Integral related to the gamma function: $\int_0^{\infty} y^{\alpha-1}e^{-y}dy=(\alpha-1)\int_0^{\infty} y^{\alpha-2}e^{-y}dy$

Question

I have difficulty with the gamma function: $\int_0^{\infty} y^{\alpha-1}e^{-y}dy=(\alpha-1)\int_0^{\infty} y^{\alpha-2}e^{-y}dy$

How do we go from left to right? Thanks!

Answer 1

Yes, this is the so-called functional equation of the Gamma Function. Use integration by parts to prove it. Have you already learned integration by parts?

Ok. Integration by parts states that $$\int ab = a\int b - \int\left(\left(\int a \right) b'\right) $$ where a and b are functions and the dash means the derivative. Try using the formula on your integral. If you wish to know how to prove the formula for integration by parts, I can show you.

Answer 2

This is standard integration by parts with $u=y^{a-1}$ and $dv=e^{-y}dy$.