0
$\begingroup$

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 $0$ ?

$\endgroup$
  • $\begingroup$ $\frac{1}{t}$ is well-defined and continuous on $(0,+\infty)$ but not integrable there, because integral over this interval doesn't converge. You have to argue why it does converge in case of your function. $\endgroup$ – Wojowu Mar 5 '16 at 17:11
  • $\begingroup$ To see it's defined in zero, look at my answer below. If you're also interested in the integral itself.. take another look! :D $\endgroup$ – Von Neumann Mar 5 '16 at 17:41
2
$\begingroup$

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:

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How do you get $\frac{\pi^2}{8}$? $\endgroup$ – drzbir Mar 5 '16 at 18:42
  • $\begingroup$ @Blid Writing the whole process would be really boring, so try to read my answer to this question math.stackexchange.com/questions/1664441/… and maybe you will find some hint to sum that series :) I will try to write here the solution, though. I will need just a bit of time because I'm out now! $\endgroup$ – Von Neumann Mar 5 '16 at 18:49
  • $\begingroup$ Hmm. I see thanks. $\endgroup$ – drzbir Mar 5 '16 at 18:56
  • $\begingroup$ @Blid Done, I've written how to calculate it, but you have to know what is the well known sum of reciprocal squares.. $\endgroup$ – Von Neumann Mar 5 '16 at 19:12
1
$\begingroup$

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 $$ \int_1^\infty\dfrac{te^t}{e^{2t}-1}\, dt \text{ is also convergernt.}$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.