x, proof}})

Proof

(∑_(k = 0)^∞)((((-1))^k)/(((2 k + 1))^2))

■

({{2 (∑_(k = 0)^∞)(1/(((4 k + 1))^2)) - 
                ((π^2)/8), proof}})

Proof

-((π^2)/8) + 
      2 (∑_(k = 0)^∞) (1/(((4 k + 1))^2))

Expand[%]

■

({{-2 (∑_(k = 0)^∞)(1/(((4 k + 1))^3))+((π^2)/8), proof}})

Proof

((π^2)/8) - 
      2 (∑_(k = 0)^∞) (1/(((4 k + 3))^2))

Expand[%]

■

({{(1/2) (∑_(k = 0)^∞)(((4^k) 
                      ((k!)^2))/((2 k)! (((2 k + 1))^2))), proof}})

Proof

(1/2) 
      (∑_(k = 0)^∞)(( (4^k) 
            ((k!)^2))/((2 k)! (((2 k + 1))^2)))

■

({{(∑_(k = 0)^∞)((((-1))^k)/(2 k + 1))(γ+ψ(k + 1))+(π/2) log(2), proof}})

Proof

(1/2) Log[2] π + 
      (∑_(k = 0)^∞)(((((-1))^k) 
            (PolyGamma[k + 1] + EulerGamma))/(2 k + 1))

Expand[%]

■

({{(∑_(k = 0)^∞)((((-1))^k)/(2 k + 1))(γ+ψ(k+(3/2)))+(π/4) log(2), proof}})

Proof

(1/4) π Log[2] + 
      (∑_(k = 0)^∞)(((((-1))^k) 
            (PolyGamma[k + (3/2)] + EulerGamma))/(2 k + 1))

Expand[%]

■

({{1 - (∑_(k = 1)^∞)(k/(4^(2 k))) ζ 
                    (2 k + 1), proof}})

Proof

1 - 
      (∑_(k = 1)^∞)(( k Zeta[2 k + 1])/(4^(2 k)))

■

({{(1/16) (∑_(k = 1
                      )^∞)((((3^k) - 1) (k + 1))/(4^k))
                           ζ (k + 2), proof}})

Proof

(1/16) 
      (∑_(k = 1)^∞)((((3^k) - 1) (k + 1) 
            Zeta[k + 2])/(4^k))

■

({{(1/8) (∑_(k = 2)^∞)((k ζ(k + 1,(3/4)))/((2^k))), proof}})

Proof

(1/8) 
      (∑_(k = 2)^∞)(( k )/(2^k)) 
          Zeta[k + 1, (3/4)]

■

({{1-(1/8) (∑_(k = 2)^∞)(k/(2^k))ζ(k + 1,(5/4)), proof}})

Proof

1 - 
      (1/8) (∑_(k = 2)^∞)(( k )/(2^k)) 
            Zeta[k + 1, (5/4)]

■

({{(( π)/2) log(2) - 
                (( π)/32) 
                  (∑_(k = 0)^∞)((((2 k + 1)!)^2)
                        /((16^k) ((k!)^4) (((k + 1))^3))), proof}})

Proof

Comparing

(1/2) π Log[2] - 
      (1/32) π 
        (∑_(k = 0)^∞)((((2 k + 1)!)^2)/((16^k) 
              ((k!)^4) (((k + 1))^3)))

and

(∫_0^1)(x - (1/2)) Sec[π x] \[DifferentialD]x

and taking into account the result of the section 15 Catalan`s constant pops up immediately .

■

({{(Sqrt[2]) (∑_(k = 0)^∞)(((2 k)!)/((8^k) 
                        ((k!)^2) (((2 k + 1))^2))) - 
                (( π)/4) log(2), proof}})

Proof

-(1/4) π Log[2] + 
      (Sqrt[2]) (∑_(k = 0)^∞)(((2 k)!)/((8^k) 
              ((k!)^2) (((2 k + 1))^2)))

We observe that \(TraditionalForm
      \`1\/\(2\ k + 1\) = 
        ∫\_0\%1\( t\^\(2\ k\)\) dt\).
Substituting it into the above series and changing the order of summation and integration, we obtain

\(TraditionalForm
      \`∑\+\(k = 0\)\%∞\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ 
              \((2\ k + 1)\)\^2\) = 
        \(∑\+\(k = 0
                \)\%∞\(\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ 
                    \((2\ k + 1)\)\)\) 
              \((∫\_0\%1\( t\^\(2  k\)\) dt)\) = 
          ∫\_0\%1
              \((∑\+\(k = 0\)\%∞\(\(\((2\ k)\)!\) 
                      t\^\(2  k\)\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\))\) 
              d\)\)

which leads to the following equality  

\(TraditionalForm
      \`\(-\(1\/4\)\)\ π\ \(log(2)\) + 
          \@2\ \(∑\+\(k = 0\)\%∞\(\((2\ k)\)!\)\/\(8\^k\ 
                  \(k!\)\^2\ \((2\ k + 1)\)\^2\)\) = 
        \(-\(1\/4\)\)\ π\ \(log(2)\) + 
          \@2\ \(∫\_0\%1
                \((∑\+\(k = 0\)\%∞\(\(\((2\ k)\)!\)\ 
                        t\^\(2\ k\)\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\))\) 
                dt\)\)

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

(∑_(k = 0)^∞)(((2 k)! (t^(2 k)))/((8^k) 
          ((k!)^2) (2 k + 1)))

PowerExpand[%,  t]

Integrate[%,  {t,  0,  1}]

Finally, we have

-(1/4) π Log[2] + (Sqrt[2]) %

FullSimplify[%]

