Is there a substitution approach to provide a closed-form solution to the following integral?
\begin{equation} \int_0^1\frac{(1-p)^{k-1}(1-p+p\ln p)}{\left(\ln p\right)^2}dp \end{equation}
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In the denominator, do you mean $\ln {(p^2)}$ or do you mean ${(\ln p)}^2$? – AryanSonwatikar Nov 09 '19 at 10:14
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1@AryanSonwatikar - it should be clear now – dsmalenb Nov 09 '19 at 10:23
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1Also, have you got all your terms correct? Cause if you add $pk\ln p -\ln p$ in the right bracket in the numerator, you will be looking at $-\frac{d}{dp} \left( \frac{p{(1-p)}^k}{\ln p}\right)$ – AryanSonwatikar Nov 09 '19 at 10:24
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1Yes, the terms are correct. Perhaps this is more about adding and subtracting terms than a substitution. – dsmalenb Nov 09 '19 at 10:28
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In that case, even Wolfram|Alpha is stuck. – AryanSonwatikar Nov 09 '19 at 10:47
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I see a reduction to a Frullani-type integral. The substitution $p=e^{-x}$ transforms it into $$I_k=\int_0^\infty e^{-x}(1-e^{-x})^k\frac{dx}{x^2}-\int_0^\infty e^{-2x}(1-e^{-x})^{k-1}\frac{dx}{x}$$ which, after integration by parts in the first term, gives $$I_k=\int_0^\infty(ke^{-2x}-e^{-x})(1-e^{-x})^{k-1}\frac{dx}{x}=(k-1)J_{k-1}-kJ_k,$$ where $$J_k=\int_0^\infty e^{-x}(1-e^{-x})^k\frac{dx}{x}=\sum_{j=0}^{k}\binom{k}{j}(-1)^{j+1}\ln(j+1)$$ is that very integral; see the middle of this answer of mine. (I don't see further reduction.)
metamorphy
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1I was not aware of this type of integral! I am glad this question pays off with more to learn! Thank you – dsmalenb Nov 09 '19 at 17:50
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+1 nice solution! I too was unaware of Frullani-type integrals. – AryanSonwatikar Nov 10 '19 at 04:14