25
$\begingroup$

How to evaluate $$I=\int_0^1\ln(1+x^2)\ln(x^2+x^3)\frac{dx}{1+x^2}?$$

It equals $\frac5{64}\pi^3-\frac92G\ln2+\frac14\pi\ln^22$ according to Mathematica, where $G$ denotes Catalan's constant.
Attempt
$$I=\frac d{ds}\int_0^1\ln(x^2+x^3)\frac{dx}{(1+x^2)^{1-s}}$$ or, $$I=\int_0^{\pi/4}2\ln\sec t\ln(\tan^2t(1+\tan t))dt$$ $$=2\int_0^{\pi/4}\left(\ln2+\sum_{n=1}^\infty\frac{(-1)^n\cos(2nx)}n\right)\left(-2\sum_{n=1}^\infty\frac{\cos(4n-2)x}{2n-1}+\ln(1+\tan x)\right)dx$$ $$=-4G\ln2+\frac14\pi\ln^22+2\sum_{n=1}^\infty\frac{(-1)^n}n\int_0^{\pi/4}\cos(2nx)\ln(\tan^2 x+\tan^3x)dx$$

$\endgroup$
  • $\begingroup$ The structure of your given answer, beside the part with Catalan's constant, reminds me of the values of the derivatives of the Betafunction. Maybe it is somehow possible to deduce $I$ back to the Betafunction with a suitable subsititution or by using IBP since we are dealing with a logarithmic integral. $\endgroup$ – mrtaurho Oct 13 '18 at 14:02
  • 3
    $\begingroup$ Please add an actual attempt. You may exploit the Fourier series of $\log\sin$ and $\log\cos$ to convert your integral into a combination of Euler sums with low weight. $\endgroup$ – Jack D'Aurizio Oct 13 '18 at 21:16
  • 2
    $\begingroup$ $$\ln(x^2+x^3)=2 \ln x+\ln(1+x)$$ $\endgroup$ – Yuriy S Oct 13 '18 at 23:52
  • $\begingroup$ Added some detail. Hope that it's useful. $\endgroup$ – Kemono Chen Oct 14 '18 at 4:23
  • $\begingroup$ Try a substitution $x=\frac {1-t}{1+t}$ $\endgroup$ – Frank W. Oct 14 '18 at 7:12
13
+50
$\begingroup$

Let $a=\ln x, b=\ln(1-x), c=\ln(1+x), d=\ln(1+x^2)$. I use the following notations: $$I_{aa} = \int_0^1 \frac{\ln^2 x}{1+x^2}dx \qquad I_{ab} = \int_0^1 \frac{\ln x \ln(1-x)}{1+x^2}dx \qquad \cdots \qquad I_{cd} = \int_0^1 \frac{\ln (1+x) \ln(1+x^2)}{1+x^2}dx$$ Hence we get $10$ integrals. My goal is to find $9$ linearly independent relations between them, so your desired value $2I_{ad}+I_{cd}$ falls out easily.


Let $x=(1-u)/(1+u)$, then $dx/(1+x^2) = du/(1+u^2)$, and we have the following transformation rules: $$\begin{aligned}a &\mapsto b-c \\ b &\mapsto \ln 2 + a - c \\ c &\mapsto \ln 2 - c \\ d &\mapsto \ln 2 + d - 2c \end{aligned}$$

For example, we apply this on $I_{aa}$,we have $$\tag{1}I_{aa} = I_{bb} - 2I_{bc} + I_{cc}$$ We can apply this transformation on each of the ten integrals, but we only yield four linearly independent relations: $$\tag{2} I_{bb}=I_{aa}-2 I_{ac}-2 G \ln 2+I_{cc}$$ $$\tag{3} I_{dd}=2 \ln (2) \left(\frac{1}{2} \pi \ln (2)-G\right)+4 I_{cc}-4 I_{cd}+I_{dd}-\frac{1}{4} \pi \ln ^2(2)$$ $$\tag{4} I_{bd}=-2 I_{ac}+I_{ad}+\ln (2) \left(\frac{1}{2} \pi \ln (2)-G\right)-G \ln (2)+2 I_{cc}-I_{cd}-\frac{1}{8} \pi \ln ^2(2)$$

