# Math Help - Today's calculation of integral #14

1. ## Today's calculation of integral #14

Find $\int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$

Clever substitutions are necessary to kill this double integral.

2. Hint: let $x=yz,$ and find another suitable substitution.

3. Originally Posted by Krizalid
Find $\int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$

Clever substitutions are necessary to kill this double integral.
Originally Posted by Krizalid
Hint: let $x=yz,$ and find another suitable substitution.
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:

$\int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$

What I did first was rewrite the integrand with that hint you gave: let $x=yz$.

$\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}$
$=\frac{(yz-y)^{2}}{y^2z\sinh(x+y)}\cdot\ln\frac{yz+y}{yz-y}$
$=\frac{(y(z-1))^{2}}{y^2z\sinh(y(z+1))}\cdot\ln\frac{y(z+1)}{y (z-1)}$
$=\frac{y^{2}(z-1)^{2}}{y^2z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}$
$=\frac{(z-1)^{2}}{z\cdot e^{y}\sinh(z+1)}\cdot\ln\frac{z+1}{z-1}$.

Here, I made another substitution: let $y=ln\frac{z+1}{z-1}$.

Now I got:

$\frac{{(z-1)^{2}y}}{z\cdot e^{y}\sinh(z+1)}$.

I have question on converting the limits of integration, and getting the correct differentials (dz and dy)

From the first substitution $x=yz$, I get dx in terms of dy and dz:

$\,dx=y\,dz+z\,dy$

From the second substitution $y=ln\frac{z+1}{z-1}$, I get dy:

$\,dy=\frac{z-1}{z+1}\cdot \frac{(z-1)(1)-(z+1)(1)}{(z-1)^{2}}dz$
$\,dy=\frac{z-1}{z+1}\cdot \frac{-2}{(z-1)^{2}}dz$
$\,dy=-\frac{2}{(z+1)(z-1)}dz$.

Substituting this into the equation for dx, I get:

$\,dx=y\,dz+z{\left(-\frac{2}{(z+1)(z-1)}dz\right)}$
$\,dx=y\,dz-\frac{2z}{(z+1)(z-1)}dz$.

Substituting, into the integral, I get:

$\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz-\frac{2z}{(z+1)(z-1)}\,dz\right)}\,dy$
$=\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y}}{z\cdot e^{y}sinh(z+1)}\right)}{\left(y\,dz\,dy-\frac{2z}{(z+1)(z-1)}\,dz\,dy\right)}$
$=\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{{(z-1)^{2}y^{2}}}{z\cdot e^{y}sinh(z+1)}\right)}\,dz\,dy-\int_{0}^{\infty}\int_{?}^{?}{\left(\frac{2y}{e^{2 y}sinh(z+1)}\right)}\,dz\,dy$.

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!

4. Originally Posted by Krizalid
Hint: let $x=yz,$ and find another suitable substitution.
Now who would think of something as crazy as that?!

5. Before to make the another substitution, we require another previous one.

6. Originally Posted by Krizalid
Before to make the another substitution, we require another previous one.
Another substitution? I have to admit, this is a very good problem...a bit challenging, though.

I'm thinking I need to make a substitution for the $sinh(z+1)$ term???

7. Ahhh, for the inner integral we require the substitution $x=yz.$ ( $z$ 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. )

8. Originally Posted by Krizalid
Ahhh, for the inner integral we require the substitution $x=yz.$ ( $z$ 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. )
So are you saying that after we differentiate x, we should get: $\,dx=y\,dz$?

9. I'm pretty tired to post a full solution so I'll post the main steps to tackle this.

Originally Posted by Krizalid

Find $\int_{0}^{\infty }\!{\int_{y}^{\infty }{\bigg\{\frac{(x-y)^{2}}{xy\sinh(x+y)}\cdot\ln\frac{x+y}{x-y}\bigg\}\,dx}\,dy}.$
Substitute $x=yz$ for the inner integral and reverse integration order, then the original one becomes $\int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\left\{ \frac{(z-1)^{2}}{z}\cdot \ln \frac{(z+1)}{(z-1)} \right\}\left\{ \frac{y}{\sinh \left( y(z+1) \right)} \right\}\,dy}\,dz}.$ Now let $\varphi =y(z+1)$ and the double integral becomes $\int_{1}^{\infty }\!\!{\int_{0}^{\infty }{\Bigg\{ \frac{(z-1)^{2}}{z(z+1)^{2}}\cdot \ln \frac{(z+1)}{(z-1)} \Bigg\}\Bigg\{ \frac{\varphi }{\sinh \varphi } \Bigg\}\,d\varphi }\,dz}.$

Consider $f(\alpha )=\int_{0}^{\infty }{\frac{xe^{-\alpha x}}{1-e^{-2x}}\,dx},\,\forall\,\alpha>0,$ then $f(\alpha )=\sum\limits_{k\,=\,0}^{\infty }{\Bigg\{ \int_{0}^{\infty }{xe^{-(2k+\alpha )x}\,dx} \Bigg\}}=\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+\alpha )^{2}}}.$ From here we have $\int_{0}^{\infty }{\frac{\varphi }{\sinh\varphi }\,d\varphi }=2\int_{0}^{\infty }{\frac{e^{-\varphi }}{1-e^{-2\varphi }}\,d\varphi }=\frac{\pi ^{2}}{4}.$

Finally, for the remaining integral substitute $\mu =\ln \dfrac{z+1}{z-1},$ and we happily get

$\frac{\pi ^{2}}{2}\int_{0}^{\infty }{\frac{\mu e^{-3\mu }}{1-e^{-2\mu }}\,d\mu}
=\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,0}^{\infty }{\frac{1}{(2k+3)^{2}}}
=\frac{\pi ^{2}}{2}\sum\limits_{k\,=\,1}^{\infty }{\frac{1}{(2k+1)^{2}}},$
hence the original double integral equals $3\zeta (2)\Bigg\{ \frac{3}{4}\zeta (2)-1 \Bigg\}.$