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

Question

Prove that $$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi \ln 2}{4}-\frac {G}{3}+\frac {2\pi}{3}\ln(2+\sqrt 3)$$ Where $G$ is the Catalan's constant.

Actually I proved this using the Feynman's trick namely by introducing the parameter $a$ such that $$\xi(a)=\int_0^{\infty} \frac {\ln(1+ax^3)}{1+x^2}dx$$

Where it is clear that $\xi(0)=0$, hence we just need $$\int_0^1 \xi'(a)da$$ which I found too. Hence proving the statement, but this method was too much lengthy because it involved heavy partial fraction decomposition and one infinite summation.

Can someone suggest some better method?

Edit: I also tried some trigonometry bashing by using the substitution $x=\tan \theta$ but got stuck midway

Answer 1

Remark. I have encountered a similar integral a few months ago and proposed it here$(I_8)$. $$\sf I_8=\int_0^1 \frac{\ln(1+x^3)}{1+x^2}dx\overset{x=\frac{1}{x}}=\int_1^\infty\frac{\ln(1+x^3)-3\ln x}{1+x^2}dx$$ $$\sf \Rightarrow 2I_8=\int_0^\infty \frac{\ln(1+x^3)}{1+x^2}dx-3\int_1^\infty \frac{\ln x}{1+x^2}dx\Rightarrow \boxed{I_8=\frac12I-\frac32G}$$

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We can solve it without using partial fraction or series in atleast two ways that I can think of. The second solution might be easier, but I like the first one more.

Solution 1. Start by letting $\sf x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt$ $$\sf I=\int_0^\infty \frac{\ln(1+x^3)}{1+x^2}dx=\int_{-1}^1 \frac{\ln\left(\frac{2(1+3t^2)}{(1+t)^3}\right)}{t^2+1}dt$$ $$\sf =2\ln 2\int_0^1\frac{1}{1+t^2}dt-3\int_{-1}^1\frac{\ln(1+t)}{1+t^2}dt+2\underbrace{\int_0^1 \frac{\ln(1+3t^2)}{1+t^2}dt}_{J}$$ $$\sf =\frac{\pi}{2}\ln 2 -\frac{3\pi}{4}\ln 2 +3G +2J$$ Where $G$ is Catalan's constant and that is quite easy to show. Now for Feynman's trick consider: $$\sf J(a)=\int_0^1 \frac{\ln((1+x^2)a+2x^2)}{1+x^2}dx\Rightarrow J'(a)=\int_0^1 \frac{1}{(1+x)^2a+2x^2}dx$$

$$\sf =\frac{1}{a+2}\int_0^1 \frac{1}{x^2+\frac{a}{a+2}}dx=\frac{1}{\sqrt{a}\sqrt{a+2}}\arctan\left(\frac{\sqrt {a+2}}{\sqrt a}\right)$$

We have $$\sf J(0)=\int_0^1 \frac{\ln 2 +\ln(x^2)}{1+x^2}dx=\frac{\pi}{4}\ln 2-2G$$ $$\sf \Rightarrow J=J(1)-J(0)+J(0)=\underbrace{\int_0^1 J'(a)da}_{X} +J(0)$$ $$\sf \text{let } \sqrt{\frac{a+2}{a}}=t\Rightarrow \frac{1}{\sqrt{a}\sqrt{a+2}}da=-\frac{2}{x^2-1}dx$$ $$\sf X=\int_0^1 \frac{1}{\sqrt{a}\sqrt{a+2}}\arctan\left(\frac{\sqrt {a+2}}{\sqrt a}\right)da=2\int_\sqrt 3^\infty \frac{\arctan x}{x^2-1}dx$$ $$\sf \overset{IBP}=\frac{\pi}{3}\ln(2+\sqrt 3)-\int_{\sqrt 3}^\infty \frac{\ln\left(\frac{x-1}{x+1}\right)}{1+x^2}dx$$ With $\sf \frac{x-1}{x+1}= t\Rightarrow x=\frac{1+t}{1-t}$ we get: $$\sf X=\frac{\pi}{3}\ln(2+\sqrt 3)-\int_{2-\sqrt 3}^1 \frac{\ln t}{1+t^2}dt$$$$\sf \overset{t=\tan x}=\frac{\pi}{3}\ln(2+\sqrt 3)-\int_0^{\frac{\pi}{4}}\ln(\tan x)dx+\int_0^\frac{\pi}{12} \ln(\tan x)dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac13G$$ See for example here in order to obtain the result from above. And finally we have:

$$\sf \boxed{J=\int_0^1\frac{\ln(1+3x^2)}{1+x^2}dx=\frac{\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2-\frac53G}$$ $$\sf \boxed{I=\int_0^\infty \frac{\ln(1+x^3)}{1+x^2}dx=\frac{2\pi}{3}\ln(2+\sqrt 3)+\frac{\pi}{4}\ln 2 -\frac13G}$$

