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

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- Apr 24th 2008, 07:19 AMragungsIntegral of x/[1+exp(x)] dx
$\displaystyle \int_0^{10}(\frac{300x}{1+e^x})dx$

- Apr 24th 2008, 07:41 AMtopsquark
- Apr 24th 2008, 07:45 AMragungs
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 AMtopsquark
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 AMAryth
$\displaystyle \int_0^{10} \left(\frac{300x}{1 + e^x}\right)dx \approx 246.59029$

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 AMragungs
ok thanks all.

- Apr 24th 2008, 01:49 PMCaptainBlack
- Apr 24th 2008, 01:53 PMMathstud28