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Show that $x_n(t)=n^2 te^{-nt}$ does not converge in $L_2(\mathbb{R_+})$.

$\int_\limits{0}^{\infty}(n^2 te^{-nt})^2dt=\int_\limits{0}^{\infty} n^4 t^2e^{-2nt} dt\geqslant\int_\limits{0}^{\infty} n^2 t^2e^{-2nt} dt$.

I tried to find a smaller integral that would diverge but I cannot lower the value more I cannot increase.

Question:

How should I prove the convergence?

Thanks in advance!

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  • $\begingroup$ That integral clearly converges for all $n$. I think you are supposed to show that the sequence of functions $(x_n)$ does not have a limit in $L^2(\Bbb{R}_+)$. In other words, that there is no function $f\in L^2$ such that $||x_n-f||\to0$ as $n\to\infty$. $\endgroup$ – Jyrki Lahtonen May 22 '18 at 17:31
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Doing the substitution $nt=x$ and $n\,\mathrm dt=\mathrm dx$, you get the integral$$n\int_0^{+\infty}x^2e^{-2x}\,\mathrm dx.$$But the integral $\int_0^{+\infty}x^2e^{-2x}\,\mathrm dx$ converges to some number greater than $0$ and therefore your sequence of integrals diverges.

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You can calculate the integral explicitly using integration by parts:

\begin{align} n^4\int_0^\infty t^2 e^{-2nt}\,dt &= n^4\left[-\frac{t^2}{2n}e^{-2nt}\Bigg|_0^\infty - \int_0^\infty 2t \left(-\frac1{2n} e^{-2nt}\right)\,dt\right]\\ &= n^3\int_0^\infty te^{-2nt}\,dt\\ &= n^3 \left[-\frac{t}{2n}e^{-2nt}\Bigg|_0^\infty - \int_0^\infty \left(-\frac1{2n} e^{-2nt}\right)\,dt\right]\\ &= \frac{n^2}2 \int_0^\infty e^{-2nt}\,dt\\ &= \frac{n}4 e^{-2nt}\Bigg|_0^\infty\\ &= \frac{n}4\\ \end{align}

which goes to $+\infty$ as $n\to\infty$.

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