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I have to solve the following question :

"By n-fold differentiation of the function : $g(x)= \int_{0}^{\infty} e^{-tx} dt$

Show that $\int_{0}^{\infty} x^{n} e^{-x} dx = n! $ "

Firstly, I tried to solve g(x) and I got this far :

$g(x)= \int_{0}^{\infty} e^{-tx} dt\\ g'(x) = \int_{0}^{\infty} \frac{\partial}{\partial x} e^{-tx} dt\\ g'(x) = \int_{0}^{\infty} -te^{-tx} dt$

I can't figure out how to integrate the final part and also link it to this equation $\int_{0}^{\infty} x^{n} e^{-x} dx = n! $

Any help would be much appreciated.

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    $\begingroup$ Try integration by parts. $\endgroup$ – Karn Watcharasupat Apr 25 '18 at 12:48
  • $\begingroup$ Thank you. I have now integrated the equation $\endgroup$ – user546917 Apr 25 '18 at 15:00
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We can show that \begin{align} g(x)&=\int_0^{\infty} e^{-tx}\,dt=\frac{1}{x} \end{align} Differentiating both sides with respect to $x$ a few times shows that \begin{align} \int_0^{\infty} -te^{-tx}\,dt\Big|_{x=1}&=-1 \\ \\ \int_0^{\infty} t^2 e^{-tx}\,dt\Big|_{x=1}&=2 \\ \\ \int_0^{\infty} -t^3 e^{-tx}\,dt\Big|_{x=1}&=-6 \end{align} We can thus speculate that $$\int_0^{\infty} x^n e^{-x}\,dx=n!$$ And you may be able to use induction on $$\left[\frac{\partial^n}{\partial t^n}\int_0^{\infty}e^{-tx}\,dx\right]\Biggr|_{t=1}$$ to prove this conjecture.

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  • $\begingroup$ Thank you for your help. Seeing the method it makes clear sense now. $\endgroup$ – user546917 Apr 26 '18 at 13:13

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