$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$

Question

We have to find the integration of

$$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4}dx$$

In this I tried to do substitution of $x=e^t$

After that got stuck .

Answer 1

May be this will be a more elementary solution.

Let us start with a the substitution $x=2y$. Then \begin{gather*} I = \int_{0}^{\infty}\dfrac{\ln(x)}{x^2+2x+4}\, dx = \int_{0}^{\infty}2\dfrac{\ln(2y)}{4y^2+4y+4}\, dy =\\[2ex] \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln(2)}{y^2+y+1}\, dy + \dfrac{1}{2}\int_{0}^{\infty}\dfrac{\ln y}{y^2+y+1}\, dy= \dfrac{\ln(2)}{2}I_1+ \dfrac{1}{2}I_2\tag{1} \end{gather*} where \begin{equation*} I_1 = \int_{0}^{\infty}\dfrac{1}{y^2+y+1}\, dy =\dfrac{4}{3}\int_{0}^{\infty}\dfrac{1}{\left(\frac{2y+1}{\sqrt{3}}\right)^{2}+1}\, dy = \dfrac{2}{\sqrt{3}}\left[\arctan\left(\dfrac{2y+1}{\sqrt{3}}\right)\right]_{0}^{\infty} = \dfrac{2\pi}{3\sqrt{3}}\tag{2} \end{equation*} and \begin{equation*} I_2 = \int_{0}^{\infty}\dfrac{\ln(y)}{y^2+y+1}\, dy =\left[y=\dfrac{1}{z}\right] = \int_{0}^{\infty}\dfrac{-\ln(z)}{z^2+z+1}\, dz = -I_2. \tag{3} \end{equation*} But (3) implies that $I_2=0$. This and (2) substituted in (1) give us $I = \dfrac{\pi\ln(2)}{3\sqrt{3}}$.

Answer 2

Let $I$ be defined by the integral

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

and let $J$ be the contour integral

$$J=\oint_{C}\frac{\log^2(z)}{z^2+2z+4}\,dz \tag2$$

where the contour $C$ is the classical "key-hole" contour for which the keyhole coincides with the branch cut along the positive real axis.

Then, we can write $(2)$ as

$$\begin{align} J&=\int_0^R \frac{\log^2(x)}{x^2+2x+4}\,dx-\int_0^R \frac{(\log(x)+i2\pi)^2}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\\\\ &=-i4\pi \int_0^R \frac{\log(x)}{x^2+2x+4}\,dx+4\pi^2\int_0^R\frac{1}{x^2+2x+4}\,dx+\int_{C_R}\frac{\log^2(z)}{z^2+2z+4}\,dz\tag 3 \end{align}$$

As $R\to \infty$ the integral over $C_R$ vanishes and we have from $(1)$

$$\lim_{R\to \infty}J=-i4\pi I +4\pi^2\int_0^\infty\frac{1}{x^2+2x+4}\,dx\tag 4$$

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Next, we apply the residue theorem to evaluate the left-hand side of $(4)$. Proceeding we find that

$$\begin{align} \lim_{R\to \infty}J&=2\pi i \text{Res}\left(\frac{\log^2(z)}{z^2+2z+4}\,dz, z=-1\pm i\sqrt{3}\right)\\\\ &=2\pi i \left(\frac{(\log(2)+i2\pi/3)^2}{i2\sqrt{3}}+\frac{(\log(2)+i4\pi/3)^2}{-i2\sqrt{3}}\right)\\\\ &=-\frac{i4\pi^2\log(2)}{3\sqrt{3}}+\frac{4\pi^3}{3\sqrt{3}}\tag 5 \end{align}$$

Equating real and imaginary parts of $(3)$ and $(5)$ reveals

$$\int_0^\infty \frac{\log(x)}{x^2+2x+4}\,dx =\frac{\pi \log(2)}{3\sqrt{3}}$$

and as a bonus

$$\int_0^\infty\frac{1}{x^2+2x+4}\,dx=\frac{\pi}{3\sqrt{3}}$$

And we are done!

