TITLE: Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even
QUESTION [66 upvotes]: I have a question:

Suppose $f$ is continuous and even on $[-a,a]$, $a>0$ then prove that
  $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$

How can I do this? Don't know how to start.

REPLY [4 votes]: I've been looking at this more closely and it's now pretty much demystified:
If $f(x)$ is even and $u_k(x)$ a sequence of uneven functions, the integrals $\int_{-a}^a$ over $f(x)\,u_k(x)$ are trivially zero and so you have
$$\int_{-a}^a \, f(x) \left( \frac{1}{2} + \sum_{k=0}^\infty u_k(x) \right) = \frac{1}{2}\int_{-a}^a f(x) \,{\mathrm d}x = \int_0^a f(x) \,{\mathrm d}x.$$
The functions 
$$\dfrac{1}{1+e^{v(x)}} = \frac{1}{2} - \frac{1}{4}v(x)+\frac{1}{48} v(x)^3+\dots$$ 
are cases of functions with such an uneven expansion, where $u_k(x)$ are uneven powers of some fixed function $v(x)$.
Even more specifically, your case is that sequence with $v(x)\propto x$.