2
$\begingroup$

Convergence of $\int _0^{\pi }\frac{1}{\sqrt{\left|\tan x\right|}}dx$

I have a trouble when doing this question. I tried to separate the integral into two integrals like this: $$\int _0^{\frac{\pi }{2}\:}\frac{1}{\sqrt{\tan x}}dx+\int _{\frac{\pi }{2}}^{\pi\:}\frac{1}{\sqrt{-\tan x}}dx$$ As for the first integral, I can prove it converges, but for the second integral I can not find a way to prove it converges. I don't know if my separation is the right approach to solve this problem. Can anyone suggest me a way to do this or another approach to solve this problem? Thank you so much.

$\endgroup$
2
  • $\begingroup$ Shouldn't your second integral be from $\pi/2$ to $\pi$ ? $\endgroup$ Apr 22, 2020 at 7:53
  • $\begingroup$ Oh yes, wrong typing sorry $\endgroup$
    – C.Elliot
    Apr 22, 2020 at 7:54

1 Answer 1

2
$\begingroup$

If the integral converges,

$$\int_0^\pi\frac{dx}{\sqrt{|\tan x|}}=2\int_0^{\pi/2}\frac{dx}{\sqrt{\tan x}}.$$

Then we can integrate the singularity separately by

$$\int_0^{\pi/2}\frac{dx}{\sqrt{\tan x}} = \int_0^{\pi/2}\left(\frac1{\sqrt{\tan x}}-\frac1{\sqrt x}\right)dx+\int_0^{\pi/2}\frac{dx}{\sqrt x} \\= \int_0^{\pi/2}\left(\frac1{\sqrt{\tan x}}-\frac1{\sqrt x}\right)dx+2\sqrt x\bigg|_0^{\pi/2}$$ where the new integrand is bounded.

enter image description here

(We used that for small $x$, $\tan x\sim x$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.