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For each $n\in\mathbb{N}$ and $x\in\mathbb{R}$ define $$ g_n(x)=2nxe^{-nx^2}\qquad h_n(x)=g_{n+1}(x)-g_n(x) $$ Prove that $$ \int\limits_0^\infty\sum_{n=0}^\infty h_n(t)dt \neq \sum_{n=0}^\infty\int_0^\infty h_n(t)dt $$ So far, I've shown that the LHS is equal to $0$ by noting that $$ \sum\limits_{n=0}^\infty h_n(t)=g_\infty (t) - g_0(t) = \lim_{n\to \infty} 2nxe^{-nx^2} =0 $$ by L'Hopital's rule. Thus, we have an integral of a constant function equal to $0$, which is $0$.

For the RHS, I've got the integral down to $$ \int\limits_0^\infty h_n(t)dt= \int\limits_0^\infty 2(n+1)xe^{-(n+1)x^2}dx - \int\limits_0^\infty 2nxe^{-nx^2}dx=[e^{-nx^2}-e^{-(n+1)x^2}]_0^\infty=0 $$

This gets me to believe that RHS equals LHS, which is directly contrary to the question. Am I missing something?

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1 Answer 1

$h_0(t)=2 x e^{-t^2}$ and $\int_0^\infty 2x e^{-t^2} dt=1$. Your argument for $\int_0^\infty h_n(t)dt$ applies just fine to the rest of the $n$, so you end up with $0=1$.

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Ahh, that makes sense. Thanks heaps. –  Lachlan Oct 7 '12 at 14:44

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