I want to find the following integral:
$$\int_0^{\infty} \frac{1}{(1+x^n)(1+x²)}.$$
I enter small values of $n$ on Wolfram and it seems like the answer should be $\pi /4$. Can anyone confirm this for any $n$? I try to do fraction decomposition but it gets ugly really fast.
(I think this should have been answered somewhere but I really can’t find it. Apologize in advance if this is a duplication)
trick : Make substituition $x=\frac{1}{t}$
ie integral becomes $$I=\int_{0}^{\infty}\frac{t^ndt}{(1+t^2)(1+t^n)}$$.
Adding this with original integral:
$$2I=\int_{0}^{\infty}\frac{1+t^n}{(1+t^2)(1+t^n)}=\int_{0}^{\infty}\frac{dt}{1+t^2}$$
Can you end it now?
$$I=\int_{0}^{\infty} \frac{dx}{(1+x^n)(1+x^2)}~~~~(*)$$ Let $x=\tan t$, then $$I=\int_{0}^{\pi/2} \frac{dt}{1+\tan^nt}~~~~~~~(1)$$ Use the property: $\int_{0}^a f(x) dx=\int_{0}^{a} f(a-x) dx~~~~(2)$ $$\implies I=\int_{0}^{\pi/2} \frac{dt}{1+\cot^n t}~~~~(3)$$ Add (1) and (3), to get $$2I=\int_{0}^{\pi/2} \left(\frac{1}{1+\tan^n t}+\frac{1}{1+\cot^n t}\right)dt=\frac{\pi}{2} \implies I=\frac{\pi}{4}$$
