# Generalization to this integral

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

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!

• Have you tried the substitution $x \mapsto 1/x$ to your original integral? This will do a magic... – Sangchul Lee Aug 22 '15 at 14:05
• Ah! My bad! Overthought it! Thanks! BTW, what about my "condensed" form? – Kartik Sharma Aug 22 '15 at 14:19
• Now that is much more nontrivial... I am working with this now, but I have no idea. – 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.
$$\int _{0}^{\infty }\!{\frac {\ln \left( x-w \right) {x}^{s}}{{x}^{2}+ 1}}{dx}= \\ \pi \, \left[ 2\,\ln \left( -w \right) w\sin \left( 1/2 \,\pi \,s \right) +2\,w\arctan \left( {w}^{-1} \right) \cos \left( 1/2 \,\pi \,s \right) +\sin \left( 1/2\,\pi \,s \right) w\ln \left( {w}^{ 2}+1 \right) +\sin \left( 1/2\,\pi \,s \right) w\ln \left( {w}^{-2} \right) - \left( -{w}^{-1} \right) ^{-s}\Phi \left( -{w}^{- 2},1,1/2-1/2\,s \right) \right]\\ \left( \cos \left( 1/2\,\pi \,s \right) \right) ^{-1} \left( \sin \left( 1/2\,\pi \,s \right) \right) ^{-1}{w}^{-1} 4^{-1}$$
• I leave non-integer values of $a$ to someone else... – GEdgar Aug 22 '15 at 15:21