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Beta function is defined as:

$$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}dt$$

for $\Re(x) , \Re(y)>0$, I want to show that:$\frac{B(x,y)}{c^y}=\int_0^\infty \frac{t^{x-1}dt}{(c+t)^{x+y}}$.

I thought of changing variables to $s=\frac{c}{t}-c$, but only for the case $c>0$, then I get the following integral: $$\int_0^\infty\frac{1}{(s+c)^{x+y}}c^xs^{y-1}c^{1-y}ds$$ which looks similar to what I need to show but not quite, any advice as to what change of variables will yield the appropriate integral?

Thanks. The reference of the question is question 8 on page 47 of the book: Special Functions by Andrews et al.

It isn't mentioned if $c>0$ or not, so I assume it can be negative.

MathematicalPhysicist
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1 Answers1

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Use:

$$ t=\frac{u}{c+u}\quad dt = \frac{c}{{(c+u)}^{2}}du $$

$$ \frac{B(x,y)}{c^y} = \int_{0}^{1} {\frac{{t}^{x-1}{(1-t)}^{y-1}}{c^y}dt}\\ \frac{B(x,y)}{c^y} = \int_{0}^{\infty}{\frac{1}{c^y}{(\frac{u}{c+u})}^{x-1}{(1-\frac{u}{c+u})}^{y-1}}\frac{c}{{(c+u)}^{2}}du\\ \frac{B(x,y)}{c^y} = \int_{0}^{\infty}{\frac{1}{c^y}{(\frac{u}{c+u})}^{x-1}{(\frac{c}{c+u})}^{y-1}}\frac{c}{{(c+u)}^{2}}du\\ = \int_{0}^{\infty}{\frac{1}{c^y}\frac{{u}^{x-1}c^y}{{(c+u)}^{x+y-2}}\frac{1}{c}\frac{c}{{(c+u)}^{2}}du}\\ =\int_{0}^{\infty}{\frac{{u}^{x-1}}{{(c+u)}^{x+y}}du}\\ $$

Oussama Boussif
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