Show that $\frac{te^t}{e^{2t}-1}$ is integrable?

Question

How to show that the function $f\mapsto\dfrac{te^t}{e^{2t}-1}$ is integrable on $(0,+\infty)$ ?

I think it suffices to say that $f\mapsto\dfrac{te^t}{e^{2t}-1}$ is well-defined and continuous on $(0,+\infty)$, then it is continuous, right? My colleague told me that is not correct, you have to approximate the function at $0$ and at $+\infty$.

EDIT I think I can show something at $+\infty$.

We have $\dfrac{te^t}{e^{2t}-1}\sim\dfrac{te^t}{e^{2t}}=te^{-t}=o(\dfrac{1}{t^2})$

How about at $0$ ?

Answer 1

You can simply do the integral to show it is integrable. Collect a $e^{2t}$ factor in the dominator:

$$\int_0^{+\infty}\frac{te^t}{e^{2t}(1 - e^{-2t})}\ \text{d}t$$

Arrange the integral, and use the Geometric series:

$$\frac{1}{1 - e^{-2t}} = \sum_{k = 0}^{+\infty} e^{-2tk}$$

Getting

$$\sum_{k = 0}^{+\infty}\int_0^{+\infty} t\ e^{-t(1+2k)}\ \text{d}t$$

The integral is trivial (you can do it by parts once) and you get in the end:

$$\sum_{k = 0}^{+\infty} \frac{1}{(1 + 2k)^2}$$

The Series does converge to

$$\boxed{\frac{\pi^2}{8}}$$

Showing the function has no problem in $0$

Use Taylor Series:

$$\frac{t(1+t)}{1 + 2t - 1} = \frac{t + t^2}{2t} = \frac{1}{2} + \frac{t}{2} = \frac{1}{2}$$

Well defined in zero.

How to show it's sum is $\frac{\pi}{8}$

If we write the first terms of the sum, we have:

$$1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \frac{1}{81} + \cdots + \frac{1}{n_{\text{disp}}^2}$$

This sum is indeed the sum of all the odd squares. This is a particular sum, and we can see it as the sum of ALL the reciprocal squares, minus the sum of the EVEN reciprocal squares, indeed we can write:

$$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \frac{1}{49} + \frac{1}{64} + \frac{1}{81} + \cdots + \frac{1}{n^2}$$

if we subtract from this the sum of all even reciprocal squares, we obtain exactly our sum. Translated into mathish it's like to have (calling OUR sum $S$)

$$S = \frac{1}{n^2} - \frac{1}{(2n)^2}$$

namely again: our sum is the whole sum of reciprocal squares, minus the sum of all the EVEN reciprocal squares. We can do that simple subtraction:

$$S = \frac{3n^2}{4n^4} = \frac{3}{4n^2}$$

This means that our sum is three quarters the value of the sum of all the reciprocal squares which is a well known series (also it's the Riemann Zeta vaulted in $2$):

$$\sum_{k = 1}^{\infty} \frac{1}{n^2} = \sum_{k = 1}^{+\infty} \frac{1}{(k+1)^2} = \frac{\pi^2}{6}$$

Since our sum is three quarters of that value we get:

$$S = \frac{3}{4}\cdot \frac{\pi^2}{6} = \frac{\pi^2}{8}$$

Answer 2

the function $$\dfrac{te^t}{e^{2t}-1} = \frac{t(1+\cdots)}{1+2t+\cdots - 1} = \frac{1}{2}+\cdots \text{ for } t = 0+\cdots$$ therefore the integral $\int_0^1\dfrac{te^t}{e^{2t}-1}\, dt$ converges. we also have $$\dfrac{te^t}{e^{2t}-1} = \frac{t}{e^t}+\cdots = te^{-t}+\cdots = \frac{1}{2}+\cdots \text{ for } t = \infty$$ and $\int_1^{\infty} te^{-t} \, dt$ convergent implies that $$ \int_1^\infty\dfrac{te^t}{e^{2t}-1}\, dt \text{ is also convergernt.}$$