Calculate the following integral: \begin{equation} \int_1^{\sqrt{2}}\frac{1}{x}\ln\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)dx \end{equation}

I am having trouble to calculate the integral. I tried to use by parts method but it didn't help. Wolfram Alpha gives me $0$ as the answer but I don't know how to get it. I also tried to search the similar question here and I got this: $$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx,$$ yet it didn't help much. Besides, I don't understand the answers there. Could anyone here please help me to calculate the integral preferably (if possible) with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.

  • $\begingroup$ Integration by parts followed by partial fractions decomposition did not help ? $\endgroup$ – Claude Leibovici Jul 18 '14 at 11:27
  • $\begingroup$ @ClaudeLeibovici No. I have tried it. It yields a creepy fraction from and $\ln x$. $\endgroup$ – Anastasiya-Romanova 秀 Jul 18 '14 at 11:30
  • $\begingroup$ The creepy fraction should just be a rational function, and the log term should be outside the integral. $\endgroup$ – zibadawa timmy Jul 18 '14 at 13:00
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    $\begingroup$ $\displaystyle\int_1^{\sqrt[3]2}f(x)~dx=-\int_{\sqrt[3]2}^{\sqrt2}f(x)~dx$. $\endgroup$ – Lucian Jul 18 '14 at 13:25
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    $\begingroup$ Hi @Anastasiya-Romanova秀! Maybe you will find this approach useful, even though it's $4$ years late :D math.stackexchange.com/a/3049039/515527 $\endgroup$ – Zacky Dec 22 '18 at 0:56

Update: Finally, a complete solution. Sorry it took so long.

Split the integral up into 3. \begin{align} I &=-\int^{\sqrt{2}}_1\frac{\log{x}}{x}dx+\int^{\sqrt{2}}_1\frac{\log{((x^2-1)^2+1)}}{x}dx-\int^{\sqrt{2}}_1\frac{\log{((x-1)^2+1)}}{x}dx\\ &=-\frac{1}{8}(\log{2})^2+\frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx-\int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx \end{align} The second integral is rather easy to evaluate. \begin{align} \frac{1}{2}\int^1_0\frac{\log(1+x^2)}{1+x}dx &=\frac{1}{2}\int^1_0\int^1_0\frac{x^2}{(1+x)(1+ax^2)}dx \ da\tag1\\ &=\frac{1}{2}\int^1_0\frac{1}{1+a}\int^1_0\frac{1}{1+x}+\frac{x-1}{1+ax^2}dx \ da\\ &=\frac{1}{2}\int^1_0\frac{\log{2}}{1+a}+\frac{\log(1+a)}{2a(1+a)}-\underbrace{\frac{\arctan(\sqrt{a})}{\sqrt{a}(1+a)}}_{\text{Let} \ y=\arctan{\sqrt{a}}}da\\ &=\frac{1}{2}\left[(\log{2})^2+\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{a}da}_{-\operatorname{Li}_2(-1)=\frac{\pi^2}{12}}-\frac{1}{2}\underbrace{\int^1_0\frac{\log(1+a)}{1+a}da}_{\frac{1}{2}(\log{2})^2}-\frac{\pi^2}{16}\right]\\ &=\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96} \end{align} The third integral can be evaluated using dilogarithms. \begin{align} \int^{\sqrt{2}-1}_0\frac{\log(1+x^2)}{1+x}dx &=\sum_{r=\pm i}\int^{\sqrt{2}-1}_0\frac{\log(r+x)}{1+x}dx\tag2\\ &=-\sum_{r=\pm i}\int^{\frac{\lambda}{\sqrt{2}}}_{\lambda}\log\left(r-1+\frac{\lambda}{y}\right)\frac{dy}{y}\tag3\\ &=-\sum_{r=\pm i}\int^{\frac{r-1}{\sqrt{2}}}_{r-1}\frac{\log(1+y)}{y}-\frac{1}{y}\log\left(\frac{y}{r-1}\right)dy\tag4\\ &=\frac{1}{4}(\log{2})^2+\sum_{r=\pm i}\mathrm{Li}_2\left(\frac{1-r}{\sqrt{2}}\right)-\mathrm{Li}_2(1-r)\tag5\\ &=\frac{1}{4}(\log{2})^2+\mathrm{Li}_2(e^{i\pi/4})+\mathrm{Li}_2(e^{-i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{i\pi/4})-\mathrm{Li}_2(\sqrt{2}e^{-i\pi/4})\\ &=\frac{1}{4}(\log{2})^2-\frac{\pi^2}{96}\tag6\\ \end{align} It follows that $$I=-\frac{1}{8}(\log{2})^2+\frac{3}{8}(\log{2})^2-\frac{\pi^2}{96}-\frac{1}{4}(\log{2})^2+\frac{\pi^2}{96}=0$$ Explanation
$(1)$: Differentiate under the integral sign
$(2)$: Factorise $1+x^2$, let $r=\pm i$
$(3)$: Let $\displaystyle y=\frac{\lambda}{1+x}$
$(4)$: Let $\lambda=r-1$
$(5)$: Recognise that $\displaystyle\int\frac{\ln(1+y)}{y}dy=-\mathrm{Li}_2(-y)+C$ and $\displaystyle\int\frac{\ln(ay)}{y}dy=\frac{1}{2}\ln^2(ay)+C$
$(6)$: Use the identities here

