$$\int_{-\pi}^{\pi}\frac{\cos^2(x)}{1+a^x}dx, a>0$$ 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?


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

  • $\begingroup$ 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

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. $$2\mathrm I=\int_{-\pi}^{\pi}\dfrac{1+a^x}{1+a^x}\cdot\cos^2x\mathrm dx=2\int_{0}^{\pi}\dfrac{1+\cos 2x}{2}\mathrm dx \implies \mathrm I=\pi/2$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.