Catalan's constant

Catalan's constant, $G$ is a mathematical constant, which is approximately equal to

$$0.9159655941772190150546035149323841107741493742816721$$

It can be expressed as $\displaystyle \beta(2) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)^2} = \dfrac1{1^2} - \dfrac1{3^2} + \dfrac1{5^2} - \dfrac1{7^2} + \cdots$,
where $\beta(\cdot)$ denotes the Dirichlet beta function.

$$G=\frac { { \psi }_{ 1 }\left( \frac { 1 }{ 4 } \right) -{ \psi }_{ 1 }\left( \frac { 3 }{ 4 } \right) }{ 16 }$$

$$\begin{aligned} \frac { { \psi }_{ 1 }\left( \frac { 1 }{ 4 } \right) -{ \psi }_{ 1 }\left( \frac { 3 }{ 4 } \right) }{ 16 } & =\frac {\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+\frac { 1 }{ 4 } \right) }^{ 2 } } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+\frac { 3 }{ 4 } \right) }^{ 2 } } } }{ 16 } \\ & =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( 4n+1 \right) }^{ 2 } } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( 4n+3 \right) }^{ 2 } } } \\ & =G \end{aligned}$$

$$G=\int_0^1 \dfrac{\arctan x}{x} dx$$

Use the Maclaurin series expansion for the arctangent, and intergrate term by term: $$\int_0^1 \dfrac{\arctan x}{x} dx=\int_0^1\sum_{n=0}^{\infty}\dfrac{(-1)^n}{2n+1}x^{2n}dx=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2}=G$$

$$G=\int_0^{\infty} \arctan\left( e^{-x}\right)dx$$

$$\int_0^{\infty} \arctan\left( e^{-x}\right)dx=-\int_0^{\infty} \dfrac{\arctan\left( e^{-x}\right)}{e^{-x}}d(e^{-x})\;\stackrel{t=e^{-x}}{=}\;\int_0^1 \dfrac{\arctan t}{t} dt=G$$

$$G=-\int_0^1 \dfrac{\ln x}{x^2+1} dx=\int_1^{\infty}\dfrac{\ln x}{x^2+1}dx$$

Integrate by parts: $$G=\int_0^1 \dfrac{\arctan x}{x} dx=\ln x\arctan x\left|\begin{array}{l}1\\0\end{array}\right.-\int_0^1 \dfrac{\ln x}{x^2+1} dx=-\int_0^1 \dfrac{\ln x}{x^2+1} dx$$ Make the change of variable $y=\dfrac{1}{x}$ in above integral $$G=-\int_0^1 \dfrac{\ln x}{x^2+1} dx=\int_1^{\infty}\dfrac{\ln y}{y^2+1}dy$$

$$G=\dfrac{1}{2}\int_0^{\infty} \dfrac{x}{\cosh x} dx$$

Write the hyperbolic cosine in terms of exponentials, then expand in powers of $e^{-x}$ using the geometric series formula. Integrate term by term. Then: $$\dfrac{1}{2}\int_0^{\infty} \dfrac{x}{\cosh x} dx=\int_0^{\infty}\dfrac{xe^{-x}}{1+e^{-2x}}dx=\sum_{n=0}^{\infty}(-1)^n\int_0^{\infty}xe^{-(2n+1)x}dx=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2}=G$$

$$G=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}\dfrac{x}{\sin x}dx$$

Let $y=\tan\dfrac{x}{2}$ in $\displaystyle G=\int_0^1 \dfrac{\arctan y}{y} dy$ and applying the double-angle formula for sine. Then: $$G=\int_0^1 \dfrac{\arctan y}{y} dy=\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\dfrac{xdx}{\tan\dfrac{x}{2}\cos^2\dfrac{x}{2}}=\dfrac{1}{4}\int_0^{\frac{\pi}{2}}\dfrac{xdx}{\sin\dfrac{x}{2}\cos\dfrac{x}{2}}=\dfrac{1}{2}\int_0^{\frac{\pi}{2}}\dfrac{x}{\sin x}dx$$

$$G=-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx$$

Use the boundedly convergent Fourier series $$-\ln|2\sin x|=\sum_{n=1}^{\infty}\dfrac{\cos2nx}{n},\qquad x\in\mathbb{R}$$ Integrate term by term, then: $$-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx=2\int_0^{\frac{\pi}{4}}\sum_{n=1}^{\infty}\dfrac{\cos2nx}{n}dx=\sum_{n=1}^{\infty}\dfrac{\sin\dfrac{n\pi}{2}}{n^2}=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2}=G$$

