Hello there I am trying to calculate $$ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx $$ NOT using mathematica, matlab, etc. We are given that $\sigma, \omega$ are complex. Note, the integral should have different values for $|\sigma \omega^{-1/2}| < 1$ and $|\sigma \omega^{-1/2}| > 1.$ I am stuck now and not sure how to approach it. Note this integral is useful since in the limit $\sigma \to \sqrt{\omega}$ and using $Li_2(-1)=-\pi^2/12$ we obtain $$ \int_0^\infty \frac{\ln(1+x)\ln(1+x^2)}{x^3}dx=\frac{\pi}{2}. $$ We also know that $$ \ln(1+x)=-\sum_{n=1}^\infty \frac{(-1)^nx^n}{n}, \ |x|\leq 1. $$ Thanks

  • $\begingroup$ Mathematica yields the condition: $\large\sigma, \omega \in {\mathbb R}$ and $\large\sigma>0$, $\large\omega > 0$. $\endgroup$ Commented Apr 1, 2014 at 2:51
  • $\begingroup$ Well I'm not quite sure, I do trust you, however the mathematics journal this came from says different. I can send it to you if you would like. Thank you @FelixMarin $\endgroup$ Commented Apr 1, 2014 at 2:57
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    $\begingroup$ I know I am late but did you try differentiating the following integral: $$I(\sigma,\omega)=\int_0^{\infty} \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}\,dx$$ first wrt $\sigma$ and then wrt $\omega$ because that blows away the $x^3$ in denominator. I tried that and got: $$\frac{d}{d\omega}\left(\frac{dI}{d\sigma}\right)=\int_0^{\infty} \frac{dx}{(1+\sigma x)(1+\omega x^2)}$$ The above can be easily evaluated by the substitution $\sqrt{\omega }x=\tan t$. Now I am not sure about the next step. Is it ok to integrate twice (i.e first wrt $\omega$ and then wrt $\sigma$)? $\endgroup$ Commented May 17, 2014 at 11:02
  • $\begingroup$ @PranavArora Please post your solution based on this method. It may be correct. It seems very clever to me, this integral has a closed form so you are probably walking down the correct path my friend:) $\endgroup$ Commented May 22, 2014 at 5:27

3 Answers 3


One may adopt the approach as in Pranav Arora's comment. But this approach involves a double integral whose calculation seems painful. So here is an indirect approach that makes calculation slightly easier (at least to me):

Let us consider the following integral: for $\alpha, \beta \in \Bbb{C}\setminus(-\infty, 0]$ and $0 < s < 1$,

$$ I = I(s,\alpha,\beta) := \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta x)}{x^{2+s}} \, dx, $$

Differentiating w.r.t. $\alpha$ and $\beta$, we have

$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \int_{0}^{\infty} \frac{dx}{x^{s}(1+\alpha x)(1+\beta x)}. $$

Using standard complex analysis technique (you man use keyhole contour), it follows that

$$ \frac{\partial^{2}I}{\partial\alpha\partial\beta} = \frac{\pi}{\sin \pi s} \frac{\beta^{s} - \alpha^{s}}{\beta - \alpha} \quad \Longrightarrow \quad I = \frac{\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{x^{s} - y^{s}}{x - y} \, dxdy. \tag{1} $$

Replace $\beta$ by $i\beta$ (with $\beta > 0$). Then (1) yields

$$ 2I(s, \alpha, i\beta) = \frac{2\pi}{\sin \pi s} \int_{0}^{\alpha}\int_{0}^{\beta} \frac{i^{s}x^{s} - y^{s}}{x + iy} \, dxdy. $$

Now assume that $\alpha, \beta > 0$. Taking real parts of the identity above and taking $s \to 1^{-}$, it follows that

\begin{align*} \tilde{I}(\alpha, \beta) &:= \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta^{2}x^{2})}{x^{3}} \, dx \\ &= \int_{0}^{\alpha}\int_{0}^{\beta} \frac{2xy \log(y/x) + \pi x^{2}}{x^{2}+y^{2}} \, dxdy. \tag{2} \end{align*}