Of course, we have explicit evaluation of $I_{aa}$, which can be our fifth linearly independent relation: $$\tag{5} I_{aa} = \frac{\pi^3}{16}$$


To find more relations, we must rely on other methods. Here I use contour integration. Let $\log_1$ denote logarithm with branch cut at negative $x$-axis, while $\log_2$ denote logarithm with cut at positive $x$-axis. Integrate the function $$\frac{(\log_1 z)^a(\log_2 (z-1))^b}{1+z^2}$$ around a contour with two keyhole, wrapping around the two cuts: $(1,\infty)$ and $(-\infty,0)$. Then we obtain $$\int_1^\infty \cdots + \int_{-\infty}^0 \cdots = 2\pi i \text{(Sum of residues)}$$ The first integral's range can be brought back to $(0,1)$ via $x\mapsto 1/x$. The second integral, we first bring it back to $(0,\infty)$, then split intervals, finally apply $x\mapsto 1/x$ for the one with range $(1,\infty)$. After all these, We have $$\int_0^1 \frac{f_{a,b}(x)}{1+x^2} dx = 2\pi i \text{(Sum of residues)}$$ where $$f_{a,b}(x) = (-\ln (x))^a \left[(\ln (1-x)-\ln (x))^b-(\ln (1-x)-\ln (x)+2 \pi i)^b\right]-\left[(-\ln (x)-\pi i)^a-(-\ln (x)+\pi i)^a\right] (\ln (x+1)-\ln (x)+\pi i)^b-\left[(\ln (x)-\pi i)^a-(\ln (x)+\pi i)^a\right] (\ln (x+1)+\pi i)^b$$

Now apply this to $a=1,b=2$: $$\int_0^1 \frac{f_{1,2}(x)}{1+x^2}dx = -\frac{17 i \pi ^4}{16}+\frac{1}{4} i \pi ^2 \ln^2(2)-\pi ^3 \ln(2)$$ Hence comparing imaginary part:$$\tag{6}-2 \pi I_{aa}+4 \pi I_{ab}-4 \pi I_{ac}+4 \pi I_{cc}-\pi ^4=\frac{1}{4} \pi ^2 \ln ^2(2)-\frac{17 \pi ^4}{16}$$ This this our sixth linearly independent relation. Apply above method again to $a=0,b=3$: $$\tag{7}-6 \pi I_{bb}-6 \pi I_{aa}+12 \pi I_{ab}+2\pi^4 =-\frac{3}{4} \pi ^2 \ln (2)$$


The final two relations come from gamma/zeta function. Note that $$\int_1^\infty \frac{\ln^2(1+x^2)}{1+x^2}dx = I_{dd}-4I_{ad}+4I_{aa}$$ Hence $$\tag{8}2I_{dd}-4I_{ad}+4I_{aa} = \int_0^\infty \frac{\ln^2(1+x^2)}{1+x^2}dx = 4\int_0^{\pi/2} \ln^2(\cos x)dx = \frac{1}{6} \left(\pi ^3+12 \pi \ln ^2 2\right)$$

The last relation is more nontrivial: $$I_{ad}+I_{ab}+I_{ac} = \int_0^1 \frac{\ln x \ln \left(1-x^4 \right)}{1+x^2}dx = \frac{\pi^3}{16}-3G\ln 2 \tag{9}$$

which uses, in a critical way, values of digamma function.


Now solve those $9$ equations, we have one free variable (this involves a new constant, see below), and that free variable cancels for $2I_{ad}+I_{cd}$, proving your claim.

