$$ \int_0^\infty \frac{\ln(1 + x^a)x^s}{1+x^2} \ dx $$

Actually the problem was $ \displaystyle \int_0^\infty \frac{\ln(1 + x^a)}{(1+x^2)\ln(x)} \ dx $.

But I guess the form of a Mellin Transform would be much better. Also, the above form would allow us to use Complex Analysis in a much easier way.

I don't know how to do this problem. I have tried using Ramanujan Master Theorem but I got stuck in finding a definite series. I have guessed that there is an easier way to approach to this problem using Contour integration but I just don't know how. Please help!

  • 2
    $\begingroup$ Have you tried the substitution $x \mapsto 1/x$ to your original integral? This will do a magic... $\endgroup$ – Sangchul Lee Aug 22 '15 at 14:05
  • $\begingroup$ Ah! My bad! Overthought it! Thanks! BTW, what about my "condensed" form? $\endgroup$ – Kartik Sharma Aug 22 '15 at 14:19
  • $\begingroup$ Now that is much more nontrivial... I am working with this now, but I have no idea. $\endgroup$ – Sangchul Lee Aug 22 '15 at 14:25

At least Maple agrees that $$ \int_0^\infty \frac{\ln(1 + x^a)x^s}{1+x^2} \, dx $$ is non-trivial. Even with $s=0$. Maple does $$ \int_0^\infty \frac{\ln(1 + x^2)}{1+x^2} \ dx\quad\text{and}\quad \int_0^\infty \frac{\ln(1 + x^4)}{1+x^2} \, dx $$ OK, but $$ \int_0^\infty \frac{\ln(1 + x^3)}{1+x^2} \, dx $$ in terms of dilogarithms. And the problem $$ \int_0^\infty \frac{\ln(1 + x^5)}{1+x^2} \, dx $$ is still running...

I stopped it. Actually $$ \int_0^\infty \frac{\ln(x-w)}{1+x^2} \, dx $$ is evaluated in terms of dilogs, and and $$ \int_0^\infty \frac{\ln(1 + x^5)}{1+x^2} \, dx $$ is a sum of 5 of those. For the full problem, $$ \int_0^\infty \frac{\ln(x-w)x^s}{1+x^2} \, dx, $$ Maple does it it terms of the Lerch Phi function.

$$ \int _{0}^{\infty }\!{\frac {\ln \left( x-w \right) {x}^{s}}{{x}^{2}+ 1}}{dx}= \\ \pi \, \left[ 2\,\ln \left( -w \right) w\sin \left( 1/2 \,\pi \,s \right) +2\,w\arctan \left( {w}^{-1} \right) \cos \left( 1/2 \,\pi \,s \right) +\sin \left( 1/2\,\pi \,s \right) w\ln \left( {w}^{ 2}+1 \right) +\sin \left( 1/2\,\pi \,s \right) w\ln \left( {w}^{-2} \right) - \left( -{w}^{-1} \right) ^{-s}\Phi \left( -{w}^{- 2},1,1/2-1/2\,s \right) \right]\\ \left( \cos \left( 1/2\,\pi \,s \right) \right) ^{-1} \left( \sin \left( 1/2\,\pi \,s \right) \right) ^{-1}{w}^{-1} 4^{-1} $$

  • $\begingroup$ I leave non-integer values of $a$ to someone else... $\endgroup$ – GEdgar Aug 22 '15 at 15:21

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