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\quad r \equiv -1 + \root{3}\ic\quad\mbox{and}\quad R > 0}$,

Hereafter, $\ds{\ln\pars{z}}$ is the principal $\ds{\ln}$-branch cut. \begin{align} \int_{0}^{4R}{\ln\pars{x} \over x^{2} + 2x + 4} & = \int_{0}^{4R}\ln\pars{x}\pars{{1 \over x - r} - {1 \over x - \bar{r}}} {1 \over r - \bar{r}}\,\dd x \\[5mm] & = {\root{3} \over 3}\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x \label{1}\tag{1} \end{align}

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Then, \begin{align} \Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x & = -\,\Im\int_{0}^{4R}{\ln\pars{r\bracks{x/r}} \over 1 - x/r}\,{\dd x \over r} = -\,\Im\int_{0}^{4R/r}{\ln\pars{rx} \over 1 - x}\,\dd x = -\,\Im\int_{0}^{R\bar{r}}{\ln\pars{rx} \over 1 - x}\,\dd x \\[5mm] & = -\,\Im\bracks{-\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} + \int_{0}^{R\bar{r}}{\ln\pars{1 - x} \over x}\,\dd x} \\[5mm] & = -\,\Im\bracks{-\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} - \,\mrm{Li}_{2}\pars{R\,\bar{r}}} \end{align}

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With $\quad$ $\ds{\,\mrm{Li}_{2}}$-Inversion Formula$\quad$ and $\ds{\quad R\,\bar{r} \not\in \left[0,1\right)}$: \begin{align} \Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x & = \Im\braces{\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} + \bracks{-\,\mrm{Li}_{2}\pars{1 \over R\,\bar{r}} - {\pi^{2} \over 6} - {1 \over 2}\,\ln^{2}\pars{-R\bar{r}}}} \\[5mm] & = \Im\braces{\ln\pars{1 - R\,\bar{r}}\ln\pars{4R} - {1 \over 2}\,\ln^{2}\pars{-R\,\bar{r}} - \,\mrm{Li}_{2}\pars{1 \over R\,\bar{r}}} \end{align}

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\begin{align} &\Im\int_{0}^{4R}{\ln\pars{x} \over x - r}\,\dd x \\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\sim}\,\,\,& \Im\braces{\ln\pars{1 + R + R\root{3}\ic}\ln\pars{4R} - {1 \over 2}\,\ln^{2}\pars{R + R\root{3}\ic}} \\[5mm] = &\ \arctan\pars{{R \over 1 + R}\root{3}}\ln\pars{4R} - {1 \over 2}\,\Im\pars{\bracks{% \ln\pars{2R} + \arctan\pars{\root{3}}\ic}^{2}} \\[5mm] = &\ \arctan\pars{{R \over 1 + R}\root{3}}\ln\pars{4R} - \ln\pars{2R}\,\arctan\pars{\root{3}} \\[5mm] \stackrel{\mrm{as}\ R\ \to\ \infty}{\to}\,\,\,& \bbx{\ds{{1 \over 3}\,\pi\ln\pars{2}}}\label{2}\tag{2} \end{align}

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With \eqref{1} and \eqref{2}: $$ \bbx{\ds{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 2x + 4} = {\root{3} \over 9}\,\pi\ln\pars{2}}} \approx 0.4191 $$

Answer 4

Let $$I = \int^{\infty}_{0}\frac{\ln x}{x^2+2x+4}dx = \int^{\infty}_{0}\frac{\ln x}{(x+1)^2+(\sqrt{3})^2}dx$$

put $\displaystyle x=\frac{1^2+(\sqrt{3})^2}{y} = \frac{4}{y}$ and $\displaystyle dx = -\frac{4}{y^2}dy$

$$I = \int^{0}_{\infty}\frac{\ln (4)-\ln (y)}{16+8y+4y^2}\cdot -\frac{4}{y^2}\cdot y^2 dy = \int^{\infty}_{0}\frac{\ln 4-\ln y}{y^2+2y+4}dy$$

$$I = \ln 4\int^{\infty}_{0}\frac{1}{(y+1)^2+(\sqrt{3})^2}dy-I$$

So $$I = \ln 2 \cdot \frac{1}{\sqrt{3}}\cdot \tan^{-1}\left(\frac{y+1}{\sqrt{3}}\right)\bigg|^{\infty}_{0} = \frac{\ln 2}{\sqrt{3}}\cdot \bigg(\frac{\pi}{2}-\frac{\pi}{6}\bigg) = \frac{\pi \cdot \ln 2}{3\sqrt{3}}$$