Find $\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$

Question

Need solve the next integral

$$\int_{0}^{\infty} \frac{\log(x) }{\sqrt{x} (x+1)^{2}}\,dx$$

Tried something with Laurent’s series, but i can’t conclude anything.

Thanks

Answer 1

METHODOLGY $1$: CONTOUR INTEGRATION

Enforce the substitution $x\mapsto x^2$ to find that

$$\int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx=4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\tag1$$

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Let $f(z)$ be given by

$$f(z)=\oint_C \frac{\log^2(z)}{(z^2+1)^2}\,dz\tag2$$

where we choose to cut the plane along the positive real axis and where $C$ is the classical keyhole contour.

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We then have from $(2)$

$$\begin{align} \int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\\\\&+2\pi i \text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\tag3\end{align}$$

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The left-hand side of $(3)$ becomes

$$\begin{align} \int_0^\infty \frac{\log^2(x)-\left(\log(x)+i2\pi\right)^2}{(x^2+1)^2}\,dx&=-i4\pi\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx\\\\& +4\pi^2\int_0^\infty \frac{1}{(x^2+1)^2}\,dx\tag4 \end{align}$$

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Note that the imaginary part of the right-hand side of $(4)$ is $-\pi$ times the integral of interest on the right-hand side of $(1)$. Thus, we find that

$$\begin{align} \int_0^\infty \frac{\log(x)}{\sqrt x(x+1)^2}\,dx&=-2\text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i\pi/2}\right)\right)\\\\ &-2 \text{Re}\left(\text{Res}\left(\frac{\log^2(z)}{(z^2+1)^2}, z=e^{i3\pi/2}\right)\right)\\\\ &=-2\left(-\frac\pi4+\frac{3\pi}{4}\right)\\\\ &=-\pi \end{align}$$

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METHODOLGY $2$: REAL ANALYSIS ONLY

We begin by enforcing the substitution $x\mapsto \tan(x)$ in the integral on the right-hand side of $(1)$ to reveal

$$\begin{align} 4\int_0^\infty \frac{\log(x)}{(x^2+1)^2}\,dx&=4\int_0^{\pi/2}\cos^2(x) \log(\tan(x))\,dx\\\\ &=4\int_0^{\pi/2}\cos^2(x) \log(\sin(x))\,dx\\\\&-4\int_0^{\pi/2}\cos^2(x) \log(\cos(x))\,dx\tag5 \\\\ &=4\int_0^{\pi/2}(2\cos^2(x)-1)\log(\sin(x))\,dx\tag6\\\\ &=4\int_0^{\pi/2}\cos(2x)\log(\sin(x))\,dx\tag7\\\\ &=4\left(-\int_0^{\pi/2}\cos^2(x)\,dx\right)\tag8\\\\ &=-\pi \end{align}$$ as expected.

In going from $(5)$ to $(6)$ we made use of the transformation $x\mapsto \pi/2 -x$ in the second integral on the right-hand side of $(5)$.

In going from $(7)$ to $(8)$, we used integration by parts with $u=\log(\sin(x))$ and $v=\sin(x)\cos(x)$

