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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 $$\frac{n!}{t^{n+1}} = \int_0^{\infty}x^ne^{-tx}$$ 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 $$t^{n+1}\int_0^{\infty}x^ne^{-tx} = \int_{0}^{\infty}x^ne^{-x}dx,$$ but I don't see how this is true. Can anyone explain this?

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1  
Hint for the last equality: Make the substitution $tx \mapsto x$. –  sos440 Oct 2 '12 at 3:51
4  
Set $t = 1$. ${}$ –  Qiaochu Yuan Oct 2 '12 at 3:52

1 Answer 1

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

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