Show that $\int^\infty_0\left(\frac{\ln(1+x)} x\right)^2dx$ converge.

Question

Show that $$\int\limits^\infty_0\left(\frac{\ln(1+x)} x\right)^2dx$$ converge.

I have utterly no clue on this integral. Please give me some hints. Thanks you.

Answer 1

Since

$$\lim_{x\to0}\left(\frac{\ln(1+x)}{x}\right)^2=1$$ then the integral $$\int_0^1\left(\frac{\ln(1+x)}{x}\right)^2dx$$ exists, moreover we have $$\ln^2(1+x)=_\infty o(x^{1/2})$$ so $$\left(\frac{\ln(1+x)}{x}\right)^2=_\infty o\left(\frac{1}{x^{3/2}}\right)$$ and then the integral $$\int_1^\infty \left(\frac{\ln(1+x)}{x}\right)^2dx$$ also exists. Conclude.

Answer 2

If you use the change of variables $\ln(1+x)=u$ things will be easier. The integral becomes

$$ \int _{0}^{\infty }\!{\frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u }}-1 \right) ^{2}}}{du}.$$

Now you can see that close to $0$ the integrand behaves as

$$ \frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u }}-1 \right) ^{2}}\sim_{u\sim 0} \frac {{u}^{2}{{\rm e}^{0}}}{ \left( (1+u)-1 \right) ^{2}} = 1 $$

which is integrable. Note that we used Taylor series the function

$$ e^{u} = 1+u+\frac{u^2}{2!}+\dots\,. $$

Try to do the same with the other end of the interval and to compare the integrand with an integrable function.

Added: For the other end you should be able to see that

$$ \frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u }}-1 \right) ^{2}}\sim_{u\sim \infty} \frac {{u}^{2}{{\rm e}^{u}}}{ ({{\rm e}^{u }}) ^{2}}=\dots\,. $$

Answer 3

The integrand function is integrable over the $I=[0,1]$ interval since it is continuous over there. Moreover, for any $x\geq 1$ we have $\log(1+x)\leq x^{3/7}$, hence the integrand function is integrable over $[1,+\infty)$ since it is positive and bounded by the integrable function $\frac{1}{x^{8/7}}$.

By exploiting the change of variable suggested by Mhenni Benghorbal, an integration by parts and a geoemtric series, you can also check that: $$\int_{0}^{+\infty}\frac{\log^2(1+x)}{x^2}\,dx = 2\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{3}.$$

Answer 4

If you don't mind, I would like to evaluate this integral. \begin{align} \int^\infty_0\frac{\ln^2(1+x)}{x^2}dx\tag1 &=2\int^\infty_0\frac{\ln(1+x)}{x(1+x)}dx\\\tag2 &=2\left[\int^1_0\frac{\ln(1+x)}{x(1+x)}dx+\int^1_0\frac{\ln\left(1+\frac{1}{x}\right)}{1+x}dx\right]\\\tag3 &=2\left[\int^1_0\frac{\ln(1+x)}{x}dx-\int^1_0\frac{\ln x}{1+x}dx\right]\\\tag4 &=2\left[-\mathrm{Li}_2(-1)-\sum_{n \ge 0}(-1)^n\int^1_0x^n\ln{x}dx\right]\\\tag5 &=2\left[\frac{\pi^2}{12}-\underbrace{\sum_{n \ge 1}\frac{(-1)^n}{n^2}}_{-\eta \ (2)}\right]\\\tag6 &=2\left(\frac{\pi^2}{12}+\frac{\pi^2}{12}\right)\\ &=\frac{\pi^2}{3} \end{align} It automatically follows that the integral converges.

Explanation:
$(1)$ Integrate by parts
$(2)$ Split the integral into $2$ and substitute $x \mapsto \frac{1}{x}$ for the second integral
$(3)$ Simplify using partial fractions and properties of logarithms
$(4)$ Expand $\frac{1}{1+x}$ as a series
$(5)$ Integrate by parts
$(6)$ $\eta (2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$

Answer 5

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_{0}^{\infty}% {\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x} =\int_{0}^{\infty}{1 \over x}\,2\ln\pars{1 + x}\,{1 \over 1 + x}\,\dd x \\[3mm]&=2\int_{0}^{\infty}\bracks{% {\ln\pars{1 + x} \over x} - {\ln\pars{1 + x} \over x + 1}}\,\dd x \end{align}

With $\ds{\Lambda > 0}$: \begin{align}\int_{0}^{\Lambda}{\ln\pars{1 + x} \over x}\,\dd x &=\int_{0}^{-\Lambda}{\ln\pars{1 - x} \over x}\,\dd x =-\int_{0}^{-\Lambda}{\rm Li}_{2}'\pars{x}\,\dd x =-{\rm Li}_{2}\pars{-\Lambda} \\[3mm]&={\rm Li}_{2}\pars{-\,{1 \over \Lambda}} + {\pi^{2} \over 6} +\half\,\ln^{2}\pars{\Lambda} \\[5mm]\int_{0}^{\Lambda}{\ln\pars{1 + x} \over x + 1}\,\dd x &=\half\,\ln^{2}\pars{1 + \Lambda} \end{align}

Then, \begin{align}\color{#c00000}{\int_{0}^{\Lambda}% {\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x} ={\pi^{2} \over 3} + 2\,{\rm Li}_{2}\pars{-\,{1 \over \Lambda}} -\bracks{\ln^{2}\pars{1 + \Lambda} - \ln^{2}\pars{\Lambda}} \end{align}

Note that $\ds{\lim_{\Lambda \to \infty}{\rm Li}_{2}\pars{-\,{1 \over \Lambda}} = 0}$ and $$ \ln^{2}\pars{1 + \Lambda} - \ln^{2}\pars{\Lambda} =\bracks{\Lambda\ln\pars{1 + {1 \over \Lambda}}}\, {\ln\pars{1 + \Lambda} + \ln\pars{\Lambda} \over \Lambda}\ \stackrel{\Lambda \to \infty}{\Huge\longrightarrow}\ {0\atop} $$

such that $$ \color{#66f}{\large\int_{0}^{\infty}% {\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x = {\pi^{2} \over 3}} \approx {\tt 3.2899} $$