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The user known as sos440 posted this: $$\begin{align*} \sum_{n=0}^\infty \frac{r^n}{n!} \int_0^\infty x^n e^{-x} \; dx & = \int_{0}^\infty \sum_{n=0}^\infty \frac{(rx)^n}{n!} e^{-x} \; dx = \int_0^\infty e^{-(1 - r)x} \; dx \\ & = \frac{1}{1 - r} = \sum_{n=0}^\infty r^n \end{align*}$$ (citing Tonelli's theorem to justify interchanging the sum and the integral), and concluded that $$ \frac{1}{n!}\int_0^\infty x^n e^{-x}\,dx=1. $$

Is this a very isolated thing or just an instance of something generally useful for some much broader class of integrals? (Obviously, what is seen below is generally true, but how generally is it useful?)

$$ \sum_{n=0}^\infty r^n \int_A f_n(x)\,dx = \int_A \sum_{n=0}^\infty \big( r^nf_n(x) \big) \, dx = \int_A g(r,x) \, dx = h(r) = \sum_{n=0}^\infty c_n r^n $$ $$ \therefore \int_A f_n(x)\,dx = c_n. $$

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    I guess this could seem like magic, but isn't this just computing the exponential generating function of the integrals? We've all seen how generating function methods can make hard tasks seem simple.2012-04-13
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    We could say $$\sum_{n=0}^\infty \frac{r^n}{n!} \mathcal{M} \{ f \}(n) = r \mathcal{L} \{ f \} (-r).$$2012-04-17
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    The method is essentially the same as using the moment generating method of calculating the moments of an exponentially distributed random variable. The same idea applied to the normal distribution gives$$\int_{-\infty}^\infty x^{2n}e^{-x^2/2}dx =\sqrt{2\pi}\frac{(2n)!}{2^nn!}.$$Unfortunately, I can't think of many simple examples, as not that many continuous distributions have simple explicit forms for both the pdf and MGF.2013-08-12
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    > how generally is it useful < Is this really an answerable question? It's useful for any integral of that form!2013-08-16

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