Catalan constant

Integral involving Catalan constant

Today we show the proof of the following result found on Twitter: $$\int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dx dy= 2\beta(2)$$

Proof

$$I = \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dxdy = \int_{0}^{\frac{\pi}{2}} y\frac{2}{\pi}\int_{0}^{\infty} \frac{w^{\frac{2y}{\pi}-1}}{w^2+1} dwdy \quad (w \mapsto x^{\frac{\pi}{2y}})$$ Recall the integral representation fo $\sec(x)$ (proof in the Appendix): $$\sec(x) = \frac{2}{\pi}\int_{0}^{\infty} \frac{t^{\frac{2x}{\pi}}}{t^2+1}dt$$ Hence $$\frac{2}{\pi}\int_{0}^{\infty} \frac{w^{\frac{2y}{\pi}-1}}{w^2+1} dw = \sec\left(y-\frac{\pi}{2}\right) = \csc(y) = \frac{1}{\sin(y)}$$ Therefore $$I = \int_{0}^{\frac{\pi}{2}} \frac{y}{\sin(y)}dy = 2\int_{0}^{\infty} \operatorname{arccot}(e^w) dw \quad \left(w \mapsto \ln\left(\cot\left(\frac{y}{2}\right)\right)\right)$$ Recall the series expansion for $\operatorname{arccot}(x)$ $$\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)} x^{-2n-1}$$ Hence $$\begin{align*} I =& 2\int_{0}^{\infty} \operatorname{arccot}(e^w) dw \\ =& 2\int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)} e^{-w(2n+1)}dw\\ =& 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)}\int_{0}^{\infty} e^{-w(2n+1)}dw\\ =& 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}\\ =& 2\beta(2) \end{align*}$$ $$\boxed{\therefore \int_{0}^{\frac{\pi}{2}} \int_{0}^{\infty} \frac{1}{\sqrt[y]{x^{\pi}}+1} dxdy = 2\beta(2)}$$

Appendix: Integral representation of $\sec x$

How do we know that $\displaystyle \sec x= \frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy$?

The proof can be obtained trough contour integration in a branch cut but here we will use a more straigthforward approach with the help of the Ramanujan's master theorem

Let $|a|<1$
Recall $$\frac{1}{1+y^2} = \int_{0}^{1} t^{y^2} dt$$ Therefore $$\begin{align*} \int_{0}^{\infty}\frac{y^a}{1+y^2}dy =& \int_{0}^{\infty}\int_{0}^{1} y^a t^{y^2} dtdy\\ =&\int_{0}^{1} \int_{0}^{\infty} y^a t^{y^2} dydt\\ =& \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt \quad (w \mapsto y^2) \end{align*}$$ The integral $\displaystyle \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw$ is the Mellin transform of $\displaystyle t^w$ at $\displaystyle \frac{a+1}{2}$
Recall that $$t^w = e^{w\ln(t)} = \sum_{n=0}^{\infty} \frac{\ln^n(t)w^n}{n!} = \sum_{n=0}^{\infty} \frac{(-\ln(t))^n(-w^n)}{n!}$$ Therefore, by Ramanujan's master theorem $$\int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dw = \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}}$$ Therefore $$\begin{align*} \frac{1}{2}\int_{0}^{1} \int_{0}^{\infty} w^{\frac{a}{2}-\frac{1}{2}} t^{w} dwdt =& \frac{1}{2}\int_{0}^{1} \Gamma\left(\frac{a+1}{2}\right)\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}\left[-\ln(t)\right]^{-\frac{a+1}{2}} dt\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\int_{0}^{1}v^{-\frac{a+1}{2}}e^{-v} dv \quad (v\mapsto -ln(t))\\ =& \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) \end{align*}$$ Finally, recall the Euler's reflection formula $$\Gamma(z)\Gamma(1-z)= \frac{\pi}{\sin \pi z} = \pi \csc \pi z$$ Therefore $$\int_{0}^{\infty}\frac{y^a}{1+y^2}dy = \frac{1}{2}\Gamma\left(\frac{a+1}{2}\right)\Gamma\left(\frac{1-a}{2}\right) = \frac{\pi}{2} \csc\left(\pi \frac{1+a}{2}\right)=\frac{\pi}{2} \sec\left(\pi \frac{a}{2}\right)$$ If we let $\displaystyle x = \pi \frac{a}{2}$ then $\displaystyle a=\frac{2x}{\pi}$. Then $$\boxed{\frac{2}{\pi}\int_{0}^{\infty}\frac{y^\frac{2x}{\pi}}{1+y^2}dy = \sec x}$$