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Integrate/hint for this definite integral

$$\int_0^\infty(\log\theta)^n\frac{1}{\theta^{k+2}}\text{d}\theta,$$

where $n$ and $k$ are positive integers. It is a simplified form of my earlier question that I posted on 19 Mar 2014. Any suggestion is very welcome. Thanks.

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  • $\begingroup$ Your integral doesn't converge. Are you sure that the lower integration limit is $0$ instead of $1$? $\endgroup$
    – Lucian
    Mar 26, 2014 at 3:41

3 Answers 3

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As it currently stands, the integral diverges. If however you meant to write $1$ instead of $0$, then its closed form expression is $$\int_1^\infty\frac{\ln^n\theta}{\theta^{k+2}}d\theta=\frac{n!}{(k+1)^{n+1}}$$ This can be deduced from Euler's initial formula for the $\Gamma$ function, followed by a small change in variable.

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You can try to apply the exponential integral function $E_n=\int_{1}^{\infty}\dfrac{e^{-xt}dt}{t^n}$ or you can try the gamma function.

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The antiderivative of the integrand is given by $$-\log ^{n+1}(x) ((k+1) \log (x))^{-n-1} \Gamma (n+1,k \log (x)+\log (x))$$ which can be simplified to $$-\log ^{n+1}(x) \text { } E_{-n}((k+1) \log (x))$$ where appears the elliptic integral. As already said by other prticipants the integral to infinity diverges if the lower bound is $0$; if the lower bound is $1$, you get Lucian's result if $\Re(n)>-1$ and $\Re(k)>-1$.

If the lower bound is $a$, the integral is given by $$\frac{\log ^n(a) ((k+1) \log (a))^{-n} (\Gamma (n+1,(k+1) \log (a))-n \Gamma (n))+\left(\frac{1}{k}+1\right)^{-n} k^{-n} \Gamma (n+1)}{k+1}$$ provided $\Re(n)>-1$ and $\Re(k)>0$.

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