  • $\begingroup$ I appreciate your effort. Let me check it first. +1 $\endgroup$ – Anastasiya-Romanova 秀 Jul 18 '14 at 14:15
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    $\begingroup$ @Downvoter I would really like to know what needs to be improved in this answer. Thanks. $\endgroup$ – SuperAbound Aug 22 '14 at 17:34

Similar problem (two answers)

$$ I = \int_1^{\sqrt{2}}\frac{1}{x}\log\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)\mathrm{d}x $$

Integrate by parts

$$ I = \left.\log(x)\log\left(\frac{2-2x^2+x^4}{2x-2x^2+x^3}\right)\right|_1^{\sqrt{2}}- \int_1^{\sqrt{2}}\log(x)\left(\frac{x^6-4 x^5+8 x^4-10 x^2+8 x-4}{x \left(x^2-2 x+2\right) \left(x^4-2 x^2+2\right)}\right)\mathrm{d}x $$

Partial fractions

$$I = \frac{1}{2} \log (2) \log \left(\frac{2}{4 \sqrt{2}-4}\right) - \int_1^{\sqrt{2}}\log(x)\left(-\frac{2 (x-1)}{x^2-2 x+2}+\frac{4 \left(x^3-x\right)}{x^4-2 x^2+2}-\frac{1}{x}\right)\mathrm{d}x$$

It should be easy enough to show that $\int_1^\sqrt{2} \log(x)/x \,\mathrm{d}x = \left.\tfrac12 \log(x)^2\right|_1^\sqrt{2} = \tfrac18\log(2)^2$.

We are left with (after simplification)

$$I = -\frac{1}{8} \log (2) \left(\log (8)-4 \log \left(1+\sqrt{2}\right)\right) - \int_1^{\sqrt{2}}\log(x)\left(-\frac{2 (x-1)}{x^2-2 x+2}+\frac{4\! \left(x^3-x\right)}{x^4-2 x^2+2}\right)\mathrm{d}x$$

I will now show how to finish the evaluation$\def\I{\mathcal{I}}$

$$I_1 = \int_1^\sqrt{2} \frac{2(x-1)\log(x)}{x^2-2x+2} \mathrm{d}x$$

$$I_2 = \int_1^\sqrt{2} \frac{4x(x^2-1)\log(x)}{x^4-2x^2+2} \mathrm{d}x$$

The technique employ involves a partial fractions into the complex numbers. The first expression has a quadratic in the denominator and the second has a quadratic in $x^2$ in the denominator. It should be clear that it is sufficient to solve the equation in terms of

$$\I_1(a) = \int \frac{\log(x)}{x-a} \mathrm{d}x$$

$$\I_2(b) = \int \frac{x\log(x)}{x^2-b^2} \mathrm{d}x$$

The technique used will be geometric series expansions.

$$\I_1(a) = \int \frac{\log(x)}{1-x/a} \mathrm{d}x = -\frac1a \sum_{n=0}^{\infty}\int\log(x)\frac{x^n}{a^n} \mathrm{d}x$$

Integrate by parts

$$\I_1(a) = -\sum_{n=0}^\infty \frac{a^{-(n+1)} x^{n+1}}{(n+1)^2}-\sum_{n=0}^\infty\frac{a^{-(n+1)} x^{n+1} \log (x)}{(n+1)}$$

The first sum is a series expansion of the polylogarithm function of order $2$. The second is also a series for the polylogarithm but it is of order $1$ so it is easily expressible in terms of the logarithm. In all,

$$\I_1(a) = \operatorname{Li}_2\!\left(\frac{x}{a}\right) + \log (x) \log \!\left(1-\frac{x}{a}\right)$$

A similar procedure shows that

$$\I_2(b) = \frac{1}{4}\! \left(\operatorname{Li}_2\left(\frac{x^2}{b^2}\right)+2\log (x) \log \left(1-\frac{x^2}{b^2}\right)\right)$$

Without adding unnecessary details, we need to find

$$2\I_2(\sqrt{1+i}) + 2\I_2(\sqrt{1-i}) -\I_1(1+i) - \I_1(1-i)$$

and then evaluate at $x = 1$ and $x = \sqrt{2}$. A large part of the evaluation goes to zero at $x = 1$ but we are not so fortunate at $x = \sqrt{2}$. What remains is a product of logarithms and a sum of several dilogarithms. These dilogarithms are very easy to evaluate because they all fit under certain dilogarithm identities which can be found here.

For the time being, I will not explicitly evaluate the final sum but that may come in the near future. With the dilogaithm identities, evaluation is more of a nuisance than anything. I will assure you that you do come across the value which zeros out the integral. Feel free to ask if you have any questions.

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    $\begingroup$ Wait!? $$ \int_1^{\sqrt{2}}\frac{2(x-1)}{x^2-2x+2}\ln x\,dx $$ and $$ \int_1^{\sqrt{2}}\frac{4(x^3-x)}{x^4-2x^2+2}\ln x\,dx $$ are not easy. I calculate them using by parts and series expansion for $\ln(1-x)$. Could you elaborate them using the easiest way? Thanks $\endgroup$ – Anastasiya-Romanova 秀 Jul 18 '14 at 17:57
  • $\begingroup$ @V-Moy Done.${}$ $\endgroup$ – Brad Jul 24 '14 at 15:47
  • $\begingroup$ Although this is not an elementary way (and I'm having some difficulties to understand it) but still thank you (>‿◠)✌ $\endgroup$ – Anastasiya-Romanova 秀 Jul 25 '14 at 8:29

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