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$$ \mbox{Show that}\quad \int^{\infty}_{0}{x^{p - 1}\ln\left(x\right) \over 1 + x} =-\pi^{2}\csc\left(p\pi\right)\cot\left(p\pi\right) $$

$$ \mbox{I guess I have to use the fact that}\quad \int^{\infty}_{0}{x^{p - 1} \over 1 + x} =\pi\csc\left(p\pi\right),\quad\mbox{please give me some idea.} $$

Thank you.

Felix Marin
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JSCB
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1 Answers1

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Lest the question remain unanswered:

We have

$$\int_0^\infty \frac{x^{p-1}\ln x}{1+x}\,dx = \int_0^\infty \frac{d}{dp}\frac{x^{p-1}}{1+x}\,dx = \frac{d}{dp} \int_0^\infty \frac{x^{p-1}}{1+x}\,dx$$

for $0 < p < 1$, since the dominated convergence theorem guarantees that differentiation under the integral is legitimate.

And

$$\frac{d}{dp} \frac{\pi}{\sin (p\pi)} = - \frac{\pi^2\cos (p\pi)}{\sin^2 (p\pi)} = -\pi^2\csc (p\pi)\cot (p\pi).$$

Daniel Fischer
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  • I may be missing something obvious, but how did you get from $\int_0^\infty \frac{x^{p-1}}{1+x},dx$ to $\frac{\pi}{\sin (p\pi)}$? – Akiva Weinberger Aug 07 '14 at 07:47
  • @columbus8myhw That was given in the question as a fact, so I took that as known. One can get that for example by the residue theorem, or by transforming it into a form one recognises as an integral representation of the beta function, specifically $B(p,1-p)$. – Daniel Fischer Aug 07 '14 at 10:54