# Integral of x/[1+exp(x)] dx

• Apr 24th 2008, 07:19 AM
ragungs
Integral of x/[1+exp(x)] dx
$\displaystyle \int_0^{10}(\frac{300x}{1+e^x})dx$
• Apr 24th 2008, 07:41 AM
topsquark
Quote:

Originally Posted by ragungs
$\displaystyle \int_0^{10}(\frac{300x}{1+e^x})dx$

There is no closed form expression for this integral. Are you looking for an approximation?

-Dan
• Apr 24th 2008, 07:45 AM
ragungs
I need the true value of this integral to calculate the error of my approximation.

I wonder if this integral can be solved by hand.
• Apr 24th 2008, 07:52 AM
topsquark
Quote:

Originally Posted by ragungs
I need the true value of this integral to calculate the error of my approximation.

I wonder if this integral can be solved by hand.

As I said there is no closed form expression. You cannot get an exact value of it.
$\displaystyle \int \frac{x}{1 + e^x}~dx = \frac{x^2}{2} - ln(1 + e^x) - \sum_{n = 1}^{\infty} \frac{e^{nx}}{n^2}$

There is no expression for that summation in terms of elementary functions. This thing can only be approximated.

-Dan

Edit: A made a mistake in copying "polylog" function from the Integrator site. I have fixed it.
• Apr 24th 2008, 07:52 AM
Aryth
$\displaystyle \int_0^{10} \left(\frac{300x}{1 + e^x}\right)dx \approx 246.59029$

I used my graphing calculator to calculate the integral... It's not exact either, but it's closer than a lot of people can get.
• Apr 24th 2008, 07:58 AM
ragungs
ok thanks all.
• Apr 24th 2008, 01:49 PM
CaptainBlack
Quote:

Originally Posted by topsquark
There is no closed form expression for this integral. Are you looking for an approximation?

-Dan

It does in terms of the polylogarithm $\displaystyle {\rm Li}_2$

RonL
• Apr 24th 2008, 01:53 PM
Mathstud28
Quote:

Originally Posted by CaptainBlack
It does in terms of the polylogarithm $\displaystyle {\rm Li}_2$

RonL

$\displaystyle -Li_2(1+e^{x})-x\ln(1+e^{x})+\frac{1}{2}x^2$?