A Tricky Integral

Let the given integral be $I$ .Therefore,
$I = \int_0^\infty \frac{\ln x}{x^2 + 2x + 4} dx$
Put $x = \frac{4}{t}$ . Therefore, $dx = -4\frac{dt}{t^2}$ .The integral can thus be rewritten as,

$I = \int_\infty^0 \frac{\ln(\frac{4}{t})}{\frac{16}{t^2} + \frac{8}{t} + 4} \bigg(-4\frac{dt}{t^2}\bigg)$

Note the change in limits (in the steps above and below). Therefore, taking lcm and rewriting the integral we get,

$I = \int_0^\infty \frac{\ln 4 - \ln t}{t^2 + 2t + 4} dt$

$\Rightarrow I = \int_0^\infty \frac{\ln 4}{t^2 + 2t + 4} dt - \int_0^\infty \frac{\ln t}{t^2 + 2t + 4}dt$

$\Rightarrow I = \int_0^\infty \frac{\ln 4}{(t+1)^2 + 3} dt - I$

$\Rightarrow 2I = \ln 4 \int_0^\infty \frac{dt}{(t+1)^2 + 3}$

It is a standard integral equal to $\frac{1}{\sqrt{3}} \arctan\bigg(\frac{t+1}{\sqrt{3}}\bigg)$ . Substituting the limits, we get,

$\Rightarrow 2I = \frac{\ln 4}{\sqrt{3}} \bigg[ \arctan(\infty) - \arctan(\frac{1}{\sqrt{3}})\bigg]$

$\Rightarrow I = \frac{\ln 4}{2\sqrt{3}} \bigg( \frac{\pi}{2} - \frac{\pi}{6} \bigg)$

$\Rightarrow I = \frac{\pi}{6\sqrt{3}} \times \ln 4 = \boxed{0.419}$

We're given that:

$\displaystyle I = \int_0^\infty \dfrac{ \ln x } { (x+1)^2 + (\sqrt 3)^2 } dx$ .

Letting $x+1 = \sqrt 3 \tan \theta$ , we get $dx = \sqrt 3 \sec^2 \theta d\theta$ . So, the integral then can be written as:

$\displaystyle I = \int_{\pi/6}^{\pi/2} \dfrac{ \ln( \sqrt3\tan \theta - 1) } { \sqrt 3 } d\theta$

Or:

$\displaystyle I = \dfrac 1 {\sqrt 3}\int_{\pi/6}^{\pi/2} \ln\left( \sqrt3\tan \left(\dfrac{2\pi} {3} - \theta\right) - 1\right) d\theta$

This can be simplified to:

$\displaystyle \begin{aligned} I &= \dfrac 1 {\sqrt 3}\int_{\pi/6}^{\pi/2} \ln\left( \dfrac 4 { \sqrt3\tan \theta - 1}\right) d\theta \\ &= \dfrac 1 {\sqrt 3}\int_{\pi/6}^{\pi/2} \ln 4 d\theta - I \\ \end{aligned}$

Which gives us $\displaystyle I = \dfrac {\pi \ln 2 } {3 \sqrt 3 }$ .

Using the property of logarithms, we can rewrite the integral as:

$\int_0^{\infty} dx\,\frac{\log\frac{x}{2} +\, \log 2}{x^2+2x+4} = I_1 + I_2$

With $I_1$ and $I_2$ as:

$I_1 = \int_0^{\infty} dx\,\frac{\log\frac{x}{2}}{x^2+2x+4} \, \qquad I_2 = \int_0^{\infty}dx\, \frac{ \log 2}{x^2+2x+4}$

Now we concentrate our attention to $I_1$ . Let $x = 2 y$ and we got:

$I_1 = \frac{1}{2} \int_0^{\infty} dy \,\frac{\log y}{y^2+y+1}$

Looking at the denominator, we think at the substitution $y = e^z$ in order to obtain an hyperbolic cosine (watch out the integration extreme):

$I_1 = \frac{1}{2} \int_{-\infty}^{\infty}dz\, \frac{z}{1+2 \cosh z}$ = 0

In fact this is an odd function integrated over an even domain, so it vanishes! We remain with $I_2$ , that is easily integrated, thus our integral value is:

$\int_0^{\infty} dx\,\frac{\log x}{x^2+2x+4} = I_1 + I_2 = I_2 = \frac{\pi \log 2}{3 \sqrt{3}} \approx 0.419$