$$G=2\int_0^{\frac{\pi}{4}}\ln(2\cos x)dx$$

Use $\displaystyle G=-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx$, write sine in terms of cosine, and apply the well-known evaluation $\displaystyle \int_0^{\frac{\pi}{2}}\ln(2\cos t)dt=0$, then: $$\begin{aligned} G&=-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx=-2\int_0^{\frac{\pi}{4}}\ln(2\cos\left(\dfrac{\pi}{2}-x\right))dx=-2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln(2\cos t)dt\\ &=-2\int_0^{\frac{\pi}{2}}\ln(2\cos t)dt+2\int_0^{\frac{\pi}{4}}\ln(2\cos t)dt=2\int_0^{\frac{\pi}{4}}\ln(2\cos t)dt \end{aligned}$$

$$G=\int_0^{\frac{\pi}{4}}\ln(\cot x) dx=-\int_0^{\frac{\pi}{4}}\ln(\tan x) dx$$

Add $\displaystyle G=-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx$ and $\displaystyle G=2\int_0^{\frac{\pi}{4}}\ln(2\cos x)dx$, then: $$G=\int_0^{\frac{\pi}{4}}\ln(\cot x) dx=-\int_0^{\frac{\pi}{4}}\ln(\tan x) dx$$

$$G=\int_0^{\frac{\pi}{2}} \sinh^{-1}(\sin x) dx=\int_0^{\frac{\pi}{2}} \sinh^{-1}(\cos x) dx$$

Write the inverse hyperbolic sine as an integral whose integrand can be expressed in to powers of sine. And employing the reduction formula then: $$\begin{aligned} \int_0^{\frac{\pi}{2}} \sinh^{-1}(\sin x) dx&=\int_0^{\frac{\pi}{2}} \ln\left(\sin x+\sqrt{1+\sin^2x}\right) dx\\ &=\int_0^{\frac{\pi}{2}}\int_0^x \dfrac{\cos ydy}{\sqrt{1+\sin^2y}}dx\\ &=\int_0^{\frac{\pi}{2}}\int_0^x \cos y\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}(\sin y)^{2n}dydx\\ &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}\dfrac{1}{2n+1}\int_0^{\frac{\pi}{2}}(\sin x)^{2n+1}dx\\ &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}\dfrac{1}{2n+1}.\dfrac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2}=G \end{aligned}$$ Replacing $x$ by $\dfrac{\pi}{2}-x$ immediately gives $\displaystyle G=\int_0^{\frac{\pi}{2}} \sinh^{-1}(\cos x) dx$

$$G=\int_0^1\int_0^1 \dfrac{dxdy}{1+x^2y^2}$$

Start with $\displaystyle G=\int_0^1 \dfrac{\arctan y}{y} dy$ and write the arctangent in terms of the integral which defines it. $$G=\int_0^1 \dfrac{\arctan y}{y} dy=\int_0^1\dfrac{dy}{y}\int_0^y\dfrac{du}{1+u^2}\stackrel{u=yx}{=}\int_0^1\dfrac{dy}{y}\int_0^1\dfrac{ydx}{1+x^2y^2}=\int_0^1\int_0^1 \dfrac{dxdy}{1+x^2y^2}$$

1. $\displaystyle G=\dfrac{1}{2}\sum_{k=0}^{\infty} \dfrac{4^k(k!)^2}{(2k)!(2k+1)^2}$

2. $\displaystyle G=\sum_{n=0}^{\infty}\dfrac{1}{2^{n+1}}\sum_{k=0}^n \binom{n}{k}\dfrac{(-1)^k}{(2k+1)^2}$

3. $\displaystyle G=1-\sum_{n=1}^{\infty} \dfrac{n\zeta(2n+1)}{16^n}$

4. $\displaystyle G=\dfrac{1}{8}\sum_{n=2}^{\infty} \dfrac{n}{2^n}\zeta\left(n+1,\dfrac{3}{4}\right)$

5. $\displaystyle G=\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{n}{2^n}\left(1-\dfrac{1}{2^n}\right)\zeta(n+1)$

6. $\displaystyle G=\dfrac{1}{2}\sum_{n=0}^{\infty} \dfrac{2^n}{\displaystyle(2n+1)\binom{2n}{n}}\sum_{k=0}^n \dfrac{1}{2k+1}$

7. $\displaystyle G=\dfrac{1}{2}\sum_{n=0}^{\infty} \dfrac{4^n}{\displaystyle(2n+1)^2\binom{2n}{n}}$

8. $\displaystyle G=\sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n } } \left( \frac { 1 }{ { \left( 6n+1 \right) }^{ 2 } } -\frac { 1 }{ { \left( 6n+3 \right) }^{ 2 } } +\frac { 1 }{ { \left( 6n+5 \right) }^{ 2 } } \right)$

• Digamma function

• Zeta function