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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

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

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?

  • 2
    Hint for the last equality: Make the substitution $tx \mapsto x$.2012-10-02
  • 14
    Set $t = 1$. ${}$2012-10-02

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