Define $ \ell_0 (x)$ by $ \ell_0 (x) := \int_1^x \frac{e^{\zeta}}{\zeta} d \zeta $ and define $\ell_1 (x)$ by $\ell_1 (x) := (1-x) \ell_0 (x) + x \ell_0 ' (x).$ Then I want to prove that $ \int_0^\infty | \ell_0 (x) |^2 e^{-x} dx \quad \text{and} \quad \int_0^\infty | \ell_1 (x) |^2 e^{-x} dx $ both diverge.
Proof that both $ \int_0^\infty | \ell_0 (x) |^2 e^{-x} dx$ and $\int_0^\infty | \ell_1 (x) |^2 e^{-x} dx $ diverge
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functional-analysis
1 Answers
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Note that $e^{\zeta/2}>\zeta$ for $\zeta\geq 2$, and so $\ell_0(x)\geq \int_2^x e^{\zeta/2}d\zeta=2e^{x/2}-2e$. Thus $\int_0^\infty|\ell_0(x)|^2e^{-x}dx= \int_0^\infty \frac{4e^x-4e^{x/2+1}+4e^2}{e^x}dx$ and since the limit of the integrand as $x\to\infty$ is clearly $4$ (hence not $0$), this integral cannot converge. I'll leave the second one to you.
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0That would fit in what I considered to be "your case". I made my comment because "not going to zero" does not mean "converging to some non-zero limit" in general. – 2012-07-31