# Proof of an integral involving even and odd functions

This note provides a proof of the following identity

$\int_{-a}^a \frac {f(x)}{1+c^{g(x)}} \ dx = \int_0^a f(x)\ dx$

where $f(x)$ is an even function, $g(x)$, an odd function, and $a$ and $c$ are constants.

Proof:

\begin{aligned} \int_{-a}^a \frac {f(x)}{1+c^{g(x)}} \ dx & = {\color{#3D99F6} \int_{-a}^0 \frac {f(x)}{1+c^{g(x)}} \ dx} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Let }u=-x \implies dx = -du \\ & = {\color{#3D99F6} - \int_a^0 \frac {f(-u)}{1+c^{g(-u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Since for even function }f(-u) = f(u) \\ & = {\color{#3D99F6} \int_0^a \frac {f(u)}{1+c^{-g(u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{ and odd function }g(-u) = - g(u) \\ & = {\color{#3D99F6} \int_0^a \frac {c^{g(u)} f(u)}{1+c^{g(u)}} \ du} + \int_0^a \frac {f(x)}{1+c^{g(x)}} \ dx & \small \color{#3D99F6} \text{Replace }u \text{ with }x \\ & = \int_0^a \frac {\left(1+c^{g(x)}\right) f(x)}{1+c^{g(x)}} \ dx \\ & = \int_0^a f(x) \ dx \quad \blacksquare \end{aligned}

Note by Chew-Seong Cheong
1 year, 6 months ago

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so g can be any function that is odd

- 1 year, 3 months ago

Yes, I use a general odd function $g(x)$ and no other condition.

- 1 year, 2 months ago

did you do this by your self

- 1 year, 3 months ago

I didn't discover it. I have seen two examples of its use and I put it into this note so that I can use it in my solutions in the future.

- 1 year, 2 months ago