Proof of an integral involving even and odd functions

This note provides a proof of the following identity

aaf(x)1+cg(x) dx=0af(x) dx\int_{-a}^a \frac {f(x)}{1+c^{g(x)}} \ dx = \int_0^a f(x)\ dx

where f(x)f(x) is an even function, g(x)g(x), an odd function, and aa and cc are constants.


aaf(x)1+cg(x) dx=a0f(x)1+cg(x) dx+0af(x)1+cg(x) dxLet u=x    dx=du=a0f(u)1+cg(u) du+0af(x)1+cg(x) dxSince for even function f(u)=f(u)=0af(u)1+cg(u) du+0af(x)1+cg(x) dx and odd function g(u)=g(u)=0acg(u)f(u)1+cg(u) du+0af(x)1+cg(x) dxReplace u with x=0a(1+cg(x))f(x)1+cg(x) dx=0af(x) dx\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
3 weeks, 4 days ago

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