I'm having some difficulties to execute this integral: $$\int_{1}^x \frac{t+\log(t)}{1+t^2} \,dt = \int_{1}^x \frac{t}{1+t^2} \,dt + \int_{1}^x \frac{\log(t)}{1+t^2} \,dt = ... = \frac{1}{2}\log|1+x^2| - \frac{1}{2}\log(2) + ... $$

Could you please give me some hints to solve the following: $$ \int_{1}^x \frac{\log(t)}{1+t^2} \,dt $$

I get something similar to the Catalan constant but I'm not sure I'm doing it in the correct way: http://en.wikipedia.org/wiki/Catalan%27s_constant

  • $\begingroup$ It's not clear to me what you are asking. I don't think you can expect an analytic solution to the integral, since you get Catalan's constant $G$ when $x=0$, and that would imply the existence of a (fairly) simple expression for $G$. But perhaps you want something else, like a nice series expansion for the integral? $\endgroup$ – Harald Hanche-Olsen Jun 1 '14 at 13:53
  • $\begingroup$ Thanks Harald, tbh I need to calculate $$ \lim_{x \to inf} \frac{\int_{1}^x \frac{t+\log(t)}{1+t^2} \,dt}{x}$$ and I was hoping to get an analytic solution or something simpler respect to what I've got. $\endgroup$ – user138859 Jun 2 '14 at 16:23
  • $\begingroup$ Ah, but that integrand is asymptotically equal to $1/t$, so the integral in the numerator should behave like $\ln t$, with the result that the limit is zero. Or to put it differently, just treat the limit directly with L'Hôpital's rule. Poof, no integral! And the limit is zero. $\endgroup$ – Harald Hanche-Olsen Jun 2 '14 at 20:01
  • $\begingroup$ Ah, that sounds a lot easier! Many thanks again Harold. $\endgroup$ – user138859 Jun 4 '14 at 6:26

It is probably impossible to obtain a closed form without dilogarithm function (or other special functions such as Lerch's function for example)

enter image description here

  • 1
    $\begingroup$ +1. Anyway, why didn't type your answer using LaTeX? All your posts always use image. $\endgroup$ – Tunk-Fey Jun 1 '14 at 21:50
  • $\begingroup$ Many thanks JJacquelin, this seems correct. Tbh I was expecting something simpler but it is what it is. $\endgroup$ – user138859 Jun 2 '14 at 18:06

There is an analytical solution for the integral but its expression involves polylogarithms $$\int_{1}^x \frac{\log(t)}{1+t^2} \,dt=C-\frac{1}{2} i \text{Li}_2(-i x)+\frac{1}{2} i \text{Li}_2(i x)+\log (x) \tan ^{-1}(x)$$ As told by Harald Hanche-Olsen, you can have a nice series expansion from the Taylor expansion of $$\frac{1}{1+t^2}=1-t^2+t^4-t^6+...$$ and then the problem reduces to $$I_n=\int_{1}^x t^n \log(t) dt=\frac{x^{n+1} ((n+1) \log (x)-1)+1}{(n+1)^2}$$

  • $\begingroup$ Many thanks Claude. $\endgroup$ – user138859 Jun 2 '14 at 18:07
  • $\begingroup$ You are very welcome ! $\endgroup$ – Claude Leibovici Jun 2 '14 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.