Compute $\displaystyle \int_{-1}^1\frac{x^2}{1+e^x}\,dx$
I can't get it!
Look at the u = -x case:
$\displaystyle \int_1^{-1}\frac{(-u)^2}{1+e^{-u}}\,(-du)$
$\displaystyle = \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx$
Now add the two integrals. (Hint: Do an integration by parts on the second integral and see what you get.)
-Dan
ok, i'm not getting anywhere with this:
$\displaystyle \int_{-1}^1 \frac{u^2}{1 + e^{-u}} \, du = \int_{-1}^1 \frac{x^2}{1 + e^{-x}} \, dx = -x^2 \ln \left( e^{-x} + 1 \right) + 2 \int x \ln \left( e^{-x} + 1 \right) ~dx$
i also did by parts on the last integral also, but it never seemed to be leading me anywhere. and even if i could get a nice result, i'm not exactly sure how you expect me to combine the two integrals when they have different variables.
This is the key, 'cause we have the same integration limits, so we can turn back the $\displaystyle u$ into $\displaystyle x$
Now
$\displaystyle \int_{-1}^1\frac{x^2}{1+e^x}\,dx+\int_{-1}^1\frac{x^2}{1+e^{-x}}\,dx=\frac13x^3\bigg|_{-1}^{\phantom{-}1}=\frac23$
Divide by two this last result and yields the desired.
Hmmm...
I plugged the integral in on my calculator and got that
$\displaystyle \int \frac{x^2}{1 + e^{-x}}dx = \frac{1}{3}x^3 - \int \frac{x^2}{1 + e^x} dx$
At a glance it looked like an integration by parts, but I guess it isn't. (Or, rather, it still may be but I can't figure how to do it.) I'll have to work on the integral a bit, but in any event you should be able to see where to go with it from here.
Edit: Ah wait! I see where Krizalid is going with this. The integrand isn't an even function, but the two are reflections of each other over the y-axis. Since the integration limits are symmetric the two integrals must be equal to each other.
-Dan