Q) $\int_2^{\infty}\left( \frac{1}{x\log^2x} \right)^pdx$
$p=1$ is simple by substitution. How can I do this in general? Thanks.
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For $p=2$ https://www.wolframalpha.com/input/?i=int%281%2F%28x+%28ln%28x%29%29%5E2%29%5E2%2C%7Bx%2C2%2Cinfty%7D%29 ... indeed this will give answers for larger $p$. – Donald Splutterwit Dec 06 '19 at 02:06
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Make $x=e^t$ to make $$I_p=\int_2^{\infty}\left( \frac{1}{x\log^2x} \right)^p\,dx=\int_{\log(2)}^{\infty}e^{(1-p) t} t^{-2 p}\,dt$$ and $$\int e^{(1-p) t} t^{-2 p}\,dt=-t^{1-2 p} E_{2 p}((p-1) t)$$ where appears the exponential integral function. Then
$$\color{blue}{I_p=\frac{ E_{2 p}((p-1) \log (2))}{\big[\log(2)\big]^{2p-1}}}$$
Claude Leibovici
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