4

The problem is to show that the improper integral $I = \int_0^{\infty} \frac{x^2}{e^x-1} dx$ converges to $2\sum_1^{\infty} \frac1{n^3}$.

Previously, I computed the following integral:

$$f(x) = -\int_0^x\frac{\log(1-t)}{t}dt = \sum_1^{\infty} \frac{x^n}{n^2}$$

which holds for $-1 \le x \le 1$. From this, I deduce the following:

$$g(x) = \int_0^x \frac{f(t)}{t}dt = \sum_1^{\infty} \frac{x^n}{n^3}$$

which also holds for $-1 \le x \le 1$. I am attempting to show that $I = g(1)$.

The first thing I do with $I$ is make the substitution $t = 1 - e^{-x}$, which yields:

$$I = \int_0^1 \frac{\log^2(1-t)}{t} dt$$

Then I attempt to integrate by parts, differentiating $\log(1-t)$ and integrating $\frac{\log(1-t)}{t}$, which yields:

$$I = -\log(1-t)f(t)|_0^1-\int_0^1\frac{f(t)}{1-t}dt $$

That doesn't work. The first term doesn't converge. The upper bound evaluates to $-\log(0)\frac{pi^2}6$ and the second term is not g(1). The denominator needs to be $t$, not $t-1$. So I am not sure what to do next.

larryh
  • 207
  • I was hoping someone would point out the error in my logic. – larryh Jan 30 '16 at 02:39
  • The answer provided by Lee Fisher below is much clearer than the ones in the referenced post. Without his answer, I did not understand the answers in the reference post. – larryh Jan 30 '16 at 03:31

3 Answers3

2

So it's well known that $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ provided $|x|\leq 1$ but a perhaps less often used variant is that if $|x|>1$ then $$\frac{1}{x-1}= \frac{1}{x}\frac{1}{1-\frac{1}{x}} = \frac{1}{x}\sum_{n=0}^\infty x^{-n} = \sum_{n=1}^\infty x^{-n}$$ This means that, for our integral, since $x\in (0,\infty)$ that means $e^x \in (1,\infty)$. So we can rewrite our integral as: $$ \int_0^\infty \frac{x^2}{e^x-1}dx = \int_0^\infty\left(\sum_{n=1}^{\infty}x^2e^{-nx}\right)dx$$ and since our sum converges within the boundary of the integral we can interchange the sum and the integral. So we consider: $$\int x^2e^{-nx}dx = \frac{-e^{-nx}(n^2x^2+2nx+2)}{n^3}$$ which is obtained by using integration by parts twice. Then plugging in boundaries gives our desired result:

$$ \int_0^\infty \frac{x^2}{e^x-1}dx = 2\sum_{n=1}^\infty \frac{1}{n^3}$$

Lee Fisher
  • 1,934
  • Dropped a minus sign edit. – Lee Fisher Jan 30 '16 at 02:07
  • I realize after quickly writing this answer, that it might not have been very helpful to you. What you were doing seemed very complicated to me, but I'll comment if I see a way to proceed with your initial approach. – Lee Fisher Jan 30 '16 at 02:53
  • That would be greatly appreciated. I would like to know how to proceed starting from the integral after I made the substitution. – larryh Jan 30 '16 at 03:02
  • I think you solved the problem the way that it was intended in the textbook. The problem just before this one was a generalized version of the integral you performed by parts. There is another problem in the textbook that has a form of the integral (after the substitution) that I was attempting to solve, so that is why I made the substitution. – larryh Jan 30 '16 at 03:26
2

Let's proceed along a similar line as in the OP. We enforce the substitution $x\to -\log(x)$ so that

$$\int_0^\infty \frac{x^2}{e^x-1}\,dx=\int_0^1\frac{\log^2(x)}{1-x}\,dx \tag 1$$

Integrating by parts the right-hand side of $(1)$ with $u=\log^2(x)$ and $v=\log(1-x)$ yields

$$\begin{align} \int_0^1\frac{\log^2(x)}{1-x}\,dx&=2\int_0^1\frac{\log(x)\log(1-x)}{x}\,dx\\\\ &=2\int_0^1\left(\int_1^x \frac{1}{t}\,dt\right)\frac{\log(1-x)}{x}\,dx\\\\ &=2\int_0^1\frac{1}{t}\left(-\int_0^t\frac{\log(1-x)}{x}\,dx\right)\,dt\\\\ &=2\int_0^1\frac{\text{Li}_2(t)}{t}\,dt\\\\ &=2\text{Li}_3(1)\\\\ &=2\zeta(3) \end{align}$$

where have made use of the relationships between $(i)$ the dilogarithm function $\text{Li}_2(x)$ and the trilogarithm function $\text{Li}_3(x)$ and $(ii)$ the polylogarithm function $\text{Li}_s(1)=\zeta(s)$ (for $\text{Re}(s)>1$).

Mark Viola
  • 179,405
  • Thank you for the reply! I almost had it, it just didn't do the integration by parts correctly. – larryh Jan 30 '16 at 12:20
  • Is there an easier way to show that $ \lim_{u \rightarrow 1} \log^2(1-u)\log(u) = 0$ and similarly $\lim_{u \rightarrow 0} \log^2(1-u)\log(u) = 0$ other than applying l'Hopitals rule four times to get rid of the logarithms? – larryh Jan 30 '16 at 13:38
  • @larryh You're welcome. My pleasure. -Mark – Mark Viola Jan 30 '16 at 15:14
  • @larryh Larry, yes there is an easier way. Note that $\log(x)=(x-1)+O((x-1)^2)$ and $\log(1-x)=-x+O(x^2)$. Then, use $\lim_{x\to 0}x\log(x)=0$. -Mark – Mark Viola Jan 30 '16 at 15:20
0

$$-\int_0^\infty x^2 * (1-e^{x}) dx = -\int_0^\infty x^2 \sum_{n=1}^\infty\begin{equation*} \binom{-1}{n} \end{equation*}(-e^{x})^n dx = \int_0^\infty x^2\sum_{n=1}^\infty e^{nx} dx= \sum_{n=1}^\infty \int_0^\infty x^2 * e^{nx}dx $$ put nx=-y dx=-1/n dy $$\sum_{n=1}^\infty 1/n^3 \int_0^\infty e^{-y}*y^2 dy= \sum_{n=1}^\infty 1/n^3 \Gamma(3)=2\sum_{n=1}^\infty 1/n^3 $$

Andrew Amgad Amin
  • 1