# Showing that a∫−af(x)1+exdx=a∫0f(x)dx\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx, when ff is even

I have a question:

Suppose $f$ is continuous and even on $[-a,a]$, $a>0$ then prove that

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

I=a∫−af(x)1+ex dx⋯(1) I=a∫−af(x)1+e−x dx[∵\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*}
$$\textbf{Note.}\textbf{Note.}$$ A similar problem, which uses result $$(2)(2)$$ can be found here: