# Help on this integral $I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$

$$I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$$

Here is my attempt

$$\frac{x}{1-x^2}dx=-\frac{1}{2}d\ln(1-x^2)$$, integration by part, we got $$I=-\frac{1}{2}P-Q$$

Where $$P=\int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)}{1+x^2}dx$$, and $$Q=\int_0^1 \frac{x\cdot\arctan(x)\cdot\ln(1-x^2)}{1+x^2}dx$$

How to proceed then? or should I switch to other ways?

• @KStarGamer yes, it is divergent at x=1
– user1026811
Commented Feb 17, 2022 at 19:02
• @KStarGamer Ah, yes, they are convergent. I didn't see that $\ln\left(\frac{1+x^2}{2}\right)$ factor, thank you.
– user1026811
Commented Feb 17, 2022 at 19:28
• @Quanto I did the integration by part, as shown in my post, I couldn't proceed. Can you give some hint?
– user1026811
Commented Feb 17, 2022 at 19:30
• @Quanto I’d be interested to see your derivation Commented Feb 17, 2022 at 19:33
• @FShrike - The integrations of P and Q are still quite involved. I’ll type up when I get a chance Commented Feb 17, 2022 at 19:55

A (revolutionary) solution by Cornel Ioan Valean

It is one of those integrals that are very resistant to the real methods and make you wonder if you ever can do anything to get a decent (real) solution! Well, with some amount of creativity it is possible to get a very elegant solution. The integral is also included in the sequel of (Almost) Impossible Integrals, Sums, and Series (an upcoming title).

Let's observe that by symmetry we have $$I=\int _0^1\int _0^1\int _0^1\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx$$ $$=\frac{1}{3}\int _0^1\int _0^1\int _0^1\biggr(\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}+\frac{y^2}{(1+y^2)(1+x^2)(1+x^2 y^2 z^2)}+\frac{z^2}{(1+z^2)(1+y^2)(1+x^2 y^2 z^2)}\biggr)dzdydx$$ $$\overset{\color{red}{\text{A key observation}}}{=}\frac{1}{3}\int _0^1\int _0^1\int _0^1\frac{\color{red}{(1+x^2)(1+y^2)(1+z^2)-(1+x^2y^2z^2)}}{(1+x^2)(1+y^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx$$ $$\small =\frac{1}{3}\biggr(\underbrace{\int _0^1\int _0^1\int _0^1\frac{dzdydx}{1+x^2 y^2 z^2}}_{\displaystyle \beta(3)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{(2n-1)^3}=\pi^3/32}-\underbrace{\int _0^1\int _0^1\int _0^1\frac{dzdydx}{(1+x^2)(1+y^2)(1+z^2)}\biggr)}_{\displaystyle \pi^3/64}=\frac{\pi^3}{192}.\tag1$$

On the other hand, we get $$\small I=\int _0^1\int _0^1\int _0^1\frac{x^2}{(1+x^2)(1+z^2)(1+x^2 y^2 z^2)}dzdydx=\int _0^1\int _0^1 \frac{x^2(\pi/4-x y\arctan(x y))}{(1+x^2)(1-x^2 y^2)}dydx$$ $$\overset{xy=t}{=}\int _0^1\int _0^x \frac{x(\pi/4-t\arctan(t))}{(1+x^2)(1-t^2)}dtdx=\int _0^1\int _t^1 \frac{x(\pi/4-t\arctan(t))}{(1+x^2)(1-t^2)}dxdt$$ $$=-\frac{\pi}{8}\underbrace{\int_0^1 \frac{\log((1+t^2)/2)}{1-t^2}\textrm{d}t}_{\displaystyle -\pi^2/16}-\frac{1}{2}\color{blue}{\int_0^1 \frac{t\arctan(t) \log(2/(1+t^2))}{1-t^2}\textrm{d}t},\tag2$$ where the first integral after the last equal sign becomes trivial with the variable change $$\displaystyle t\mapsto \frac{1-t}{1+t}$$.

Combining $$(1)$$ and $$(2)$$, we arrive at the desired result, $$\color{blue}{\int_0^1 \frac{t\arctan(t) \log(2/(1+t^2))}{1-t^2}\textrm{d}t=\frac{\pi^3}{192}}.$$

End of story

More details, more results (and connections to other tough integrals) involving such symmetry ideas in three dimensions will be found in the sequel of (Almost) Impossible Integrals, Sums, and Series.

Here is a problem proposed by Cornel years ago in RMM that works great using such ideas as above $$\int_0^{\pi/4}\int_0^{\pi/4}\frac{(\tan^2(x)+\tan^2(y))\log(\tan(x))\log(\tan(y))\operatorname{Ti}_2(\tan(x) \tan(y))}{\tan(x)\tan(y)}\textrm{d}x\textrm{d}y$$ $$=\frac{1}{368640}\psi^5\left(\frac{1}{4}\right)-\frac{21}{32}\zeta(6)-\frac{2}{3}G^3,$$ where $$\displaystyle \operatorname{Ti}_2(x)$$ is the inverse tangent integral, $$G$$ represents the Catalan's constant, and $$\psi^{(n)}$$ denotes the Polygamma function.

• Awesome solution! How can this be constructed?! I mean how to start from that blue-color integral to find such a symmetry construction, too amazing! or maybe they begin with that 3D integral and finally find this byproduct integral in blue color?
– user1026811
Commented Feb 18, 2022 at 0:28
• @GGplay In this particular case, the extraction started from the investigation of the special multiple integral. However, depending on the context, it may be way more challenging, but at the same time very exciting, to explore ways of transforming a single integral to a useful multiple integral that to further calculated by means of symmetry or other creative ways. Doing this is ART, and very possible with much work, practice! Commented Feb 18, 2022 at 0:58
• This solution made the question look so trivial. Congrats on such beauty!! (+1) Commented Feb 18, 2022 at 17:18
• @AliShadhar Thank you so much, Ali! :-) Commented Feb 18, 2022 at 17:54
• what is "revolutinary" about this method? evaluating double integrals in two different ways is clever but by no means groundbreaking. Here a nice example where this technique is employed: math.stackexchange.com/questions/2454168/… Commented Feb 18, 2022 at 20:33

Here is a more general solution. Use the contour setup by Sangchul Lee, it can be shown that

$$\int_0^1 \frac{x \arctan a x}{1-x^2}\ln\frac{1+a^2}{1+a^2x^2}dx=\frac13 \arctan^3 a$$ Let $$a=1$$ to obtain $$\int_0^1 \frac{x \arctan x}{1-x^2}\ln\frac{2}{1+x^2}dx=\frac13 \arctan^3 (1)=\frac{\pi^3}{192}$$

• thank you! Is there a way to solve it without using contour integrals?
– user1026811
Commented Feb 17, 2022 at 20:36
• @GGplay - You can integrate P and Q using elementary techniques, which are quite lengthy Commented Feb 17, 2022 at 20:48
• @GGplay - see, for example, math.stackexchange.com/q/542741/686284 Commented Feb 17, 2022 at 20:50
• @Quanto I like the beauty of contour integrations but frankly my skill in practical integration is not very high... how elementary are you talking? Can P and Q be integrated -by-parts and substituted (at length) and that suffices? Or would one need to introduce some series manipulations or regularisation techniques etc. Commented Feb 17, 2022 at 20:50
• @FShrike - Q is linked above and P can be split into $$2P=\int_0^1 \frac{\ln^2(1-x^2)}{1+x^2}dx+ \int_0^1 \frac{\ln^2\frac{1+x^2}{2}}{1+x^2}dx+ -\int_0^1 \frac{\ln^2\frac{2(1-x^2)}{1+x^2}}{1+x^2}dx$$ Commented Feb 17, 2022 at 21:04