let-0-lt-a-lt-1-find-the-valueof-0-t-a-1-1-t-2-dt-

Question Number 63661 by mathmax by abdo last updated on 06/Jul/19

l e t 0 < a < 1 f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t^{2}} d t

Commented by mathmax by abdo last updated on 07/Jul/19

c h a n g e m e n t t^{2} = x g i v e \int_{0}^{\infty} \frac{t^{a - 1}}{1 + t^{2}} d t = \int_{0}^{\infty} \frac{{(x^{\frac{1}{2}})}^{a - 1}}{1 + x} \frac{1}{2} x^{- \frac{1}{2}} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{a - 1}{2} - \frac{1}{2}}}{1 + x} d x = \frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{a}{2} - 1}}{1 + x} d x = \frac{1}{2} \frac{π}{s i n (\frac{π a}{2})} \Rightarrow

\int_{0}^{\infty} \frac{t^{a - 1}}{1 + t^{2}} d t = \frac{π}{2 s i n (\frac{π a}{2})} .