Calculate integration with Euler integration

Question

Calculate: $$\int^{\pi /2}_{0}\tan ^{\alpha} x dx$$ with $| \alpha|<1$

Answer: $\pi/(2 \cos \frac{\alpha \pi}{2})$

In $B$ function or integral, we have $B(x,y)=\int^{1}_{0}t^{x-1}(1-t)^{y-1}dt$ which is defined on interval $(0,1)$, but the problem listed above seems out of range with $\alpha$ uncertain. Thank you for your help.

Edit: The previous one seems broken... thanks for @curiousguest

You don't need Euler's beta function to evaluate this one: $\int\tan\alpha x,dx=-\frac{1}{\alpha}\ln|\cos\alpha x|+C$. Maybe you mean something else? — CuriousGuest, Jun 30 '14 at 10:17
Oh, I know what's wrong. I can get $- \frac{1}{\alpha} \ln(|\cos \frac{\pi}{2} \alpha|)$ as result. But how could it be the same as the answer listed above? — Zhen Zhang, Jun 30 '14 at 10:57
It's actually not the same, that's why I asked if you have written the right integral. — CuriousGuest, Jun 30 '14 at 11:00
Yes that's what is given on book. Maybe we should ignore it? Thank you anyway :) — Zhen Zhang, Jun 30 '14 at 11:03
Oh, I got it. The function should be $\tan^\alpha x$, then your answer is right. — CuriousGuest, Jun 30 '14 at 11:06

score 2 · Accepted Answer · answered Jun 30 '14 at 11:19

First, show that $$\int_0^{\pi/2}\sin^\alpha x\cos^\beta x\,dx=\frac12\mathrm{B}\left(\frac{1+\alpha}{2},\frac{1+\beta}{2}\right)$$ by substitution $t=\sin^2 x$. Then your integral is $$I=\frac12\mathrm{B}\left(\frac{1+\alpha}{2},\frac{1-\alpha}{2}\right)=\frac{\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{1-\alpha}{2})}{2\Gamma(1)}=\frac12\Gamma\left(1-\frac{1-\alpha}{2}\right)\Gamma\left(\frac{1-\alpha}{2}\right)=\frac{\pi}{2\sin\pi(\frac{1-\alpha}{2})}$$ by Euler's reflection formula.

Calculate integration with Euler integration

1 Answers1