The new constant comes from $$\tag{10} I_{bb} = \int_0^1 \frac{\ln^2 x}{x^2-2x+2}dx = 2 \Im\left[\text{Li}_3\left(\frac{1+i}{2}\right)\right]$$

This follows directly from the indefinite integration: $$\int \frac{\ln^2 x}{x-a} = -2 \text{Li}_3\left(\frac{x}{a}\right)+2 \ln (x) \text{Li}_2\left(\frac{x}{a}\right)+\ln^2(x) \ln\left(1-\frac{x}{a}\right)$$

To consummate this approach, we obtain simultaneous evaluation of all $10$ integrals, all are nontrivial (except $I_{aa}, I_{bb}$) when considered individually. $$\begin{aligned} \int_0^1 \frac{\ln^2(1+x)}{1+x^2} dx &= -2 G \ln (2)-4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{7 \pi ^3}{64}+\frac{3}{16} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln^2(1+x^2)}{1+x^2} dx &= -2 G \ln (2)+4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{7 \pi ^3}{96}+\frac{7}{8} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln x \ln(1-x)}{1+x^2} dx &= \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{\pi ^3}{128}-\frac{1}{32} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln x \ln(1+x)}{1+x^2} dx &= -2 G \ln (2)-3 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{11 \pi ^3}{128}+\frac{3}{32} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln x \ln(1+x^2)}{1+x^2} dx &= -G \ln (2)+2 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{\pi ^3}{64}-\frac{1}{16} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln (1-x) \ln(1+x)}{1+x^2} dx &= -G \ln (2)-\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{3 \pi ^3}{128}+\frac{3}{32} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln (1-x) \ln(1+x^2)}{1+x^2} dx &= -\frac{1}{2} G \ln (2)+4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)-\frac{5 \pi ^3}{64}+\frac{1}{8} \pi \ln ^2(2) \\ \int_0^1 \frac{\ln (1+x) \ln(1+x^2)}{1+x^2} dx &= -\frac{5}{2} G \ln (2)-4 \Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\frac{7 \pi ^3}{64}+\frac{3}{8} \pi \ln ^2(2) \end{aligned}$$

The Mathematica input is:

{aa -> \[Pi]^3/16, bb -> 2 Im[PolyLog[3, 1/2 + I/2]], cc -> (7 \[Pi]^3)/64 - 4 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] + 3/16 \[Pi] Log[2]^2, dd -> -((7 \[Pi]^3)/96) + 4 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] - 1/8 \[Pi] Log[2]^2 + 1/4 \[Pi] Log[4]^2, ab -> -(\[Pi]^3/128) + Im[PolyLog[3, 1/2 + I/2]] - 1/32 \[Pi] Log[2]^2, ac -> (11 \[Pi]^3)/128 - 3 Im[PolyLog[3, 1/2 + I/2]] - 2 Catalan Log[2] + 3/32 \[Pi] Log[2]^2, ad -> -(\[Pi]^3/64) + 2 Im[PolyLog[3, 1/2 + I/2]] - Catalan Log[2] - 1/16 \[Pi] Log[2]^2, bc -> (3 \[Pi]^3)/128 - Im[PolyLog[3, 1/2 + I/2]] - Catalan Log[2] + 3/32 \[Pi] Log[2]^2, bd -> -((5 \[Pi]^3)/64) + 4 Im[PolyLog[3, 1/2 + I/2]] - 1/2 Catalan Log[2] + 1/8 \[Pi] Log[2]^2, cd -> (7 \[Pi]^3)/64 - 4 Im[PolyLog[3, 1/2 + I/2]] - 5/2 Catalan Log[2] + 3/8 \[Pi] Log[2]^2}
$\endgroup$
  • $\begingroup$ If you assume $\int \frac{\ln^2 x}{x-a} = -2 \text{Li}_3\left(\frac{x}{a}\right)+2 \ln (x) \text{Li}_2\left(\frac{x}{a}\right)+\ln^2(x) \ln\left(1-\frac{x}{a}\right)$ for $a$ complex you need to prove only the 5th and last equality. It's not a big deal IMHO. $\endgroup$ – FDP Oct 28 '18 at 14:05
  • $\begingroup$ @FDP You can obtain $I_{aa}, I_{bb}, I_{cc}$ from this indefinite integral. But I have diffculty in seeing how to obtain the remaining $7$ integrals and the one given by OP. Could you elaborate more? $\endgroup$ – pisco Oct 28 '18 at 16:44
  • $\begingroup$ I think the first two integrals permit us to deduce the others. Let me think more carefully and if i find something i will post an answer. Anyway i work also on an "elementary" solution. $\endgroup$ – FDP Oct 28 '18 at 17:18

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.