8
$\begingroup$

Does this integral converge to any particular value? $$\int_{0}^{+\infty} \frac{x}{1+e^x}\,dx$$ If the answer is yes, how should I calculate its value?
I tried to use convergence tests but I failed due to the complexity of the integral itself.

$\endgroup$
2
  • 2
    $\begingroup$ Please look here. It's almost the same. $\endgroup$
    – Galc127
    Commented Feb 20, 2016 at 17:51
  • 1
    $\begingroup$ In general, $$\int_0^\infty\frac{x^n}{e^x-1}~dx~=~n!~\zeta(n+1),$$ and $$\int_0^\infty\frac{x^n}{e^x+1}~dx~=~n!~\eta(n+1).$$ See the Riemann $\zeta$ and Dirichlet $\eta$ function for more information. $\endgroup$
    – Lucian
    Commented Feb 21, 2016 at 1:02

3 Answers 3

11
$\begingroup$

Sure it does.

Collect $e^x$ in the denominator and you will get

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

Since the integral range is from $0$ to infinity, you can see the fraction in this way:

$$\int_0^{+\infty}x e^{-x}\frac{1}{1 + e^{-x}}\ \text{d}x$$

and you can make use of the geometric series for that fraction:

$$\frac{1}{1 + e^{-x}} = \frac{1}{1 - (-e^{-x})} = \sum_{k = 0}^{+\infty} (-e^{-x})^k$$

thence thou have

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

Namely

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

This is trivial, you can do it by parts getting

$$\int_0^{+\infty} x e^{-x(1+k)}\ \text{d} = \frac{1}{(1+k)^2}$$

Thence you have

$$\sum_{k = 0}^{+\infty}(-1)^k \frac{1}{(1+k)^2} = \frac{\pi^2}{12}$$

Which is the result of the integration

If you need more explanations about the sum, just tell me!

HOW TO CALCULATE THAT SERIES

There is a very interesting trick to calculate that series. First of all, let's write it with some terms, explicitly:

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

The first terms of the series are:

$$-\left(-1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} - \frac{1}{25} + \frac{1}{36} - \frac{1}{64} + \frac{1}{128} - \cdots\right)$$

namely

$$\left(1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36} + \frac{1}{64} - \frac{1}{128} + \cdots\right)$$

Now let's call that series $S$, and let's split it into even and odd terms:

$$S = \left(1 + \frac{1}{9} + \frac{1}{25} + \frac{1}{49} + \cdots\right) - \left(\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + \frac{1}{64} + \cdots\right) ~~~~~ \to ~~~~~ S = A - B$$

Where obviously $A$ and $B$ are respectively the odd and even part.

now the cute trick

take $B$, and factorize out $\frac{1}{4}$:

$$B = \frac{1}{4}\left(1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + \cdots\right)$$

Now the series in the bracket is a well known series, namely the sum of reciprocal squares, which is a particular case of the Zeta Riemann:

$$\zeta(s) = \sum_{k = 1}^{+\infty} \frac{1}{k^s}$$

which is, for $s = 2$

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

Thence we have:

$$B = \frac{\zeta(2)}{4} = \frac{\pi^2}{24}$$

Are you seeing where we want to go? But this is not enough since we don't know what $A$ is. To do that, we can again split $B$ into even and odd terms! But doing so, we will find again the initial $A$ and $B$ series:

$$B = \frac{1}{4}\left(\left[1 + \frac{1}{9} + \frac{1}{25} + \cdots\right] + \left[\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \cdots\right]\right) ~~~ \to ~~~ B = \frac{1}{4}\left(A + B\right)$$

This means:

$$4B - B = A ~~~~~ \to ~~~~~ A = 3B$$

So

$$A = 3\cdot \frac{\pi^2}{24} = \frac{\pi^2}{8}$$

Not let's get back to the initial series $S$ we wanted to compute, and substitution this we get:

$$S = A - B = \frac{\pi^2}{8} - \frac{\pi^2}{24} = \frac{\pi^2}{12}$$

$\endgroup$
2
  • $\begingroup$ Could you please explain more about the sum and how you got the final answer? $\endgroup$
    – FreeMind
    Commented Feb 20, 2016 at 18:01
  • 1
    $\begingroup$ @FreeMind DONE! ^^ $\endgroup$
    – Enrico M.
    Commented Feb 20, 2016 at 18:42
6
$\begingroup$

Comparison test:

$$\frac x{1+e^x}\le\frac x{e^x}\;$$

and you can directly do integration by parts in the rightmost function to check it converges.

$\endgroup$
2
$\begingroup$

It is possible to generalize the result.

Claim: $$I=\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx=-\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right) $$ where $\textrm{Li}_{n}\left(x\right) $ is the polylogarithm and $a>0$.

Consider $$\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\frac{\left(-1\right)^{n}e^{-nb}}{n^{a+1}}\int_{0}^{\infty}y^{a}e^{-y}dy=\frac{\Gamma\left(a+1\right)\left(-1\right)^{n}e^{-bn}}{n^{a+1}} $$ and recalling that $$\textrm{Li}_{k}\left(x\right)=\sum_{n\geq1}\frac{x^{n}}{n^{k}} $$ we have $$\Gamma\left(a+1\right)\textrm{Li}_{a+1}\left(-e^{-b}\right)=\sum_{n\geq1}\left(-1\right)^{n}\int_{0}^{\infty}x^{a}e^{-n\left(x-b\right)}dx=\int_{0}^{\infty}x^{a}\sum_{n\geq1}\left(-1\right)^{n}e^{-n\left(x-b\right)}dx= $$ $$=-\int_{0}^{\infty}\frac{x^{a}}{e^{x-b}+1}dx.$$ Maybe it's interesting to note that the function ${x^{a}}/{e^{x-b}+1}$ is the Fermi-Dirac distribution.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .