Closed form for $\int \frac{1}{a^x + x^a} dx$

Question

Is there a closed form for the following integral? I can't seem to find it online, nor do I have any idea how to approach this problem..... $$\int\frac{1}{a^x + x^a}dx$$ where $a \ne 0 ; a\in R $

Answer 1

Assuming $a>1$, the primitive can be written in terms of a series whose terms depend on the values of an incomplete $\Gamma$ function. For instance:

$$\int_{0}^{+\infty}\frac{dx}{e^x+x^e}=\int_{0}^{+\infty}\sum_{n\geq 0}(-1)^nx^{ne}e^{-(n+1)x}\,dx=\sum_{n\geq 0}\frac{(-1)^n\,\Gamma(en+1)}{(n+1)^{en+1}}$$ and: $$\frac{\Gamma(en+1)}{(n+1)^{en+1}}\approx \sqrt{2\pi e/n}\left(\frac{n}{n+1}\right)^{en+1} \approx e^{-e}\sqrt{\frac{2\pi e}{n}}.$$

Answer 2

Case $1$: $a\in\mathbb{R}^+$ , $a\neq1$ and $x^aa^{-x}\leq1$

Then $\int\dfrac{1}{a^x+x^a}dx$

$=\int\dfrac{a^{-x}}{1+x^aa^{-x}}dx$

$=\int\sum\limits_{n=0}^\infty(-1)^nx^{an}a^{-(n+1)x}~dx$

$=\int\sum\limits_{n=0}^\infty(-1)^n(n+1)^{-an-1}(\ln a)^{-an-1}\gamma(an+1,(n+1)x\ln a)+C$