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.
Contents
Trigamma Representations
\[ G=\frac { { \psi }_{ 1 }\left( \frac { 1 }{ 4 } \right) -{ \psi }_{ 1 }\left( \frac { 3 }{ 4 } \right) }{ 16 } \]
\[\begin{equation} \begin{split} \frac { { \psi }_{ 1 }\left( \frac { 1 }{ 4 } \right) -{ \psi }_{ 1 }\left( \frac { 3 }{ 4 } \right) }{ 16 } & =\frac {\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+\frac { 1 }{ 4 } \right) }^{ 2 } } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( n+\frac { 3 }{ 4 } \right) }^{ 2 } } } }{ 16 } \\ & =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( 4n+1 \right) }^{ 2 } } } -\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { \left( 4n+3 \right) }^{ 2 } } } \\ & =G \end{split} \end{equation} \]
Integral Representations
\[G=\int_0^1 \dfrac{\arctan x}{x} dx\]
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: \[\begin{align} G&=-2\int_0^{\frac{\pi}{4}}\ln(2\sin x)dx=-2\int_0^{\frac{\pi}{4}}\ln(2\cos\left(\dfrac{\pi}{2}-x\right))dx=-2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\ln(2\cos t)dt\\ &=-2\int_0^{\frac{\pi}{2}}\ln(2\cos t)dt+2\int_0^{\frac{\pi}{4}}\ln(2\cos t)dt=2\int_0^{\frac{\pi}{4}}\ln(2\cos t)dt \end{align} \]
\[G=\int_0^{\frac{\pi}{4}}\ln(\cot x) dx=-\int_0^{\frac{\pi}{4}}\ln(\tan x) dx\]
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: \[\begin{align} \int_0^{\frac{\pi}{2}} \sinh^{-1}(\sin x) dx&=\int_0^{\frac{\pi}{2}} \ln\left(\sin x+\sqrt{1+\sin^2x}\right) dx\\ &=\int_0^{\frac{\pi}{2}}\int_0^x \dfrac{\cos ydy}{\sqrt{1+\sin^2y}}dx\\ &=\int_0^{\frac{\pi}{2}}\int_0^x \cos y\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}(\sin y)^{2n}dydx\\ &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}\dfrac{1}{2n+1}\int_0^{\frac{\pi}{2}}(\sin x)^{2n+1}dx\\ &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n}\dfrac{1}{2n+1}.\dfrac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdots(2n+1)}\\ &=\sum_{n=0}^{\infty} \dfrac{(-1)^n}{(2n+1)^2}=G \end{align}\] 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. \[G=\int_0^1 \dfrac{\arctan y}{y} dy=\int_0^1\dfrac{dy}{y}\int_0^y\dfrac{du}{1+u^2}\stackrel{u=yx}{=}\int_0^1\dfrac{dy}{y}\int_0^1\dfrac{ydx}{1+x^2y^2}=\int_0^1\int_0^1 \dfrac{dxdy}{1+x^2y^2}\]
\[\int _{ 0 }^{ 1 }{ \Gamma \left(1+\frac { x }{ 2 } \right)\Gamma \left(1-\frac { x }{ 2 } \right) \, dx } =\frac { aG }{ b\pi } \]
If the equation holds true for coprime positive integers \(a\) and \(b\), find \(a+b\).
\[ \]
Notations:
\( \Gamma(\cdot) \) denotes the Gamma function.
\(G\) denote Catalan's constant, \(\displaystyle G = \sum_{n=0}^\infty \dfrac{ (-1)^n}{(2n+1)^2} \approx 0.916 \).
In finite Series Representations
\(\displaystyle G=\dfrac{1}{2}\sum_{k=0}^{\infty} \dfrac{4^k(k!)^2}{(2k)!(2k+1)^2}\)
\(\displaystyle G=\sum_{n=0}^{\infty}\dfrac{1}{2^{n+1}}\sum_{k=0}^n \binom{n}{k}\dfrac{(-1)^k}{(2k+1)^2}\)
\(\displaystyle G=1-\sum_{n=1}^{\infty} \dfrac{n\zeta(2n+1)}{16^n}\)
\(\displaystyle G=\dfrac{1}{8}\sum_{n=2}^{\infty} \dfrac{n}{2^n}\zeta\left(n+1,\dfrac{3}{4}\right)\)
\(\displaystyle G=\dfrac{1}{2}\sum_{n=1}^{\infty} \dfrac{n}{2^n}\left(1-\dfrac{1}{2^n}\right)\zeta(n+1)\)
\(\displaystyle G=\dfrac{1}{2}\sum_{n=0}^{\infty} \dfrac{2^n}{\displaystyle(2n+1)\binom{2n}{n}}\sum_{k=0}^n \dfrac{1}{2k+1}\)
\(\displaystyle G=\dfrac{1}{2}\sum_{n=0}^{\infty} \dfrac{4^n}{\displaystyle(2n+1)^2\binom{2n}{n}}\)
\(\displaystyle G=\sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n } } \left( \frac { 1 }{ { \left( 6n+1 \right) }^{ 2 } } -\frac { 1 }{ { \left( 6n+3 \right) }^{ 2 } } +\frac { 1 }{ { \left( 6n+5 \right) }^{ 2 } } \right) \)
Brilli the Bug has set out on a journey of infinite steps starting at the origin of the \(xy\)-plane. It moves in the following manner:
- After each, \(n\)th step, it turns \(90^\circ\) counter-clockwise
- Each \(n\)th step is of length \(D_{n}\) where \( D_{n}\) is given by \(D_{n} = \dfrac{2}{(n+1)^{2}} \) for \( n \geq 0 \).
If the final displacement of brilli from the starting is given by \(\dfrac{1}{\alpha}\sqrt{ \beta G^{\gamma} + \pi^{\eta}} \), find \( \alpha +\beta +\gamma +\eta \).
Notation: \(G\) denotes the Catalan's constant.