Answer 2

Make the substitution $x=e^z$, $dx=e^z dz$: $$ I=\int _{0}^{\infty} \frac{\log(x)}{\sqrt{x}(x+1)^2}\,dx $$ $$ \Rightarrow \int _{-\infty}^{\infty} \frac{z}{\sqrt{e^z}(e^z+1)^2}\cdot e^z\,dz = \int _{-\infty}^{\infty} \frac{ze^{z/2}}{(e^z+1)^2}\,dz $$Let $R>0$. Consider the rectangular contour $\gamma:R\to R+2\pi i \to -R + 2\pi i \to -R \to R$. In the interior of $\gamma$, the integrand is singular at $\pi i$. By the Residue Theorem, $$ \oint_{\gamma} f(z)\,dz = 2\pi i \text{Res}(f;\pi i) $$ First, let's calculate the residue: $$ \frac{z e^{z/2}}{(e^z+1)^2}(z-\pi i) = \left(-\frac{\pi }{48}+\frac{i}{24}\right) (z-i \pi )+\frac{\frac{\pi }{2}+i}{z-i \pi }-\frac{\pi }{(z-i \pi )^2}-\frac{\pi }{24}-\frac{i}{2}+\cdots \Rightarrow \text{Res}(f(z);\pi i ) = i+\pi/2 $$Call $\gamma_1,\gamma_3$ the vertical parts of $\gamma$. We have $$ \left|\int _{\gamma_1} f(z)\,dz \right|=\left| \int _{0}^{2\pi} \frac{({R+i y})e^{({R+i y})/2}}{(e^{R+i y}+1)^2}\cdot 2\pi i\,dy\right| $$ $$ \leq \int _{0}^{2\pi} \left |\frac{({R+i y})e^{({R+i y})/2}}{(e^{R+i y}+1)^2}\cdot 2\pi i\right |\,dy $$By standard bounds, this is bounded by $$ \leq 2\pi \int _{0}^{2\pi} \frac{(R+y)e^{R/2}}{(e^R+1)^2}\,dy = \frac{4 \pi ^2 e^{R/2} (R+\pi )}{\left(e^R+1\right)^2} $$This approaches $0$ as $R\to\infty$. Similarly, as $R\to \infty$, $\lim\limits_{R\to\infty}|\int_{\gamma_3}f(z)\,dz|=0$. Now consider the contour $\gamma_2:R+2\pi i \to -R+2\pi i$ (with $\gamma_4$ the corresponding part on the real axis). $$ \int _{\gamma_2} f(z)\,dz = -\int _{-R}^{R} \frac{(z+2\pi i)e^{(z+2\pi i)/2}}{(e^{(z+2\pi i)}+1)^2}\,dz $$ $$ = -\int _{-R}^{R} \frac{ze^{(z+2\pi i)/2}}{(e^{(z+2\pi i)}+1)^2}\,dz-2\pi i\int _{-R}^{R} \frac{e^{(z+2\pi i)/2}}{(e^{(z+2\pi i)}+1)^2}\,dz $$ $$ = \int _{-R}^{R} \frac{ze^{z/2}}{(e^{(z)}+1)^2}\,dz+2\pi i\int _{-R}^{R} \frac{e^{z/2}}{(e^{(z)}+1)^2}\,dz $$ $$ = \int _{\gamma_4}f(z)\,dz+2\pi i\left(\arctan\left(e^{R/2}\right)-\arctan\left(e^{-R/2}\right)\right) $$So, in the limit, we have $$ 2 I + 2\pi i (\pi/2) = 2\pi i( i + \pi/2) $$ $$ \therefore I = -\pi $$

Answer 3

Substituting $ \small\left\lbrace\begin{aligned}u&=\frac{1}{\sqrt{x}}\\ -2\frac{\mathrm{d}u}{u^{2}}&=\frac{\mathrm{d}x}{\sqrt{x}}\end{aligned}\right. $, we get : \begin{aligned} \int_{0}^{+\infty}{\frac{\ln{x}}{\sqrt{x}\left(1+x\right)^{2}}\,\mathrm{d}x}&=-4\int_{0}^{+\infty}{\frac{y^{2}\ln{y}}{\left(1+y^{2}\right)^{2}}\,\mathrm{d}y}\\ &=2\left[\frac{y\ln{y}}{1+y^{2}}\right]_{0}^{+\infty}-2\int_{0}^{+\infty}{\frac{1+\ln{y}}{1+y^{2}}\,\mathrm{d}y}\\ &=-2\int_{0}^{+\infty}{\frac{\mathrm{d}y}{1+y^{2}}}-2\int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}\\ &=-\pi \end{aligned}

Because $ \int_{0}^{+\infty}{\frac{\ln{y}}{1+y^{2}}\,\mathrm{d}y}=0 $, and that can be proved substituting $ \small\left\lbrace\begin{aligned}u&=\frac{1}{y}\\ \mathrm{d}y&=-\frac{\mathrm{d}u}{u^{2}}\end{aligned}\right. \cdot $

Answer 4

Let's start by getting rid of the square root, writing $x=u^2$, so that

$$\int_0^\infty{\log x\over\sqrt x(x+1)^2}\,dx=4\int_0^\infty{\log u\over(u^2+1)^2}\,du$$

Next, split $\int_0^\infty$ into $\int_0^1$ and $\int_1^\infty$ and note that

$$\int_1^\infty{\log u\over(u^2+1)^2}\,du=\int_1^0{\log(1/u)\over((1/u)^2+1)^2}\,d(1/u)=-\int_0^1{u^2\log u\over(u^2+1)^2}\,du$$

It follows that

$$\int_0^\infty{\log u\over(u^2+1)^2}=\int_0^1{(1-u^2)\log u\over(u^2+1)^2}\,du$$

Finally, note that for $|u|\lt1$ standard manipulations of geometric series tell us

$${1-u^2\over(u^2+1)^2}=(1-u^2)(1-2u^2+3u^4-4u^6+\cdots)=1-3u^2+5u^4-7u^2+\cdots$$

and integration by parts says

$$\int_0^1u^{k-1}\log u\,du=-{1\over k^2}$$

(for $k\ge1$). So

$$\int_0^1{(1-u^2)\log u\over(u^2+1)^2}\,du=-\left(1-{1\over3}+{1\over5}-{1\over7}+\cdots \right)=-{\pi\over4}$$

(invoking the Leibniz formula for the alternating sum), and thus

$$\int_0^\infty{\log x\over\sqrt x(x+1)^2}\,dx=4\int_0^\infty{\log u\over(u^2+1)^2}\,du=4\int_0^1{(1-u^2)\log u\over(u^2+1)^2}\,du=-\pi$$

Answer 5

As previous answers show, there are many ways to adress the problem of this integral.

