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I'd like to calculate $\int_0^\infty \frac{\ln(1+x)}{x^{1+a}}dx$ for $a \in (0,1).$

I don't know how to start. Would you give me any hint for this problem? Thanks in advance!

04170706
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    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps. – robjohn Nov 15 '18 at 08:19
  • @robjohn Okay. I will do it next time. In this case, I just want a hint for the beginning, so that I can use residue theorem for calculating the integral. Thanks for your comment. – 04170706 Nov 15 '18 at 08:24
  • This doesn't seem like an integral that uses a Cauchy principal value. That usually indicates a singularity around which one removes a symmetric region and limits that region to a point. – robjohn Nov 15 '18 at 08:27
  • I would integrate by parts to get rid of the log, then apply the Beta Function and Euler's Reflection Formula. – robjohn Nov 15 '18 at 08:29
  • If you want to use contour integration to compute the integral after integration by parts, see this answer. – robjohn Nov 15 '18 at 08:37
  • @robjohn: no need for integration by parts for contour integration. – Ron Gordon Nov 15 '18 at 13:59
  • @RonGordon: perhaps I'm missing something, but on first inspection, with the branch cuts for $\log(1+z)$ and $z^{1+a}$ to deal with, the contour looks difficult. – robjohn Nov 15 '18 at 15:45
  • @robjohn: yeah, I worked it out, and you actually (in principle) need two contour integrations. I never said it would be better! ;) (You end up needing to evaluate Integrate[Log[x]/(1+x)^(1+a),{x,0,Infinity}], which uses a similar contour as the original integral.) – Ron Gordon Nov 15 '18 at 15:52

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Hint. Note that the integrand is positive and the improper integral is convergent. Moreover $$\begin{align}\int_0^{\infty}\frac{\ln(1+x)}{x^{1+a}}\,dx&=-\frac{1}{a}\int_0^{\infty}\ln(1+x)\cdot d(x^{-a})\\&=-\frac{1}{a}\left[\frac{\ln(1+x)}{x^a}\right]_{0^+}^{+\infty}+\frac{1}{a}\int_0^{\infty} \frac{dx}{x^a(1+x)}.\end{align}$$ For the integral on the right side take a look at Cauchy Theorem application. What do you obtain for the first term? What is the final result?

Robert Z
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