Thirty-Three 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 analytic settings. This notebook presents thirty-three 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 of simply evaluating the representation directly in Mathematica. In other cases, explicit symbolic manipulation in Mathematica is needed to develop a proof.

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

Proof

(∫_0^1) ArcTan[x]/x \[DifferentialD]x

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

Proof

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

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

Proof

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

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

Proof

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

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

Proof

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

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

Proof

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

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

Proof

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

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({{(∫_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] // 
      FullSimplify

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({{-(∫_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] // FullSimplify

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({{\(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\) + \(7\ Zeta[3]\)\/\(4\ π\) // FullSimplify\)

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({{(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 // FullSimplify

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({{-(∫_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 // FullSimplify

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({{-(∫_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 // FullSimplify

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({{(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] // FullSimplify

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

Proof

-((π^2)/4)(∫_0^1)(x - (1/2)) 
            Sec[π x] \[DifferentialD]x

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

Proof

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

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

Proof

We use an integral representation for the elliptic function:
 E(
          x) = (∫_0^((π/2)))(Sqrt[1 - 
                  x (sin^2)(t)]) dt
 Then
\(TraditionalForm\`\(-\(1\/2\)\) + ∫\_0\%1\( E(
                x\^2)\) dx\  = \(-\(1\/2\)\) + ∫\_0\%\(π\/2\)\(∫\_0\%1\(\@\( 1 - 
                      x\^2\ \(\(sin\^2\)(
                          t)\)\)\) dx\ dt\)\)
Now we use Mathematica to do the double integration.

Integrate[(Sqrt[1 - (x^2) Sin[t]^2]), {x, 0, 1}, 
      GenerateConditions -> False]

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

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

Proof

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

Immediate corollaries

It immediately follows  that
C = (∫_0^((π/2)))(sinh^(-1))(
              cos(x)) dx
C = (∫_0^((π/2)))(csch^(-1))(
              csc(x)) dx
C = (∫_0^((π/2)))(csch^(-1))(
              sec(x)) dx

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({{(∫_0^1)(∫_0^1)\(\(1\/\(\@\(1 - x\)\ \@\(1 - y\)\ \((x + 
                              y)\)\)\) dx dy\), proof}})

Proof

We consider the more general integral (1/4)(∫_0^1)(∫_0^1)((x^λ)/((Sqrt[1 - x]) (Sqrt[1 - y]) (x + 
                          y)))dx\(dy\), where λ is an arbitrary parameter. We can evaluate this integral with Mathematica.

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

To find the original integral we need to find the limit of this expression as λ tends to 0. This in turn means that we need to find the asymptotic expansion of \(TraditionalForm\`\(\( \)\(\_3\)\) F\_2\). For that we convert the hypergeometric function into a series and look at the general term.

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[%]

We expand this into a series with respect to λ around zero.

Series[%,  {λ,  0,  1}] // Normal // FunctionExpand // 
      FullSimplify

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

Now we sum the series with this as the general term.

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

Finally, in the result of the integration at the beginning, we replace \(TraditionalForm\`\(\( \)\(\_3\)\) F\_2\) by this asymptotic expansion.

int /. _HypergeometricPFQ  ->  asymp

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

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

Proof

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

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

Proof

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

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

Proof

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

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({{\(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)))

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({{(∑_(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

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({{(∑_(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

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

Proof

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

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({{(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))

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({{(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)]

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({{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)]

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

Proof

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

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({{\(\@2\ \(∑\+\(k = 0\)\%∞\(\((2\ k)\)!\)\/\(8\^k\ \(k!\)\^2\ \((2\ k + 1)\)\^2\)\) - \(\(\ \)\(π\)\)\/4\ \(log(
                    2)\)\), proof}})

Proof

The direct approach does not work.

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

We observe that (1/(2 k + 1)) = (∫_0^1) 
                (t^(2 k)) dt.
Substitute this into the series and change the order of summation and integration.
\(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\ \((2\ k + 1)\)\))\) dt\)\)
We evauate the series and integral on the right side with Mathematica and then form the original expression.

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

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

-(1/4) π Log[2] + (Sqrt[2]) %

FullSimplify[%]

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({{\(π\/8\ \(log(\@3 + 2)\) + 
                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))) // Simplify

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

Proof

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

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