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An Integral March 25, 2011

Posted by Arjun in : calculus , trackback

Find $$\int^{\frac{\pi}{2}}_0 \frac{1}{\sqrt{\tan{x}}} \,dx$$

Comments»

1. Will B. - March 25, 2011

Should also be able to do this one with residues… I’m having some issue with the pole order or something weird.

2. Anish Tondwalkar - March 26, 2011

You can do it with residues, but not immediately. If you try now, you’ll get a pole of order 1/2, which you can’t do anything with. After you make a substitution to scale the limits and get rid of the sqrt, you get nice 4th order poles. I’ll leave it to Arjun to post the solution, but the answer should be $\frac{\pi}{\sqrt{2}}$ (or $\frac{\pi}{2\sin{(\pi/4)}}$ for the trigonometrically impaired).

3. Will B. - March 30, 2011

With the Beta Function:

$$t = \sin^2{x}$$
$$\begin{eqnarray}
I &=& \frac{1}{2} \int_{0}^{1} t^{-3/4} (1-t)^{-1/4} dt \\
&=& B \left(\frac{1}{4}, \frac{3}{4}\right) \\
&=& \frac{\Gamma(\frac{1}{4})\Gamma(\frac{3}{4})}{\Gamma(1)} \\
&=& \boxed{\frac{\pi}{\sqrt{2}}}
\end{eqnarray}$$