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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0Note that the fact that "the limit of the integral is not zero implies that the integral doesn't converge", is not true in general, because there are integrals such that the integrand doesn't go to zero, but the integral is still convergent, such as $$ \int_0^{\infty} \cos(x^2) \, dx. $$ But it is definitely true in your case. – 2012-07-31
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0@PatrickDaSilva But if the limit *exists* and is not $0$, then it fails to converge. – 2012-07-31
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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