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$$\int_{-\pi}^{\pi}\frac{\cos^2(x)}{1+a^x}dx, a>0$$ I've tried doing $z = e^{ix}$ and evaluating the resulting contour integral, but this introduced a branch cut that goes through the contour ($a^x$ becomes $a^{-i\log(z)})$.

The answer is $\frac{\pi}{2}$ (independent of $a$), but I don't know how one would get to that answer.

Could anyone help me with this integral?

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Hint: what is $\frac{1}{1+a^x}+\frac{1}{1+a^{-x}}$?

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  • $\begingroup$ This is amazing, thank you. Can I ask what your motivation was to consider this sum? $\endgroup$ – user403069 May 15 '19 at 22:33
  • $\begingroup$ I’m not sure actually. I “felt” like $(1+a^x)^{-1}$ term destroyed all attempts at finding a closed form antiderivative, integrate by parts or change variables. So I wondered if there might be a way to recombine the integral differently. Given that $\cos$ is even, this sum was the first thing to try. $\endgroup$ – Mindlack May 15 '19 at 22:38
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Using the reflection property: $\int_{a}^{b}f(x)\mathrm dx=\int_{a}^{b}f(a+b-x)\mathrm dx$ $$x\mapsto -x \implies \mathrm I =\int_{-\pi}^{\pi}\dfrac{a^x\cos^2 x}{1+a^x}\mathrm dx$$

Add the initial integral to the transformed integral to get $2\mathrm I$ and consequently the value of the integral at hand. $$2\mathrm I=\int_{-\pi}^{\pi}\dfrac{1+a^x}{1+a^x}\cdot\cos^2x\mathrm dx=2\int_{0}^{\pi}\dfrac{1+\cos 2x}{2}\mathrm dx \implies \mathrm I=\pi/2$$

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