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
aaf(x)1+exdx=a0f(x)dx

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

Answer

You have

I=aaf(x)1+ex dx(1) I=aaf(x)1+ex dx[


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

Attribution
Source : Link , Question Author : James , Answer Author : Larry

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