Concerning the antiderivative, let first $x=t^2$ to make $$I=\int \frac{\log(x)}{\sqrt x(x+1)^2}\,dx=4\int \frac{\log(t)}{(t^2+1)^2}\,dt$$ Now, using partial fraction decomposition $$\frac{4}{(t^2+1)^2}=\frac{i}{t+i}-\frac{1}{(t+i)^2}-\frac{i}{t-i}-\frac{1}{(t-i)^2}$$ $$\int \frac{\log (t)}{t+k} \,dt=\text{Li}_2\left(-\frac{t}{k}\right)+\log (t) \log \left(\frac{k+t}{k}\right)$$ $$\int \frac{\log (t)}{(t+k)^2} \,dt=\frac{\log (t)}{k}-\frac{\log (t)}{k+t}-\frac{\log (k+t)}{k}$$ Combining everything and simplifying, we end with $$I=i \big(\text{Li}_2(i t)-\text{Li}_2(-i t)\big)+\frac{2 t \log (t)}{t^2+1}+2 (\log (t)-1) \tan ^{-1}(t)=$$

Now, using asymptotics

$$\color{blue}{J(a)=4\int_0^a \frac{\log(t)}{(t^2+1)^2}\,dt=-\pi+4\sum_{n=1}^\infty(-1)^n\,n\,\frac{ (2 n+1) \log (a)+1}{(2 n+1)^2\, a^{2 n+1}}}$$

Answer 6

$$\int_0^\infty\frac{\ln x}{\sqrt{x}(1+x)^2}dx=\frac{\partial}{\partial a}\int_0^\infty\frac{x^{a-1}}{(1+x)^2}dx\Bigg|_{a\to 1/2}$$

$$=\frac{\partial}{\partial a}\text{B}(a,2-a)\Bigg|_{a\to 1/2}=\text{B}(a,2-a)\left[\psi(a)-\psi(2-a)\right]\Bigg|_{a\to 1/2}$$

$$=\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)}\left[\psi(1/2)-\psi(3/2)\right]=\frac{\pi}{2}[-2]=-\pi$$

Note that $\psi(n+1)-\psi(n)=\frac1n$.

Proof: By $\psi(n+1)=-\gamma+\int_0^1\frac{1-x^n}{1-x}dx$ we have

$$\psi(n+1)-\psi(n)=\int_0^1\frac{x^{n-1}-x^{n}}{1-x}dx=\int_0^1 x^{n-1}dx=\frac1n$$

Answer 7

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Lets consider \begin{equation} \int_{\mathcal{C}}{z^{\mu} \over z - a}\,\dd z\,,\qquad \mu \in \pars{-1,0}\,,\quad a > 0\label{1}\tag{1} \end{equation} $\ds{\mathcal{C}}$ is a key-hole contour where $\ds{z^{\mu}}$ is its principal branch.

Then, \begin{align} 2\pi\ic a^{\mu} & = -\int_{-\infty}^{0}{\pars{-x}^{\mu}\expo{\ic\pi\mu} \over x - a}\,\dd x - \int_{0}^{-\infty}{\pars{-x}^{\mu}\expo{-\ic\pi\mu} \over x - a}\,\dd x \\[5mm] & = \expo{\ic\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x - \expo{-\ic\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x \\[5mm] & = 2\ic\sin\pars{\pi\mu}\int_{0}^{\infty}{x^{\mu} \over x + a}\,\dd x \\[5mm] \implies & \bbx{\int_{0}^{\infty}{x^{\mu} \over x + a }\,\dd x = {\pi a^{\mu} \over \sin\pars{\pi\mu}}}\label{2}\tag{2} \end{align}

• Derive (\ref{2}) respect of $\ds{a}$: $\ds{\int_{0}^{\infty}{x^{\mu} \over \pars{x + a}^{\, 2}}\,\dd x = -\pi\,{\mu a^{\mu - 1} \over \sin\pars{\pi\mu}}}$

• Set $\ds{a = 1}$: $\ds{\int_{0}^{\infty}{x^{\mu} \over \pars{x + 1}^{\, 2}}\,\dd x = -\,{\pi\mu \over \sin\pars{\pi\mu}}}$

• Derive respect of $\ds{\mu}$: $\ds{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{x + a}^{\, 2}}\,\dd x = -\pi\bracks{1 - \pi\mu\cot\pars{\pi\mu}\csc\pars{\pi\mu}}}$

• Takes the limit $\ds{\mu \to -1/2}$: $$ \bbox[15px,#ffd,border:1px solid navy]{ \int_{0}^{\infty}{\ln\pars{x} \over \root{x}\pars{x + a}^{\, 2}}\,\dd x = {\large -\pi}}\\ $$