Thread: integral with log and sinh

1. integral with log and sinh

Give this one a go if you're so inclined. A clever sub will do it.

$\displaystyle \int_{0}^{\infty}\int_{y}^{\infty}\frac{(x-y)^{2}log(\frac{x+y}{x-y})}{xysinh(x+y)}dxdy$

2. Originally Posted by galactus
Give this one a go if you're so inclined. A clever sub will do it.

$\displaystyle \int_{0}^{\infty}\int_{y}^{\infty}\frac{(x-y)^{2}log(\frac{x+y}{x-y})}{xysinh(x+y)}dxdy$
I thought I had seen this before...and I searched, and was correct!

I remember working on this to no avail...plus, it turned out to be beyond the scope of what I knew anyway... >_>

--Chris

3. Here.

4. Here's another way.

An alternative sub is to let $\displaystyle x=\frac{u+t}{2}, \;\ y=\frac{u-t}{2}$

Then, we get:

$\displaystyle I=2\int_{0}^{\infty}\frac{1}{sinh(u)}\int_{0}^{u}\ frac{t^{2}log(\frac{u}{t})}{u^{2}-t^{2}}dtdu$

Then, in the inner integral, let t=uw to get the product of two known integrals.

$\displaystyle I=2\int_{0}^{\infty}\frac{udu}{sinh(u)}\int_{0}^{1 }\left(1-\frac{1}{1-w^{2}}\right)log(w)dw$

And you should get:

$\displaystyle \frac{{\pi}^{2}({\pi}^{2}-8)}{16}$