# Integrating $x^2e^{-x}$ using Feynman's trick? [duplicate]

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In the second episode of season $8$ of "The Big Bang Theory," which aired yesterday night, it is stated that one can integrate $x^2e^{-x}$ by using Feynman's trick of differentiating under the integral. Is this actually true, and if so, how to do it? And is it "better," in any sense, than the usual way of doing it by integration by parts?

## marked as duplicate by Semiclassical, Adam Hughes, Hans Lundmark, Mark Fantini, Najib IdrissiSep 24 '14 at 11:11

• I wasn't the only one who immediately went to try this when I heard that?! – Robin Goodfellow Sep 23 '14 at 16:48
• This is the second question today that comes from this episode. Funny. – Asaf Karagila Sep 23 '14 at 16:52
• What's the first? – Nishant Sep 23 '14 at 19:54
• – Jonas Dahlbæk Sep 23 '14 at 20:22

If we just compute without thinking about problems such as interchanging integration and differentiation, the derivation is very simple. $$\int_0^{\infty} x^2e^{-x}dx=\left.\left(\frac{d}{d\alpha}\right)^2\right|_{\alpha=1}\int_0^{\infty} e^{-\alpha x}dx=\left.\left(\frac{d}{d\alpha}\right)^2\right|_{\alpha=1} \frac{1}{\alpha}=\left.\frac{2}{\alpha^3}\right|_{\alpha=1}=2.$$
If you want limits other than the positive real axis, the same trick allows you to arrive at $$\int_a^{b} x^2e^{-x}dx=\left.\left(\frac{d}{d\alpha}\right)^2\right|_{\alpha=1}\int_a^{b} e^{-\alpha x}dx=\left.\left(\frac{d}{d\alpha}\right)^2\right|_{\alpha=1} \frac{1}{\alpha}(e^{-a\alpha}-e^{-b\alpha}).$$ Of course, this can also be evaluated explicitly by using the Leibniz rule, but I will leave that up to you.
• Gives: $a^2 e^{-a}-b^2 e^{-b} + 2 \left(e^{-a}-e^{-b}\right)-2 \left(b e^{-b}-a e^{-a}\right)$ – amcalde Sep 23 '14 at 17:11
• @UserX It is a two-fold application of Theorem 5 in the notes I linked to. The integrals are uniformly convergent because you have a bound of the type $x^n e^{-\alpha x}\leq C_n e^{-\alpha x/2}$ on the positive axis. As for setting $\alpha=1$, this is an application of Theorem 4 in the notes I linked to. – Jonas Dahlbæk Sep 26 '14 at 10:01
• Almost right. We use $$\frac{\partial^2}{\partial a^2} e^{-ax} =x^2 e^{-ax}.$$ – Jonas Dahlbæk Sep 26 '14 at 11:09