Evaluating a double integral

Question

I was trying to evaluate the following integral $$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx.$$ I have a guess that the value of this integral is $\frac{\pi^4}{8}$. But I am unable to prove it. Could someone please help me in evaluating this integral?

Or, can we show the following identity holds without much calculation?

$$\int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{2y \ln y \ln x}{(x^2+ y^2)( 1+y^2)} dy dx= \int_{x=0}^{\infty}\int_{y=0}^{\infty}\frac{y (\ln y )^2}{(x^2+ y^2)( 1+y^2)} dy dx.$$

Any help or hint would be appreciated. Thanks in advance.

Answer 1

We shall use an approach via Mellin transforms:

We aim to find $$I=2\int_{0}^{\infty}\frac{y\ln y}{1+y^2}\int_{0}^{\infty}\frac{\ln x}{x^2+y^2}\, dx\, dy$$

We shall begin by evaluating the Mellin transform $$\int_{0}^{\infty}\frac{x^{s-1}}{x^2+y^2}\, dx\stackrel{x\to u y}{=}y^{s-2}\int_{0}^{\infty}\frac{u^{s-1}}{u^2+1}\, du=\frac{y^{s-2}\pi}{2}\csc\left(\frac{\pi s}{2}\right)$$ which is a classical result that can be obtained via the beta function or contour integration for example: see here.

Now \begin{align*}\int_{0}^{\infty}\frac{\ln x}{x^2+y^2}\, dx &=\frac{\partial}{\partial s} \left[\frac{y^{s-2}\pi}{s}\csc\left(\frac{\pi s}{2}\right)\right]_{s=1}\\ &=\lim_{s\to 1}\left[-\frac{\pi}{4} y^{s-2}\csc\left(\frac{\pi s}{2}\right)\left(\pi\cot\left(\frac{\pi s}{2}\right)-2\ln y\right)\right]\\&=\frac{\pi\ln y}{2y}\end{align*}

This means $$I=\pi\int_{0}^{\infty}\frac{\ln^2 y}{1+y^2}\, dy$$ and using the same method $$\int_{0}^{\infty}\frac{y^{s-1}}{1+y^2}\, dy=\frac{\pi}{2}\csc\left(\frac{\pi s}{2}\right)$$ so \begin{align*}I&=\pi\frac{d^2}{ds^2}\left[\frac{\pi}{2}\csc\left(\frac{\pi s}{2}\right)\right]_{s=1}\\&=\pi\lim_{s\to 1}\left[\frac{\pi^3}{16}(\cos(\pi s)+3)\csc^3\left(\frac{\pi s}{2}\right) \right]\\&=\frac{\pi^4}{8}\end{align*}

as desired. $\square$

Answer 2

We can evaluate a more general integral. Let's denote $$I(k)=\int_0^{\infty}\int_0^{\infty}\frac{y \ln ^ky \ln x}{(x^2+ y^2)( 1+y^2)} dy dx$$ where $k=0,1,2, ...$

Then $$I(k)\overset{t=y^2}{=}\frac{1}{2^{k+1}}\int_0^\infty\int_0^\infty\frac{\ln^kt\ln x}{(x^2+t)(1+t)}dxdt\overset{x=\sqrt t \,z}{=}\frac{1}{2^{k+1}}\int_0^\infty\frac{\sqrt t\,\ln^kt}{t(1+t)}dt\int_0^\infty\frac{\ln(\sqrt t\,z)}{1+z^2}dz$$ Given that $\displaystyle \int_0^\infty\frac{\ln z}{1+z^2}dz=0$ and $\displaystyle \int_0^\infty\frac{dz}{1+z^2}dz=\frac{\pi}{2}$ $$I(k)=\frac{\pi}{2^{k+3}}\int_0^\infty\frac{\ln^{k+1}t}{\sqrt t(1+t)}dt\overset{t=x^2}{=}\frac{\pi}{2}\int_0^\infty\frac{\ln^{k+1}x}{1+x^2}dx\overset{x=e^{\pi t}}{=}\frac{\pi^{k+3}}{4}\int_{-\infty}^\infty\frac{x^{k+1}}{\cosh\pi x}dx\tag{1}$$ Let's denote $\displaystyle J(\beta)=\int_{-\infty}^\infty\frac{e^{\beta x}}{\cosh\pi x}dx$, then $\displaystyle\int_{-\infty}^\infty\frac{x^k}{\cosh\pi x}dx=\frac{d^k}{d\beta^k}J(\beta)\,\bigg|_{\beta=0}$

A straightforward evaluation of $J(\beta)$ via integration in the complex plane along a rectangular contour gives $$J(\beta)=\frac{1}{\cos\frac{\beta}{2}};\quad\int_{-\infty}^\infty\frac{x^k}{\cosh\pi x}dx=\frac{1}{2^k}\frac{d^k}{dx^k}\frac{1}{\cos x}\,\bigg|_{x=0}\tag{2}$$ Putting (2) into (1) $$I(k)=\Big(\frac{\pi}{2}\Big)^{k+3}\frac{d^{k+1}}{dx^{k+1}}\frac{1}{\cos x}\,\bigg|_{x=0}\tag{3}$$ $$I(k=1)=\Big(\frac{\pi}{2}\Big)^4$$ Using the presentation $\displaystyle \frac{1}{\cos x}=\sum_{n=0}^\infty(-1)^n\frac{E_{2n}}{(2n)!}x^{2n}$, where $E_{2n}$ denotes Euler' numbers, $$I(2n-1)=(-1)^nE_{2n}\Big(\frac{\pi}{2}\Big)^{2(n+1)};\quad E_2=-1; \,E_4=5,...$$ For even $k=2n$ we get zero.

