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Log integral

$$\int_{0}^{\infty}{\ln[x(1+x)]\ln\left(x\over 1+x\right)\over (x+2)^2}\mathrm dx={1\over 2}\ln^2(2)\tag1$$

Making an attempt:

Following from my previous post making $u={x\over 1+x}$ it doesn't worked, so I try applying binomial series

$(1)$ becomes

$$\sum_{n=0}^{\infty}(-1)^n\cdot{n+1\over 2^n}\int_{0}^{\infty}[x^n\ln^2(x)-x^n\ln^2(1+x)]\mathrm dx=2\ln^2(2)\tag2$$

Applying IBP to these indefinite integrals

$$\int x^n\ln^2(x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(x)-x^2\cdot{2\ln x-1\over 2(n+1)}+C\tag3$$

$$\int x^n\ln^2(1+x)\mathrm dx={x^{n+1}\over n+1}\cdot \ln^2(1+x)-{2\over n+1}\int\color{red}{{x^{n+1}\over 1+x}\ln(1+x)\mathrm dx}\tag4$$

$$\color{red}{\int{x^{n+1}\over 1+x}\ln(1+x)}\mathrm dx={x^{n+1}\over 2}\ln^2(1+x)-{n+1\over 2}\int x^n\ln^2(1+x)\mathrm dx\tag5$$

If I put $(5)$ into $(4)$ we get zero! What I am doing here doesn't seem to be working.

How can we go about to tackle $(1)?$

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gymbvghjkgkjkhgfkl
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  • Please include additional context in this kind of question, such as the source of the problem and its motivation, application, or interest. While an attempt does show how you have tried to solve the problem, it doesn't show anyone else why the problem might be of interest. – Carl Mummert Apr 12 '17 at 17:36

2 Answers2

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$$\int_{0}^{+\infty}\frac{\log^2(x)-\log^2(x+1)}{(x+2)^2}\,dx = I_1-I_2 = \int_{0}^{+\infty}\frac{\log^2(x)}{(x+2)^2}\,dx-\int_{1}^{+\infty}\frac{\log^2(x)}{(x+1)^2}\,dx$$ is quite simple to compute. By integration by parts $I_2 = 2\zeta(2)=\frac{\pi^2}{3}$, while through the same technique we get that $I_1$ depends on $\text{Li}_2\left(\frac{1}{2}\right)$. The dilogarithm reflection formula hence settles the question.

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Jack D'Aurizio
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  • Can I ask you a question concerning the the infinity symbol $\infty$. I often noticed that you used $+\infty$ instead of $\infty$ why is that? Is there a difference meaning between them? – gymbvghjkgkjkhgfkl Apr 12 '17 at 19:31
  • @Latte': of course. If $f(z)$ is an entire function the two integrals $$\int_{0}^{+\infty}f(z),dz\qquad\text{and}\qquad \int_{0}^{-\infty}f(z),dz$$ can be quite different! $\infty$ is not a number, it is a shortcut for denoting some limit, and there isn't a single point "arbitrarily far from the origin" in $\mathbb{R}$ or $\mathbb{R}^2$. – Jack D'Aurizio Apr 12 '17 at 19:57
  • $I_2=\zeta(2)$? – FDP Apr 12 '17 at 20:56
  • I'm afraid of the use of integration by parts is prohibited here. – FDP Apr 12 '17 at 22:08
  • @FDP: prohibited by who?! – Jack D'Aurizio Apr 12 '17 at 22:20
  • You're right, i'm wrong, $\displaystyle\int_0^{+\infty}\dfrac{2\ln x}{(1+x)x}dx$ doesn't converge but $\displaystyle \int_1^{+\infty}\dfrac{2\ln x}{(1+x)x}dx$ does. – FDP Apr 13 '17 at 06:28
  • Each integral is a derivative of Euler's famous beta function with regard to one of its parameters. – Lucian Sep 01 '17 at 22:02
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On the path of Jack D'Aurizio,

$I_1=\displaystyle \int_0^{+\infty} \dfrac{(\ln(x))^2}{(x+2)^2}dx$

In $I_1$ perform the change of variable $y=\dfrac{x}{2}$,

$\begin{align} I_1&=2\int_0^{+\infty} \dfrac{(\ln(2x))^2}{(2x+2)^2}dx\\ &=\dfrac{1}{2}\int_0^{+\infty} \dfrac{(\ln 2+\ln x)^2}{(1+x)^2}dx\\ &=\dfrac{1}{2}(\ln 2)^2\int_0^{+\infty}\dfrac{1}{(1+x)^2}dx+\ln 2\int_0^{+\infty}\dfrac{\ln x}{(1+x)^2}dx+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\ &=\dfrac{1}{2}(\ln 2)^2\left[-\dfrac{1}{1+x}\right]_0^{+\infty}+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx+\\ &\ln 2\int_1^{+\infty} \dfrac{\ln x}{(1+x)^2}dx\\ \end{align}$

In the latter integral perform the change of variable $y=\dfrac{1}{x}$,

$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx-\ln 2\int_0^1 \dfrac{\ln x}{(1+x)^2}dx\\ &=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_0^{1}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\ \end{align}$

In the latter line, in the first integral perform the change of variable $y=\dfrac{1}{x}$,

$\begin{align}I_1&=\dfrac{1}{2}(\ln 2)^2+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx+\dfrac{1}{2}\int_1^{+\infty}\dfrac{(\ln x)^2}{(1+x)^2}dx\\ &=\dfrac{1}{2}(\ln 2)^2+I_2 \end{align}$

Therefore,

$\boxed{I_1-I_2=\dfrac{1}{2}(\ln 2)^2}$

FDP
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