Calculate $\int_0^{\infty}y^{-\frac{1}{2}}(1+y)^{-m}dy$

Question

Calculate $\int_0^{\infty}y^{-\frac{1}{2}}(1+y)^{-m}dy$

This looks like a $\beta $ distribution to me, however I'm unsure of how to approach the distribution when we have $1+y$ instead of $1-y$, my attempt.

Generalising $-1/2$ for $-n$

$$\begin{align}I(-n,-m)&=\int_0^{\infty}y^{-n}(1+y)^{-m}dy \\ &\implies du = -ny^{-(n+1)}, v = \frac{(1+y)^{1-m}}{1-m} \\ &= \left(\lim_{y \to \infty}\frac{y^{-n}(1+y)^{1-m}}{1-m}\right) + \frac{n}{1-m}I(-(n+1), 1-m) \\ &= \frac{n!}{(1-m)!(m-n)!}I(0, m-n) \\ &\implies I(0,k)_{k=m-n}=\frac{1}{n-m+1} \\ &\implies I(-n,-m)=\frac{\Gamma(n+1)}{\Gamma(m)\Gamma(n-m+2)} \end{align}$$

However, when I plug in $n=1/2$ I do not get the expected answer that should be the LHS $$\frac{\Gamma(m)}{\Gamma(\frac{1}{2})\Gamma(m-\frac{1}{2})} \ne \frac{\Gamma(\frac{1}{2}+1)}{\Gamma(m)\Gamma(\frac{1}{2}-m+2)}$$

Answer 1

Substitute $\;y=-1+\frac1t=\frac{1-t}t\implies dy=-\frac{dt}{t^2}\;$ and the integral becomes

$$-\int_1^0\frac{dt}{t^2}\frac{(1-t)^{-1/2}}{t^{-1/2}}t^m=\int_0^1t^{m-3/2}(1-t)^{-1/2} dt=B\left(m-\frac12\,,\,\frac12\right)=\frac{\Gamma\left(m-\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(m\right)}$$

Maybe I made a mistake or maybe it is yours, but it is the LHS of your answer upside down....Check this.

Answer 2

We have that your integral is $$\int_0^\infty \frac{y^{-1/2}}{(1+y)^m}\,dy =\beta(1/2, m-1/2)=\frac{\Gamma(1/2)\Gamma(m-1/2)}{\Gamma(m)}.$$ Using the Legendre duplication formula, you find that this expression is equal to $$\pi\,{2m \choose m}2^{-2m}\frac{m}{2m-1}$$ valid for $m>1/2$.