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Solution 2. This should be like a follow up from your edit attempt. Rewrite the integral as $$\sf I=\int_0^\infty \frac{\ln(1-x+x^2)}{1+x^2}dx+\int_0^\infty \frac{\ln(1+x)}{1+x^2}dx$$ The second one is pretty easy $\sf I_2=\frac{\pi}{4}\ln 2+G$, and for the first integral let $\sf x=\tan t$. $$\sf I_1=\int_0^\frac{\pi}{2} \ln(1-\sin t\cos t )dt-2\int_0^\frac{\pi}{2} \ln(\cos t)dt=\frac{\pi}{2} \ln 2 +\int_0^\frac{\pi}{2}\ln(2-\sin t)dt$$ And now combining with Solution 2 we get, keeping the same notation found there: $$\sf B=\frac{1}{2}\left((A+B)-(A-B)\right)=\frac12\left(\pi\ln(2+\sqrt 3)-\pi \ln 2+\frac{\pi}{3}\ln(2+\sqrt 3)-\frac83G\right)$$ And the result follows.

Answer 2

Note

$$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx = \underset{= \frac\pi4\ln2+G}{ \int_0^{\infty} \frac {\ln(1+x)}{1+x^2}dx} + \underset{=K}{\int_0^{\infty} \frac {\ln(1-x+x^2)}{1+x^2}dx} \tag1 $$ To compute $K$, let $J(a)= \int_0^{\infty} \frac {\ln\left(\frac12 (1+x^2)\sec a-x\right)}{1+x^2}dx $ $$J’(a) = \int_0^\infty \frac{\tan a}{(x-\cos a)^2 + \sin^2a}dx=(\pi-a)\sec a $$ \begin{align} K&=\>J(\frac\pi3) =J(0)+\int_0^{\frac\pi3} J’(a)da =-2G +\int_0^{\frac\pi3} (\pi -a)\sec a\>da\\ &\overset{\text{ibp}}=-2G +(\pi-a)\tanh^{-1}(\sin a)\bigg|_0^{\frac\pi3}+\int_0^{\frac\pi3} \tanh^{-1}(\sin a)\>da\\ &=-2G +\frac{2\pi}3\ln(\sqrt3+2)+\frac23G \end{align}

where $\int_0^{\frac\pi3} \tanh^{-1}(\sin a)\>da=\frac23G $. Substitute into (1) to obtain $$\int_0^{\infty} \frac {\ln(1+x^3)}{1+x^2}dx=\frac {\pi}{4}\ln2+\frac {2\pi}{3}\ln(\sqrt 3+2)- \frac {G}{3} $$

Answer 3

Noticing that

$$ \begin{aligned} & \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x =\underbrace{\int_{0}^{\infty} \frac{\ln (1+x)}{1+x^{2}}}_{\frac{\pi}{4} \ln 2+G} d x+\underbrace{\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}}}_{K} d x \end{aligned} $$ where the first integral comes from my post.

Now we going to find the integral $K$ by observing $$ \begin{aligned}K=\int_{0}^{\infty} \frac{\ln \left(1-x+x^{2}\right)}{1+x^{2}} = \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x^2+x^{4}\right)}{1+x^{2}} d x}_{\pi\ln (2+\sqrt 3)}- \underbrace{\int_{0}^{\infty} \frac{\ln \left(1+x+x^{2}\right)}{1+x^{2}} d x}_{ \frac{\pi}{3} \ln (2+\sqrt{3})+\frac{4}{3}G} \end{aligned} $$

where the first integral comes from my post and @Quanto’s post .

Now we can conclude that $$\boxed{ \int_{0}^{\infty} \frac{\ln \left(1+x^{3}\right)}{1+x^{2}} d x=-\frac{G}{3}+\frac{\pi}{4} \ln 2 +\frac{2 \pi}{3}(2+\sqrt{3})} $$