■

({{(π/8) log((Sqrt[3]) + 2) + 
                (3/8) (∑_(k = 0)^∞)(((k!)^2)
                        /((2 k)! (((2 k + 1))^2))), proof}})

Proof

(1/8) π Log[(Sqrt[3]) + 2] + 
      (3/8) (∑_(k = 0)^∞)((( k!)^2)/((2 k)! 
              (((2 k + 1))^2)))

In the the same manner as in the previous section replacing \(TraditionalForm
      \`\(1\/\(2\ k + 1\)\) by\ \ 
        \(∫\_0\%1\( t\^\(2\ k\)\) dt\)\) and changing the order of summation and integration, we obtain

\(TraditionalForm
      \`1\/8\ π\ \(log(\@3 + 2)\) + 
          3\/8\ \(∑\+\(k = 0\)\%∞\( k!\)\^2
                  \/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\^2\)\)\  = \ 
        1\/8\ π\ \(log(\@3 + 2)\) + 
          \(3\/8\) 
            \(∫\_0\%1
                \(∑\+\(k = 0
                    \)\%∞\(\(\(\( k!\)\^2\) 
                        t\^\(2\ k\)\)\/\(\(\((2\ k)\)!\)\ \((2\ k + 1)\)\)\) 
                  dt\)\)\).

Now using Mathematica we evaluate the sum and the integral in the right side

(∑_(k = 0)^∞)(((( k!)^2) 
          (t^(2 k)))/((2 k)! (2 k + 1)))

PowerExpand[%,  t]

Integrate[%,  {t,  0,  1}]

Thus, the given sum is

(1/8) π Log[(Sqrt[3]) + 2] + ((3 %)/8)

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

% /. {
        PolyLog[2, -(((-1))^(1/6))] -> 
          -((11 (π^2))/72) - PolyLog[2, (((-1))^(5/6))], 
        PolyLog[2, (((-1))^(1/6))] -> 
          ((4 I)/3)  Catalan + ((π^2)/72) - 
            PolyLog[2, (((-1))^(5/6))]}

FullSimplify[%]

% /. 
      ρ_ Log[α_]  +  ρ_ Log[β_]  :>  
        ρ Log[α β // Expand]

■

({{(1/4) (∑_(k=0)^∞)(((2^k) ((k!)^2))/((2 k+1)!)) ψ(k + (3/2))+(π/4)  log(2)+(π/8) γ, proof}})

Proof

((π EulerGamma)/8) + (1/4) π Log[2] + 
      (1/4) (∑_(n = 0)^∞)(( (2^n) ((n!)^2) 
              PolyGamma[(3/2) + n])/((1 + 2 n)!))

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

(ψ^((0)))(n + (3/2))=(ψ^((0)))(n + (1/2))+(2/(2 n + 1))

we divide the given sum in two sums:

(∑_(n = 0)^∞)(((2^n) ((n!)^2) (ψ^((0)))(n + (3/2)))/((2 n + 1)!))=2 (∑_(n = 0)^∞)(( (2^n) 
                    ((n!)^2))/((2 n + 1)! (2 n + 1)))+(∑_(n = 0)^∞)(((2^n) ((n!)^2) (ψ^((0)))(n + (1/2)))/((2 n + 1)!))

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

\(TraditionalForm
      \`∑\+\(n = 0\)\%∞\( 2\^n\ 
              \(n!\)\^2\)\/\(\(\((2\ n + 1)\)!\)\ \((2\ n + 1)\)\) = 
        ∫\_0\%1
            \((∑\+\(n = 0\)\%∞\( 2\^n\ \(n!\)\^2\ 
                    t\^\(2\ n\)\)\/\(\((2\ n + 1)\)!\))\) 
            dt\)

and then evaluated

(∑_(n = 0)^∞)(( (2^n) ((n!)^2) 
          (t^(2 n)))/((2 n + 1)!))

(∫_0^1) PowerExpand[%, t] \[DifferentialD]t

FullSimplify[%]

Let us name the first sum as  α:

α = 2 %;

Now we consider the second sum:
(∑_(n = 0)^∞)(((2^n) ((n!)^2) (ψ^((0)))(n + (1/2)))/((2 n + 1)!))
At first we need to introduce values of of Tan and Cot at (π/8):

Unprotect[Tan,  Cot];

Tan[(π/8)] = (Sqrt[2]) - 1; Cot[(π/8)] = (Sqrt[2 ]) + 1;

Protect[Tan,  Cot];

Then with use of Mathematica we calculate that sum :

FullSimplify[
      (∑_(n = 0)^∞)(( (2^n) ((n!)^2) 
            PolyGamma[0, n + (1/2)])/((2 n + 1)!))]

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

% /. {
        PolyLog[2, (1/2) - (I/2)] -> 
          -I Catalan + ((π^2)/48) + 
            (1/32) (((π + 2 I Log[2]))^2), 
        PolyLog[2, (1/2) + (I/2)] -> 
          I Catalan + ((π^2)/48) + 
            (1/32) (((π - 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])))] 
          -> -((π^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])))] -> 
          -((π^2)/6) - 
            (1/2) Log[(((-1))^(3/4)) (1 + (Sqrt[2]))]^2 - 
            PolyLog[2, ((((-1))^(1/4))/(1 + (Sqrt[2])))]};

We name the second sum β:

β =  FullSimplify[%]

Finally, combining the values of α and  β, we get

FullSimplify[
      ((EulerGamma π)/8) + (1/4) π Log[2] + 
        ((α + β)/4)]

Now performing the following chain of simplification rules we obtain the desirable result:

% /. ArcCot[1  -  Sqrt[2]] -> -ArcCot[1  +  Sqrt[2]] - Pi/4;

Collect[%, ArcCot[1  +  Sqrt[2]], FullSimplify]

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

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

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

Expand[%]

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