How to calculate the integral
$ \int_b^c \! (c-s)^{-3/4} (s-b)^{-3/4}\,\mathrm{d}s $
Thanks.
How to calculate the integral
$ \int_b^c \! (c-s)^{-3/4} (s-b)^{-3/4}\,\mathrm{d}s $
Thanks.
Let $I=\displaystyle\int_b^c (c-s)^{-3/4}(s-b)^{-3/4}ds$. By substitution $t=\dfrac{s-b}{c-b}$ we have $s-b=(c-b)t$, $c-s=(c-b)(1-t)$, $ds=(c-b)dt$. Hence $ I=\int_0^1 (c-b)^{-3/4}(1-t)^{-3/4}(c-b)^{-3/4}t^{-3/4}(c-b)dt=(c-b)^{-1/2}B\left(1/4,1/4\right), $ where $B$ is Beta function, defined as $B(u,v)=\displaystyle\int_0^1 (1-t)^{v-1}t^{u-1}dt$ for every positive $u$ and $v$.