How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis

Question

This particular integral evaluates to, $$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\frac{\pi }{8}\ln \left(2\right)-\frac{3\pi }{8}+\frac{\pi }{3}\ln \left(2+\sqrt{3}\right)-\frac{G}{6}$$ And its been proven here. But i'd like to know how to evaluate this without complex analysis.

One of the answers uses differentiation under the integral sign directly and partial fraction decomposition on a similar integral, but doing it that way doesnt help me with this case here I tried to evaluate this way but got stuck, $$\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx=\int _0^1\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx+\int _1^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx\:\:\:\:\:\: \text{then sub}\:\:x=\frac{1}{t}\:\:\text{for the 2nd integral}$$ $$=\int _0^1\frac{\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt+\int _0^1\frac{t^2\ln \left(t^3+1\right)}{\left(t^2+1\right)^2}\:dt-3\int _0^1\frac{t^2\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$ $$=\int _0^1\frac{\ln \left(t^3+1\right)}{t^2+1}\:dt+3G+3\int _0^1\frac{\ln \left(t\right)}{\left(t^2+1\right)^2}\:dt$$ I managed to evaluate the last integral expanding the denominator but i cant think of a way to evaluate the 1st integral, please help me.

Using $1+t^3=(1+t)(1-t+t^2)$, your integral reduces to $\int_0^1\frac{\ln(1-t+t^2)}{1+t^2}dt$ and by subbing $t=\frac{1-x}{1+x}$, the latter integral reduces to calculating $\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx$ which seems tough too. — Ali Shadhar, Jul 19 '20 at 04:34
Following @AliShather's path you can use $$\int _0^1\frac{\ln \left(a+bx^2\right)}{1+x^2}:dx=\frac{\pi }{2}\ln \left(\sqrt{a}+\sqrt{b}\right)+\text{Ti}_2\left(\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)-G$$ to evaluate that remaining integral. — Dennis Orton, Jul 19 '20 at 04:40
@Dennis Orton very nice. I missed Feynman method. I think solution is complete now. — Ali Shadhar, Jul 19 '20 at 04:44
@AliShather. Using Feynman trick from the beginning leads to something tedious but doable. Why don't you write an answer using it with your appraoch (I did not want to do it). Cheers :-) — Claude Leibovici, Jul 19 '20 at 06:03
Does this answer your question? Evaluate $\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx$ — Zacky, Jul 19 '20 at 06:12
@Claude Leibovici I thought the ideas in the comments are good enough but I will provide full solution soon :) — Ali Shadhar, Jul 19 '20 at 10:57

score 2 · Accepted Answer · answered Jul 19 '20 at 11:23

With subbing $t=\frac{1-x}{1+x}$ we have

$$\int_0^1\frac{\ln(1+t^3)}{1+t^2}dt=\int_0^1\frac{\ln\left(\frac{2(1+3x^2)}{(1+x)^3}\right)}{1+x^2}dx$$

$$=\ln2\int_0^1\frac{dx}{1+x^2}+\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx-3\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$

where the first integral

$$\int_0^1\frac{dx}{1+x^2}=\arctan(1)=\frac{\pi}{4},$$

and the second integral is already calculated in the comments by Dennis or it can be found calculated in details by Zacky in this solution (see the integral $J$);

$$\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2-\frac53G.$$

For the third integral, let $x\mapsto \frac{1-x}{1+x}$

$$\int_0^1\frac{\ln(1+x)}{1+x^2}dx=\int_0^1\frac{\ln\left(\frac{2}{1+x}\right)}{1+x^2}dx=\ln2\int_0^1\frac{dx}{1+x^2}-\int_0^1\frac{\ln(1+x)}{1+x^2}dx$$

$$\Longrightarrow \int_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi}{8}\ln2$$

Combine the three results we have

$$\int_0^1\frac{\ln(1+t^3)}{1+t^2}dt=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{8}\ln 2-\frac53G.$$

All credit goes to Zacky and Dennis for evaluating the second integral as its the key to crack the integral in the question.

score 0 · Answer 2 · answered Jul 19 '20 at 06:00

As said in comments, consider $$I(a)=\int _0^{\infty }\frac{\log \left(ax^3+1\right)}{\left(x^2+1\right)^2}\,dx$$ $$I'(a)=\int _0^{\infty }\frac{x^3}{\left(x^2+1\right)^2 \left(a x^3+1\right)}\,dx$$ Using partial fraction decomposition, the integrand is $$\dfrac{ a(2a^2-1)x^2+3a^2x-a(a^2+2)}{(a^2+1)^2\left(ax^3+1\right)}-\dfrac{\left(2a^2-1\right)x+3a}{(a^2+1)^2\left(x^2+1\right)}-\dfrac{x-a}{\left(a^2+1\right)\left(x^2+1\right)^2}$$ which is not too bad. This leads to $$I'(a)=\frac{-8 \sqrt{3} \pi a^{8/3}+24 \sqrt{3} \pi a^{4/3}+16 \sqrt{3} \pi a^{2/3}+9 \pi a^3-18 a^2+12 \left(2 a^2-1\right) \log (a)-45 \pi a-18}{36 \left(a^2+1\right)^2}$$ This is not the most pleasant part but it leads to the result.

Lai · Answer 3 · 2022-08-15T02:27:04.927

I am going to prove it by integration by parts and partial fractions. $$ \begin{aligned} \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{\left(1+x^{2}\right)^{2}} d x &=-\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{2 x} d\left(\frac{1}{1+x^{2}}\right)\\ &=-\left[\frac{\ln \left(1+x^{3}\right)}{2 x\left(1+x^{2}\right)}\right]_{0}^{\infty}+\frac{1}{2} \int_{0}^{\infty} \frac{1}{1+x^{2}} \cdot \frac{\frac{3 x^{3}}{1+x^{3}}-\ln \left(1+x^{3}\right)}{x^{2}} d x\\ &=\frac{3}{2} \underbrace{ \int_{0}^{\infty} \frac{x}{\left(1+x^{2}\right)\left(1+x^{3}\right)} d x}_{\frac{1}{36}(8 \sqrt{3}-9) \pi}-\frac{1}{2} \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{x^{2}\left(1+x^{2}\right)} d x\\&= \frac{\sqrt{3} \pi}{3}-\frac{3 \pi}{8} -\frac{1}{2} \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{x^{2}} d x}_{\frac{2 \sqrt{3} \pi}{3}} +\frac{1}{2} \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}}}_{\frac{\pi \ln 2}{4}-\frac{G}{3}+\frac{2 \pi}{3} \ln (2+\sqrt{3})} \end{aligned} $$ where the last integral comes from the post.

Sum them up gives the result

$$ I=-\frac{3 \pi}{8}+\frac{\pi \ln 2}{8}-\frac{G}{6}+\frac{\pi}{3} \ln (2+\sqrt{3}) $$

How to evaluate $\int _0^{\infty }\frac{\ln \left(x^3+1\right)}{\left(x^2+1\right)^2}\:dx$ without complex analysis

3 Answers3