Thread: Today's calculation of integral #18

1. Today's calculation of integral #18

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

2. 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 I better go sleep some ..

3. It's okay. That was exactly the idea.

4. 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