In what ways can I evaluate $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$$ I tried a few methods but none work or simplify things and i cant think of substitutions that could turn things better, while the substitution $x=1/x$ gets $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^{\infty }\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$ i dont know how to proceed, splitting the integral at point 1 also doesnt help much.

The integral can also be expressed as $$I=\int _0^1\frac{\left(1+x^2\right)\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$ that first integral as mentioned in the comments has been evaluated and is $$I=\frac{\pi }{6}\ln \left(2+\sqrt{3}\right)-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$ But how to tackle the second one?

  • $\begingroup$ Have you tried Feynman's Trick? $\endgroup$ – Mark Viola Aug 16 '20 at 23:03
  • $\begingroup$ It's equal to $\frac{\pi}{6}\log\left(2+\sqrt{3}\right)-\frac{\pi^2}{12\sqrt{3}}+\frac{2G}{3}$ with $G$ the Catalan constant. See here for example. $\endgroup$ – Edward H Aug 16 '20 at 23:06
  • $\begingroup$ I have tried it, but it doesnt seem to help at first glance. $\endgroup$ – robert garcia Aug 16 '20 at 23:12
  • $\begingroup$ For your new question, see here. $\endgroup$ – Edward H Aug 17 '20 at 0:06
  • 1
    $\begingroup$ This is one the proposal of professor Vasile Mercia Popa and you may access here solved due to Zaharia Burghelea $\endgroup$ – Naren Aug 17 '20 at 7:25


\begin{align} I& =\int _0^{\infty }\frac{\ln (1+x)}{1-x^2+x^4}dx= \int _0^{1}\frac{\ln (1+x)}{1-x^2+x^4}dx +\int _1^{\infty }\frac{\ln (1+x)}{\underset{x\to 1/x} {1-x^2+x^4}}dx\\ &= \int _0^1\frac{(1+x^2)\ln (1+x)-x^2\ln x}{1-x^2+x^4}dx\\ \end{align}

Integrate by parts via

$$d\left( \cot^{-1}\frac x{x^2-1}\right)=\frac{1+x^2}{1-x^2+x^4}dx$$ $$d\left( \frac12\tan^{-1}\frac x{1-x^2} - \frac1{2\sqrt3}\tanh^{-1}\frac {\sqrt3x}{1+x^2}\right)= \frac{x^2}{1-x^2+x^4}dx $$ to express the integral as \begin{align} I= I_1 -\frac1{2\sqrt3}I_2 +\frac12I_3\tag1 \end{align}

where $$ I_1 = \int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {x}{1-x^2},\>\>\>\>\> I_2 = \int_0^1 \frac{dx}{x} \tanh^{-1}\frac {\sqrt3x}{1+x^2}\\ I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2} $$

To evaluate $I_1$, let $J_1(a) =\int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}$ \begin{align} J_1’(a) &= \int_0^1 \frac{2\cos a (x-x^2)dx}{(x^2+1)^2-(2x\cos a)^2}= - \frac\pi4\tan\frac a2+\frac12\left( a\>{\csc a}+ \ln\tan\frac a2\right) \end{align} \begin{align} I_1 &=J_1(\frac\pi6) = J_1(0)+\int_0^{\frac\pi6}J_1’(a)da \\ &= \frac\pi2\ln2-\frac\pi4 \int_0^{\frac\pi6}\tan\frac a2 da+\frac12\int_0^{\frac\pi6} d\left( a\ln\tan\frac a2\right)\\ &=\frac\pi2\ln2 -\frac\pi2\ln\cos\frac\pi{12}-\frac\pi{12}\ln\tan\frac\pi{12}= \frac\pi6\ln(2+\sqrt3)\tag2 \end{align}

To evaluate $I_2$, let $J_2(a) =\int_0^1 \frac{dx}{x} \tanh^{-1}\frac {2ax}{1+x^2}$ \begin{align} J_2’(a) &= \int_0^1 \frac{2 (1+x^2)}{(x^2+1)^2-(2ax)^2}dx = \frac\pi2\frac1{\sqrt{1-a^2}} \end{align} \begin{align} I_2 &=J_2(\frac{\sqrt3}2) = \int_0^{\frac{\sqrt3}2}J_2’(a)da = \frac\pi2 \int_0^{\frac{\sqrt3}2} \frac{da}{\sqrt{1-a^2}}=\frac{\pi^2}6\tag3 \end{align} To evaluate $I_3$ $$I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2} = \int_0^1 \frac{\tan^{-1}x}x dx+ \int_0^1\underset{x^3\to x}{\frac{\tan^{-1}x^3}x}dx\\ = \frac43\int_0^1 \frac{\tan^{-1}x}x dx= \frac43G\tag4 $$

Plug (2), (3) and (4) into (1) to obtain

$$I=\frac\pi6\ln(2+\sqrt3) -\frac{\pi^2}{12\sqrt3} +\frac23G $$


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