I found this integral in an old book of mine :
$$\int_{-a}^{a} \frac{1}{1+x^{2x}} dx$$
where $|a| \lt 1$
A hint was given that we could split this integral into even and odd functions but I don't know how to. I tried using WolframAlpha for the same but it doesn't help me out. If the exponent of $x$ would have been $2n$ then it wouldn't have been a problem but that $2x$ in the exponent really drives me out. I don't actually want an answer to this problem, I just want the even and odd functions, the integral should be split into. Any help regarding the same is appreciated 🙂
Thanks in advance
Best Answer
Interpreting $x^{2x} = (x^2)^x$ we have \begin{align*}\frac{1}{1+x^{2x}} &= \frac{1}{2}\left( \frac{1}{1+x^{2x}}+\frac{1}{1+x^{-2x}}\right) + \frac{1}{2}\left( \frac{1}{1+x^{2x}}-\frac{1}{1+x^{-2x}}\right) \\ &= \frac{1}{2}+ \cdots\end{align*} where $\cdots$ denotes the odd part. So $$\int_{-a}^a \frac{1}{1+x^{2x}}\mathrm{d}x = 2\int_0^a \frac{1}{2}\mathrm{d}x = a.$$ Note that the assumption $|a|<1$ is so that $x^{2x}$ is separated from $-1$, so the integral converges.