Find $\displaystyle \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$
Clever substitutions are necessary to kill this double integral.
I'm trying to work it out, but I got stuck. I'll put what I have. Maybe you can help me out a little bit:
$\displaystyle \int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$
What I did first was rewrite the integrand with that hint you gave: let $\displaystyle x=yz$.
$\displaystyle \frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}$
$\displaystyle =\frac{(yz-y)^{2}}{y^2z\sinh(x+y)}\cdot\ln\frac{yz+y}{yz-y}$
$\displaystyle =\frac{(y(z-1))^{2}}{y^2z\sinh(y(z+1))}\cdot\ln\frac{y(z+1)}{y (z-1)}$
$\displaystyle =\frac{y^{2}(z-1)^{2}}{y^2z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}$
$\displaystyle =\frac{(z-1)^{2}}{z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}$.
Here, I made another substitution: let $\displaystyle y=ln\frac{z+1}{z-1}$.
Now I got:
$\displaystyle \frac{{(z-1)^{2}y}}{z\cdot e^{y}\sinh(z+1)}$.
I have question on converting the limits of integration, and getting the correct differentials (dz and dy)
From the first substitution $\displaystyle x=yz$, I get dx in terms of dy and dz:
$\displaystyle \,dx=y\,dz+z\,dy$
From the second substitution $\displaystyle y=ln\frac{z+1}{z-1}$, I get dy:
$\displaystyle \,dy=\frac{z-1}{z+1}\cdot \frac{(z-1)(1)-(z+1)(1)}{(z-1)^{2}}dz$
$\displaystyle \,dy=\frac{z-1}{z+1}\cdot \frac{-2}{(z-1)^{2}}dz$
$\displaystyle \,dy=-\frac{2}{(z+1)(z-1)}dz$.
Substituting this into the equation for dx, I get:
$\displaystyle \,dx=y\,dz+z{\left(-\frac{2}{(z+1)(z-1)}dz\right)}$
$\displaystyle \,dx=y\,dz-\frac{2z}{(z+1)(z-1)}dz$.
Substituting, into the integral, I get:
$\displaystyle \int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz-\frac{2z}{(z+1)(z-1)}\,dz\right)}\,dy$
$\displaystyle =\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz\,dy-\frac{2z}{(z+1)(z-1)}\,dz\,dy\right)}$
$\displaystyle =\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y^{2}}}{z\cdot e^{y}sinh(z+1)}\right)}\,dz\,dy-\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{2y}{e^{2 y}sinh(z+1)}\right)}\,dz\,dy$.
I'm still stuck. I made a couple changes and cleaned it up a little bit. How do I determine the limits for dz?
...at least I gave it a shot!
Ahhh, for the inner integral we require the substitution $\displaystyle x=yz.$ ($\displaystyle z$ is the new variable, so we use simple differentiation here and not product rule. After this substitution you may have to invoke another one, and then use... figure out. )
I'm pretty tired to post a full solution so I'll post the main steps to tackle this.
Substitute $\displaystyle x=yz$ for the inner integral and reverse integration order, then the original one becomes $\displaystyle \int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\left\{ \frac{(z-1)^{2}}{z}\cdot \ln \frac{(z+1)}{(z-1)} \right\}\left\{ \frac{y}{\sinh \left( y(z+1) \right)} \right\}\,dy}\,dz}.$ Now let $\displaystyle \varphi =y(z+1)$ and the double integral becomes $\displaystyle \int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\Bigg\{ \frac{(z-1)^{2}}{z(z+1)^{2}}\cdot \ln \frac{(z+1)}{(z-1)} \Bigg\}\Bigg\{ \frac{\varphi }{\sinh \varphi } \Bigg\}\,d\varphi }\,dz}.$
Consider $\displaystyle f(\alpha )=\int_{0}^{\infty }{\frac{xe^{-\alpha x}}{1-e^{-2x}}\,dx},\,\forall\,\alpha>0,$ then $\displaystyle f(\alpha )=\sum\limits_{k\,=\,0}^{\infty }{\Bigg\{ \int_{0}^{\infty }{xe^{-(2k+\alpha )x}\,dx} \Bigg\}}=\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+\alpha )^{2}}}.$ From here we have $\displaystyle \int_{0}^{\infty }{\frac{\varphi }{\sinh\varphi }\,d\varphi }=2\int_{0}^{\infty }{\frac{e^{-\varphi }}{1-e^{-2\varphi }}\,d\varphi }=\frac{\pi ^{2}}{4}.$
Finally, for the remaining integral substitute $\displaystyle \mu =\ln \dfrac{z+1}{z-1},$ and we happily get
$\displaystyle \frac{\pi ^{2}}{2}\int_{0}^{\infty }{\frac{\mu e^{-3\mu }}{1-e^{-2\mu }}\,d\mu}
=\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+3)^{2}}}
=\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,1}^{\infty }{\frac{1}{(2k+1)^{2}}},$ hence the original double integral equals $\displaystyle 3\zeta (2)\Bigg\{ \frac{3}{4}\zeta (2)-1 \Bigg\}.$