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Here is a question of real analysis. I am not sure whether my conjecture is true or not? Any help will be appreciated.

Suppose $\psi : \mathbb{R}\to \mathbb{R}$ is a real valued, even and sufficiently smooth (I think we have to assume higher-order derivatives exist) function. Suppose $$\int_{\mathbb{R}}\exp(-\psi(t))dt=1\; \text{and}\; \int_{\mathbb{R}}|t|\exp(-\psi(t))dt<\infty.$$ I want to prove that $\lim_{|z|\to \infty}z\exp(-\psi(z))=0$. I have no idea how to proceed. Thank you for your help.

De vinci
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    You don't need to require any smoothness on $\psi$ besides continuity perhaps. A necessary (but insufficient) condition for the convergence of an improper integral to infinity is that the integrand goes to $0$. Turning that around, if $\int_\mathbb{R} f dx < \infty$ then $f(x)$ must go to $0$ – Ninad Munshi Nov 14 '19 at 20:48
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    @NinadMunshi It was pointed out by GEdgar that this statement is actually false (you can consider $f$ to be triangles of area $1/2^n$ as a counter example) – Yanko Nov 14 '19 at 20:57
  • this is a counter example https://math.stackexchange.com/a/2434794/396668 – De vinci Nov 14 '19 at 20:58
  • @Yanko good point, I'll leave up my comment for anyone else that falls into the trap – Ninad Munshi Nov 14 '19 at 21:00
  • I don't believe the conjecture in the OP is true, for exactly the reason that Yanko/GEdgar point out: one can make the integrand essentially the union of ever-narrower bumps so that both integrals still converge, but the integrand has no pointwise upper bound. Piecewise-linear paths are one way to accomplish this, and any such example can be approximated arbitrarily closely by real-analytic examples. – Greg Martin Nov 15 '19 at 00:40

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