Integration using hypergeometric function

Question

I would like to calculate the definite integral $\int_0^\infty \frac{dx}{x(x^a + x^{-b})}$. From Wolfram Alpha, the indefinite integral is $x^b\ _2F_1(1;b/(a+b);b/(a+b)+1;x^{a+b})/b$ Where $_2F_1$ is the hypergeometric function. But I don’t know how to evaluate this at $x=0,\infty$. I tried to use equalities for hypergeometric function on wolfram, but they didn’t help.

I am also interested in the asymptotic behavior. What does $\log(x^b\ _2F_1(1;b/(a+b);b/(a+b)+1;x^{a+b})/b)$ look like?

Using this answer (or any of the answers in Approach0), one gets that $$\int_0^{\infty} \frac{1}{x(x^a+x^{-b})}dx=\int_0^{\infty} \frac{x^{b-1}}{x^{a+b}+1}dx=\frac{\pi}{(a+b)\sin(\pi b/(a+b))},$$ for $a,b>0$. — projectilemotion, Jan 10 '23 at 00:06
Note that hypergeometric function is not good for evaluating this integral, since this ${}_2F_1$ series has radius of convergence $1$. — GEdgar, Jan 10 '23 at 00:20

score 2 · Accepted Answer · answered Jan 10 '23 at 04:14

Letting $y=\frac{1}{x^{a+b}+1}$ transforms the integral into a Beta function $$ \begin{aligned} I & =\frac{1}{a+b} \int_0^1 y^{\frac{a}{a+b}-1}(1-y)^{-\frac{a}{a+b}} d y \\ & =\frac{1}{a+b} B\left(\frac{b}{a+b}, \frac{a}{a+b}\right) \end{aligned} $$ Using the reflection property of Beta function: $B(x, 1-x)=\pi \csc (\pi x) \textrm{ for } x\notin \mathbb{Z}$, we have

$$ I=\frac{1}{a+b} \pi \csc \left(\frac{b \pi}{a+b}\right) $$

Wish it helps!

Nice and simple solution ! (+1) – Claude Leibovici Jan 10 '23 at 05:44 — Claude Leibovici, Jan 10 '23 at 05:44

Integration using hypergeometric function

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