2

I want to evaluate the integral $$I(a, b) = \int_{0}^{\infty}\frac{1}{(1+x^a)(1+x^b)} dx \tag 1$$

where $a, b$ are nonnegative real numbers. I am most interested in integer $a, b$ but would prefer a more general answer.

In this answer to another MathSE question, the $b = 2$ case was addressed and it was stated that $I(a, 2) = \pi/4$. By plugging in some numbers, I found that $I(a, a) = \frac{1}{a}\Gamma \left( \frac{1}{a} \right) \Gamma \left( 2-\frac{1}{a} \right)$ and $I(a, 0) = \frac{1}{a}\Gamma \left( \frac{1}{a} \right) \Gamma \left( 1-\frac{1}{a} \right) = \frac{\pi}{a}\csc \left( \frac{\pi}{a} \right)$. I didn't manage to solve this problem completely, but I'll include the work for the representations I got.

By making the substitution $u = 1/x$, we get $$I(a, b) = \int_0^{\infty} \frac{1}{(1+u^{-a})(1+u^{-b})(u^2)} du = \int_0^{\infty}\frac{u^{a+b-2}}{(1+u^a)(1+u^b)} du \tag 2$$

Now after making the substitution $v = u^a$, we get $$\int_0^{\infty} \frac{(v^{1/a})^{b-1}}{(v+1)(1+(v^{1/a})^b)a} dv = \int_0^{\infty} \frac{v^{(b-1)/a}}{a(v+1)(1+v^{b/a})} dv \tag 3$$

From $(2)$, by making the substitution $v = u^{a+b}$, we get $$\int_0^{\infty} \frac{1}{(a+b)v^{1/(a+b)}(1+v^{a/(a+b)})(1+v^{b/(a+b)})} dv = \int_0^{\infty} \frac{1}{(a+b)v^{1/(a+b)}(1+v^{a/(a+b)}+v^{b/(a+b)}+v)} \tag 4$$

Any help in solving the integral would be appreciated.

Varun Vejalla
  • 8,873
  • 1
    The integral gives a Fox H-function $H_{2, 2}^{2, 2}$: $$I(a, b) = H_{2, 2}^{2, 2} {\left( 1 \middle| {(0, a), (1 - \frac 1 a, b) \atop (0, a), (1 - \frac 1 a, b)} \right)},$$ or a Meijer G-function $G_{m + n, m + n}^{m + n, m + n}$ when $b/a = m/n$. – Maxim Dec 27 '19 at 06:33
  • @Maxim I don't think the Meijer G-function is correct because that would imply that $I(a, b) = I(ka, kb)$. – Varun Vejalla Dec 28 '19 at 03:15
  • The indices give the number of arguments. The actual arguments involve $a$ and $b$ separately. – Maxim Dec 28 '19 at 04:58

1 Answers1

3

The case of integer powers can be handled with a keyhole/pacman contour and multiplying the integrand by $\ln(x)$. Doing so gives:

$$\int_0^\infty\frac{\mathrm dx}{(1+x^a)(1+x^b)}=-\Re\sum_\omega\mathop{\rm Res}_{z=\omega}\frac{\ln(z)}{(1+z^a)(1+z^b)}$$

Which can be easily computed. Supposing all roots have multiplicity 1, we get

$$\int_0^\infty\frac{\mathrm dx}{(1+x^a)(1+x^b)}=\frac1a\Re\sum_{\omega^a=-1}\frac{\omega\ln(\omega)}{1+\omega^b}+\frac1b\Re\sum_{\omega^b=-1}\frac{\omega\ln(\omega)}{1+\omega^a}$$

Simply Beautiful Art
  • 74,685