Definite integral involving a lot of exponentials.

[caffeinemachine]

Problem: Evaluate
$$
\int_{0}^\infty \frac{e^{3x} - e^x}{x(e^x + 1)(e^{3x} + 1)}\ dx
$$

Attempt.

I substituted $y=e^x$, thus $dx = dy/y$, which turns the above integral to

$$
\int_{1}^\infty \frac{y^2 - 1}{(\log y)(y+1)(y^3+1)}\ dy = \int_{1}^\infty \frac{y-1}{(\log y) (y^3+1)} \ dy
$$

I am unable to make progress.

Thanks.

 

[MountEvariste]

1. Sub $y \mapsto \frac{1}{y}$ to change the bounds.

2. Note that $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t $.

3. Switch the order of the double integral

4. Expand the geometric series

5. Switch the order of integral and series

6. You're left with an infinite series.

This series can be calculated by turning it into a product by putting it inside the log and using

$\displaystyle \cos\pi z=\prod_{n \ge 0 }\left(1-\frac{4z^2}{(2n+1)^2}\right)$

I'm too tired to write out detailed answer now, but that should be the gist of it.

 

[MountEvariste]

Letting $y \mapsto \frac{1}{y}$ we get $\displaystyle I = \int_0^1 \frac{y-1}{(y^3+1)\log{y}}\,\mathrm{d}y$; since $\displaystyle \frac{y-1}{\log{y}}= \int_0^1 y^t \, \mathrm{d}t$, we have

$$\begin{aligned} I & = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}t\,\mathrm{d}y = \int_0^1 \int_0^1 \frac{y^t}{y^3+1}\,\mathrm{d}y\,\mathrm{d}t\end{aligned}$$

Since $\displaystyle \frac{1}{1+y^3} = \sum_{k \ge 0} y^{3k}$ uniformly over $y \in (0, 1)$, we have

$$\begin{aligned} I & = \int_0^1 \int_0^1 \sum_{k \ge 0}(-1)^k y^{3k+t}\,\mathrm{d}y\,\mathrm{d}t \\& = \sum_{k \ge 0}(-1)^k\int_0^1 \int_0^1 y^{3k+t}\,\mathrm{d}y\,\mathrm{d}t \\& = \sum_{k \ge 0} (-1)^k \log\left(\frac{3k+2}{3k+1}\right) \\& =\log\left(\prod_{k \ge 0} \bigg[\frac{3k+2}{3k+1}\bigg]^{(-1)^k}\right) \\&\end{aligned}$$

To calculate the product we have we have:

$$\begin{aligned}\displaystyle \prod_{k \ge 0} \bigg[\frac{3k+2}{3k+1}\bigg]^{(-1)^k} & = \prod_{k \ge 0} \bigg[\frac{6k+2}{6k+1}\bigg]\bigg[\frac{6k+4}{6k+5}\bigg] \\& = \prod_{k \ge 0} \frac{(6k+2)(6k+4)}{(6k+1)(6k+5)} \\& = \frac{\displaystyle \prod \left[1-4/6^2/(2k+1)^2\right]}{\displaystyle \prod \left[1-4/3^2/(2k+1)^2\right]} \\& = \frac{\cos{\frac{1}{6}\pi}}{\cos{\frac{1}{3} \pi}} \\& = \sqrt{3}.\end{aligned} $$


Hence $I = \log{\sqrt{3}}.$