# Today's calculation of integral #18

• May 22nd 2008, 02:28 PM
Krizalid
Today's calculation of integral #18
Find $\displaystyle \int_{0}^{\pi /2}{\frac{\tan ^{\alpha }x-1}{\ln (\tan x)}\,dx},\,0<\alpha <1.$
• May 22nd 2008, 03:03 PM
wingless
Quote:

Originally Posted by Krizalid
Find $\displaystyle \int_{0}^{\pi /2}{\frac{\tan ^{\alpha }x-1}{\ln (\tan x)}\,dx},\,0<\alpha <1.$

$\displaystyle \int_{0}^{\pi /2}{\frac{\tan^{n}x-1}{\ln (\tan x)}\,dx}$

$\displaystyle \frac{\tan ^{n}x-1}{\ln (\tan x)} = \int_0^n \tan^y x~dy$

It becomes,

$\displaystyle \int_{0}^{\pi /2} \int_0^n \tan^y x~dy ~dx$

$\displaystyle \int_0^n \int_{0}^{\pi /2} \tan^y x~dx ~dy$

I'm not in a condition to go further. It's been too late (Yawn) I better go sleep some .. (Sleepy)
• May 22nd 2008, 03:05 PM
Krizalid
It's okay. That was exactly the idea.
• May 22nd 2008, 03:13 PM
Mathstud28
Quote:

Originally Posted by Krizalid
It's okay. That was exactly the idea.

Haha...I cant remember the last integral you put up that wasn't double integration (Giggle)