In particular, when $\beta = \alpha$, by symmetry we retrieve the following formula

$$ \tilde{I}(\alpha, \alpha) = \pi \int_{0}^{\alpha}\int_{0}^{\alpha} \frac{x^{2}}{x^{2}+y^{2}} \, dxdy = \frac{\pi}{2} \int_{0}^{\alpha}\int_{0}^{\alpha} dxdy = \frac{\pi}{2}\alpha^{2}. $$

which also follows from the formula in OP's posting. In general, using polar coordinates shows that we have

$$ \tilde{I}(\alpha, \beta) = \beta^{2}J(\alpha/\beta) - \alpha^{2}J(\beta/\alpha) + \frac{\pi \alpha \beta}{2} + \frac{\pi^{2}\beta^{2}}{4} - \frac{\pi(\alpha^{2}+\beta^{2})}{2}\arctan(\beta/\alpha), \tag{3} $$

where $J$ is defined by

$$ J(x) = \int_{0}^{x} \frac{t \log t}{1+t^{2}} \, dt. $$

This function can be written in terms of elementary functions and dilogarithm.

Remark. Though we have derived this formula for positive $\alpha, \beta$, by the principle of analytic continuation (3) continues to hold on the region containing $(0, \infty)^{2}$ where both sides of (3) are holomorphic.


I really do not know how much this could help you but, using a CAS, I obtained for the integral the following result $$\frac{1}{24} \left(6 \left(\sigma ^2+\omega \right) \text{Li}_2\left(-\frac{\sigma ^2}{\omega }\right)+6 \left(\sigma ^2+\omega \right) (2 \log (\sigma )-\log (\omega )) \log \left(\frac{\sigma ^2+\omega }{\omega }\right)+12 \pi \left(\sigma ^2+\omega \right) \tan ^{-1}\left(\frac{\sigma }{\sqrt{\omega }}\right)+\sigma \left(\pi \left(12 \sqrt{\omega }-5 \pi \sigma \right)-3 \sigma (\log (\omega )-2 \log (\sigma ))^2\right)\right)$$ I suppose that there are restrictions but I have not been able to find the ensemble of them.

  • $\begingroup$ I have solutions as well to the integral for the two cases I provided in the question. It is indeed a closed form result, I am looking for a proof however. Thanks though. These integrals are from old math journals and are typical for bulgarian math contests $\endgroup$ Commented Mar 16, 2014 at 14:28
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    $\begingroup$ I should really just be unable to provide any proof of anything in this problem ! Cheers. $\endgroup$ Commented Mar 16, 2014 at 14:40
  • $\begingroup$ Well thanks for the interest. As of now I am unable to provide a proof as well! $\endgroup$ Commented Mar 16, 2014 at 22:01

why not use the $\Gamma$ function ? it seems that this question is the special case!

$\int_0^\infty \frac{\ln(1+ x)\ln(1+x^2)}{x^3}dx$$=\frac{1}{2}(\int_0^\infty \frac{e^{-x}}{\sqrt{x}}dx)^{2}$

$ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+\omega x^2)}{x^3}dx $$=$ $\int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{x^3}dx $

$\Longrightarrow$$ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{(\sigma{x})^3}dx$$=$$\frac{\sigma^{2}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$

$\Longrightarrow$ $ \int_0^\infty \frac{\ln(1+\sigma x)\ln(1+(\sigma x)^2)}{{x}^3}dx$$=$$\frac{\sigma^{5}}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{\sigma x}}dx)^{2}$

$=\frac{1}{2}(\int_0^\infty \frac{e^{-\sigma x}}{\sqrt{x}}dx)^{2}$$\cdot{\sigma^{4}}$

therefore, it seems that you just need to multiply $\sigma^{3}$

and the positive and negative of your equation depends on :

$|\sigma \omega^{-1/2}| < 1$ or $|\sigma \omega^{-1/2}| > 1$


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