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

You have

\begin{align*} I &=\int\limits_{-a}^{a}\frac{f(x)}{1+e^{x}} \ dx \qquad\qquad \cdots (1)\\\ I &= \int\limits_{-a}^{a} \frac{f(x)}{1+e^{-x}} \ dx \qquad\qquad \Bigl[ \small\because \int\limits_{a}^{b}f(x)\ dx = \int\limits_{a}^{b}f(a+b-x)\ dx \ \Bigr] \quad \cdots (2) \\\ \Longrightarrow 2I &= \int\limits_{-a}^{a} \biggl[ \frac{f(x)}{1+e^{x}} + \frac{e^{x}\cdot f(x)}{1+e^{x}} \biggr] \ dx \quad\qquad \cdots (1) + (2)\\\ &=\int\limits_{-a}^{a} f(x) \ dx = 2 \int\limits_{0}^{a} f(x) \ dx \qquad \Bigl[ \small \text{since}\ f \ \text{is even so} \ \int\limits_{-a}^{a} f(x) = 2\int\limits_{0}^{a} f(x) \Bigr] \end{align*}

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$\textbf{Note.}$ A similar problem, which uses result $(2)$ can be found here:

• Integration of a trigonometric function

This works because the even part of $\displaystyle{\frac{1}{1+e^x}}$ is $\frac{1}{2}$.

If $g:[-a,a]\to \mathbb R$ is a function, then $g$ has a unique representation as a sum of an even and an odd function, $g=h+k$, with $h(-x)=h(x)$ and $k(-x)=-k(x)$. If $f:[-a,a]\to\mathbb R$ is even, then $g(x)f(x)=h(x)f(x)+k(x)f(x)$ has even part $h(x)f(x)$ and odd part $k(x)f(x)$. Since the integral of an odd function on $[-a,a]$ is zero and the integral of an even function on $[-a,a]$ is twice the integral on $[0,a]$, this yields

$$\int_{-a}^a g(x)f(x)dx=\int_{-a}^ah(x)f(x)dx=2\int_0^a h(x)f(x)dx.$$

As has been seen in previous questions on this site (like this one) the formula for $h$ is $h(x)=\frac{1}{2}(g(x)+g(-x))$. In this problem, $\displaystyle{g(x)=\frac{1}{1+e^x}}$, and $h(x)=\frac{1}{2}$.

Another way of looking at this, where you are naturally led to the result: Since $f(x)$ is even, what you need to show is that $$\int_{-a}^a {f(x) \over 1 + e^x}\,dx = {1 \over 2}\int_{-a}^a f(x)\,dx$$ The difference between the left hand side and the right hand side is $$\int_{-a}^a f(x)\left({1 \over 1 + e^x} - {1 \over 2}\right)\,dx$$ $$= {1 \over 2}\int_{-a}^a f(x) \frac{1 - e^x}{1 + e^x}\,dx$$ Since this is to be zero for all even $f(x)$, you'd expect the function $\frac{1 - e^x}{1 + e^x}$ to be odd, so that the product $f(x) \frac{1 - e^x}{1 + e^x}$ would be odd and thus the integral becomes zero. And sure enough, one can verify readily that this function is in fact odd, so that the above integral is always zero.