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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.

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  • $\begingroup$ Try a numerical integration $\endgroup$
    – hwood87
    Commented Jul 19, 2020 at 3:21
  • $\begingroup$ 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. $\endgroup$ Commented Jul 19, 2020 at 4:34
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    $\begingroup$ 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. $\endgroup$ Commented Jul 19, 2020 at 4:40
  • $\begingroup$ @Dennis Orton very nice. I missed Feynman method. I think solution is complete now. $\endgroup$ Commented Jul 19, 2020 at 4:44
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    $\begingroup$ Does this answer your question? Evaluate $\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx$ $\endgroup$
    – Zacky
    Commented Jul 19, 2020 at 6:12

3 Answers 3

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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.

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  • $\begingroup$ Very nice, thanks guys $\endgroup$
    – user809806
    Commented Jul 29, 2020 at 23:27
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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.

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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

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

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