Answer 3

$$I=\int\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy $$ is not bad since $$A=\frac{y }{(x^2+ y^2)( 1+y^2)}=\frac{y}{(y-i) (y+i) (y-i x) (y+i x)}$$ Using partial fraction decomposition $$A=\frac 1{2(x^2-1)}\left(\frac 1{y-i} +\frac 1{y+i}-\frac 1{y-ix}-\frac 1{y+ix}\right)$$ and $$\int \frac {\log(y)}{y+a}\,dy=\text{Li}_2\left(-\frac{y}{a}\right)+\log (y) \log \left(1+\frac{y}{a}\right)$$ Therefore $$I=\frac 1{2(x^2-1)}\left(\frac{1}{2} \left(\text{Li}_2\left(-y^2\right)-\text{Li}_2\left(-\frac{y^2 }{x^2}\right)\right)+\log (y) \log \left(\frac{x^2 \left(y^2+1\right)}{x^2+y^2}\right) \right)$$ and then $$J=\int_0^\infty\frac{y \log(y) }{(x^2+ y^2)( 1+y^2)} dy =\frac{\log ^2(x)}{2 \left(x^2-1\right)}$$

Then, what remains is the computation of $$K=\int \frac{\log ^3(x)}{x^2-1}\,dx=\frac 12\left(\int \frac{\log ^3(x)}{x-1}\,dx- \int \frac{\log ^3(x)}{x+1}\,dx\right)$$ which is not too bad using a few integrations by parts.

Answer 4

To show your identity holds, make the one-dimensional change of variables $x=ty$ to get

\begin{align*} \int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &= \int_{0}^\infty \int_0^\infty \frac{y^2 \ln(y) \ln (ty)}{(y^2+1)(t^2+1) y^2} \ dy \ dt \\ &=\int_{0}^\infty \int_0^\infty \frac{ \ln(y) \ln (t)}{(y^2+1)(t^2+1)} \ dy \ dt \\ &\qquad + \int_{0}^\infty \int_0^\infty \frac{ \ln^2(y)}{(y^2+1)(t^2+1)} \ dt \ dy \\ &= \left(\int_0^\infty \frac{ \ln (t)}{t^2+1} \ dt \right)^2 + \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy. \end{align*} Note upon a change of variables $t=1/u,$ we have \begin{align*} \int_0^1\frac{ \ln (t)}{t^2+1} \ dt &= -\int_1^\infty \frac{ \ln (t)}{t^2+1} \ dt, \end{align*} which implies $\int_0^\infty\frac{ \ln (t)}{t^2+1} \ dt =0.$ Hence, \begin{align*} \int_{0}^\infty \int_0^\infty \frac{y \ln(y) \ln (x)}{(y^2+1)(x^2+y^2)} \ dy \ dx &= \frac{\pi}{2} \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy. \end{align*}

In order to evaluate the right hand side, consider the triple integral $$I=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dy \ dx \ dz.$$ By carrying out the integration directly, we can find $I=\frac{\pi^3}{8}.$ On the other hand, using Fubini's Theorem to integrate with respect to $z$ first, we get by partial fractions (or Mathematica) \begin{align*} I &=\int_0^\infty \int_0^\infty \int_0^\infty \frac{xz}{(x^2+1)(z^2+1)(x^2y^2z^2+1)} \ dz \ dx \ dy \\ &= \int_0^\infty \int_0^\infty \frac{x \ln \left(xy\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\ &= \int_0^\infty \int_0^\infty \frac{x \ln \left(x\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dy \ dx + \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\ &= \int_0^\infty \int_0^\infty \frac{x \ln \left(y\right)}{\left(x^2+1\right) \left( x^2 y^2-1\right)} \ dx \ dy \\ &= \int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy, \end{align*} in which the first double integral in the third to last equality has Cauchy Principal Value $0$ and thus vanishes, and we carried out the double integral with respect to $x$ in the second to last equality using partial fractions once again.

Hence, $$I=\int_0^\infty \frac{\ln^2(y)}{y^2+1} \ dy=\frac{\pi^3}{8},$$ and so the desired integral is equal to $$\frac{\pi}{2} I=\frac{\pi^4}{16}.$$

Remark: These type of integral calculations are prominent in my joint AMS paper with Daniele Ritelli. See here: https://www.ams.org/journals/qam/2018-76-03/S0033-569X-2018-01499-3/