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

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 .

• $$\int ^{\infty}_0 \frac{\ln x}{x^2 + 2x+ 4} \mathrm dx = \int_{-\infty}^{\infty} \frac{u e^u}{e^{2u} + 2e^u+ 4} \mathrm du$$ – DHMO Apr 14 '17 at 4:52
• @DHMO how does this help? – gt6989b Apr 14 '17 at 4:54
• @gt6989b Not really. I'm just expanding his attempt. – DHMO Apr 14 '17 at 4:55
• You can use contour integration if that is suitable for you. – Zaid Alyafeai Apr 14 '17 at 5:03
• thats defiantly the easiest way to approach it also that substitution doesn't look right ^^ should just be a u on top no? any chance partial fractions would help? – Faust Apr 14 '17 at 5:04

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 $I = \dfrac{\pi\ln(2)}{3\sqrt{3}}$.

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 $(1)$

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

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!

• +1 Thank you very much for providing such a good answer . But I am unable to understand it . As I have never studied countor integration . I am just a high school student . – Koolman Apr 14 '17 at 5:43
• @Koolman You're very welcome. And apology if this solution was beyond the scope of the current level of your courses. Was this a problem given in high school? -Mark – Mark Viola Apr 14 '17 at 5:44
• It was my mistake , I should have already mention that . Yes it is high school problem . – Koolman Apr 14 '17 at 5:47
• What tools have you learned in integration techniques? – Mark Viola Apr 14 '17 at 5:48
• Integration as the inverse process of differentiation, indefinite integrals of standard functions, definite integrals and their properties, Fundamental Theorem of Integral Calculus. Integration by parts, integration by the methods of substitution and partial fractions, application of definite integrals to the determination of areas involving simple curves. Formation of ordinary differential equations, solution of homogeneous differential equations, separation of variables method, linear first order differential equations. – Koolman Apr 14 '17 at 5:49

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


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}

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