8

We first split the integrand into 3 parts as \begin{aligned} \int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &= \underbrace{\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x}_J+\underbrace{\int_0^1 \frac{\ln (1+x)}{1+x^2} d x}_K+ \underbrace{\int_0^1 \frac{\ln (1-x)}{1+x^2} d x}_L \end{aligned}

Denotes the Catalan’s constant by $G$.

By my post, $$J=\frac{\pi}{2}\ln2-G$$

Dealing with the last $2$ integrals, we use a powerful substitution $x=\frac{1-t}{1+t} ,$ then $dx=-\frac{2dt}{(1+t)^2}.$

$$ \begin{aligned} K&=\int_1^0 \frac{\ln 2-\ln (1+t)}{\frac{2+2 t^2}{(1+t)^2}} \frac{-2 d t}{(1+t)^2} \\ &=\ln 2 \int_0^1 \frac{d t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} \end{aligned} $$

Hence $$K=\frac{\pi}{8} \ln 2 $$

$$ \begin{aligned} L=& \int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\ &=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t . \\ &=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\ &=\frac{\pi}{8} \ln 2-G \end{aligned} $$

Combining them to get $$ \begin{aligned} I &=\left(\frac{\pi}{2} \ln 2-G\right) +\frac{\pi}{8} \ln 2 +\left(\frac{\pi}{8} \ln 2-G\right)\\ &=\frac{3 \pi}{4} \ln 2-2 G \end{aligned} $$ I do want to know if it can be solved by any other elegant methods. Your comments and methods are highly appreciated.

Lai
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  • Related? – Accelerator Nov 18 '22 at 06:02
  • 1
    Thank you ! How are they related? – Lai Nov 18 '22 at 06:04
  • I was thinking my query is "related" to yours in that mine just somewhat looks like yours. I have no idea if it helps though because I was using harmonic numbers in my query. – Accelerator Nov 18 '22 at 06:06
  • It is a very nice solution. (+1) – Claude Leibovici Nov 18 '22 at 06:12
  • I was thinking contour integration with the logarithm trick, so I substituted $u=1-x^4$ to use the logarithm trick with $f(z)=\frac{\ln^{n+1}(z)}{\text{whatever}}$ but sadly the denominator does not seem very nice afterwards, so im not sure if this substitution is doable. If you manage to get a good substitution im guessing complex analysis could work but imo your splitting integral is the best method for this one – Max0815 Nov 18 '22 at 06:38
  • Thank you for your helpful opinion. – Lai Nov 18 '22 at 06:43
  • Just a thought: Maybe the Feynman trick works here. E. g. You could try to differentiate the function $\alpha \mapsto \int_0^1 \frac{\ln(1-\alpha x^4)}{1+x^2}dx$. – Bruno Krams Nov 18 '22 at 08:51

3 Answers3

7

Substitute $x=\tan t$. Then

$$1-x^4=4x^2\frac{1-x^2}{1+x^2}\left(\frac{1+x^2}{2x}\right)^2 =4\tan^2t\ \frac{\cos 2t}{\sin^2 2t} $$ and \begin{aligned} &\int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x \\ =& \int_0^{\pi/4}\ln8+2\ln \tan t +\ln (2\cos 2t) -2\ln (2\sin 2t) \ d t\\ =&\ \frac{3\pi}4\ln2 - 2G \end{aligned}

where $\int_0^{\pi/4} \ln \tan t\ dt=-G$ and $\int_0^{\pi/4}\ln (2\cos 2t)dt = \int_0^{\pi/4}\ln (2\sin 2t) d t=0$.

Quanto
  • 97,352
6

This approach is somewhat similar to yours. It is more of a comment than an answer, but it is too long to fit in the comments section.

Let the integral in question be $I$. Letting $x \to \dfrac{1-x}{1+x}$ at the beginning yields

$$I = 2\int_{0}^{1}\frac{\ln\left(1-\left(\frac{1-x}{1+x}\right)^{4}\right)}{1+\left(\frac{1-x}{1+x}\right)^{2}}\left(\frac{1}{\left(1+x\right)^{2}}\right)dx$$

which simplifies down to

$$3\ln\left(2\right)\int_{0}^{1}\frac{1}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(x\right)}{1+x^{2}}dx+\int_{0}^{1}\frac{\ln\left(1+x^{2}\right)}{1+x^{2}}dx-4\int_{0}^{1}\frac{\ln\left(1+x\right)}{1+x^{2}}dx.$$

Evaluating the first integral trivially and using the assumptions from your query, we get

$$I = \frac{3}{4}\pi\ln\left(2\right) - G + \frac{\pi}{2}\ln\left(2\right)-G - 4\left(\frac{\pi}{8}\ln\left(2\right)\right)$$

Therefore, the integral $I$ is

$$\int_{0}^{1}\frac{\ln\left(1-x^{4}\right)}{1+x^{2}}dx = \frac{3\pi}{4}\ln\left(2\right)-2G.$$

Accelerator
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2

Letting $x=\tan \theta$ yields $$ \begin{aligned} \int_0^1 \frac{\ln \left(1-x^4\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4} } \ln \left[\left(1+\tan ^2 \theta\right)\left(1-\tan ^2 \theta\right)\right] d \theta \\ &=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta+\ln \left(\frac{\cos^2 \theta-\sin ^2 \theta}{\cos ^2 \theta}\right) d \theta \\ &=-4 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta+\int_0^{\frac{\pi}{4}} \ln (\cos 2 \theta) d \theta \\ &=-4\left(-\frac{\pi}{4} \ln 2+\frac{G}{2}\right)-\frac{\pi}{4} \ln 2 \\ &=\frac{3 \pi}{4} \ln 2-2G \end{aligned} $$ where the first integral refers to the post.

Lai
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