# Differentiation under integral sign (Gamma function)

This might be a silly question, but I'm reading this article about differentiation under the integral sign, and I'm stumped by something that's written early on. The author is giving a derivation of the formula for $n!$ in terms of the gamma function. He shows how you can get

$$\frac{n!}{t^{n+1}} = \int_0^{\infty}x^ne^{-tx}\,dx$$

by differentating under the integral sign of $\int_0^{\infty}e^{-tx}dx$. He then says that the above "immediately implies" the formula

$$n! = \int_0^\infty x^ne^{-x}\,dx.$$

However, I can't for the life of me see how this follows. Multiplying the first equation by $t^{n+1}$ gives $n! = t^{n+1}\int_0^{\infty}x^ne^{-tx}$, so apparently

$$t^{n+1}\int_0^\infty x^ne^{-tx} \, dx = \int_0^\infty x^ne^{-x} \, dx,$$

but I don't see how this is true. Can anyone explain this?

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Hint for the last equality: Make the substitution $tx \mapsto x$. – Sangchul Lee Oct 2 '12 at 3:51
Set $t = 1$. ${}$ – Qiaochu Yuan Oct 2 '12 at 3:52

Making the change of variables $y=xt$ gives the desired result $$t^{n+1}\int_0^{\infty}x^ne^{-tx}dx = t^{n+1}\int_{0}^{\infty}\frac{y^n}{t^n}e^{-y}\frac{dy}{t} = \int_0^{\infty}x^n e^{-x} dx \,.$$