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Could anyone give me any little hints on how to show the following please?

Is $(\log x)e^{-x}$ Lebesgue integrable on $(0, \infty)$ ?

I cannot see how to do this. I've tried Comparison Test.

I believe it may require me to use that if $g$ is integrable and $h$ is bounded and measurable, $gh$ is integrable. $e^{-x}$ is bounded and measurable but how is $\log x$ integrable? Comparison again doesn't work and neither does looking at the modulus (since $\log x$ is unbounded at $0$ and $\infty$)

Thanks for any hints.

JohnD
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Mt123
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2 Answers2

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You can show that $f$ is integrable on $(0,\varepsilon)$, $\varepsilon>0$, by comparison with $\frac{1}{\sqrt x}$.

You can show that $f$ is integrable on $(1,\infty)$ by comparison with $xe^{-x}$, or by noting that $\log(x)e^{-x}=(\log(x)e^{-x/2})e^{-x/2}$ and showing that $\log(x)e^{-x/2}$ is bounded on $(1,\infty)$.

Jonas Meyer
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Hint: The inequality, valid for all real $x$, $1+x\le e^x$ is often useful.

This is equivalent to $\left(1+\frac xn\right)^n\le e^x$ and $\log(x)\le n(x^{1/n}-1)$.

robjohn
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