1

For $\gamma>1$ consider $$ g(\gamma) := \int_0^{\infty}\frac{1}{1+x^\gamma}dx. $$ The integral is finite for any $\gamma>1$. My question is the following:

  1. Is $g$ a decreasing function of $\gamma$? That is, for $\gamma'\geq \gamma>1$, is it true that $g(\gamma')\leq g(\gamma)$?
  2. Is it true that $\lim_{\gamma\to\infty}g(\gamma) = 1$?

I tried both claims on Wolfram alpha and they seem to be true. Unfortunately, I don't know how to prove them.

Yining Wang
  • 1,279
  • 7
  • 23
  • You realize, for $\gamma > 1$, $g(\gamma) = \frac{\pi}{\gamma} \csc \left( \frac{\pi}{\gamma} \right)$? – Eric Towers Sep 17 '17 at 02:31
  • @EricTowers This closed form is awesome! I found the derivation of the integral here: https://math.stackexchange.com/questions/247866/show-that-int-0-infty-frac11xn-dx-frac-pi-n-sin-pi-n-wh so the problem is solved thank you! – Yining Wang Sep 17 '17 at 02:36

0 Answers0