Evaluation of Definite Integral of Logarithmic Function

To evaluate the given definite integral:

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

We apply the substitution $x = 2t$, which gives $dx = 2 \, dt$.

When $x = 0$, $t = 0$, and as $x \to \infty$, $t \to \infty$. Substituting these into the integral gives:

$I = \int_0^\infty \frac{\log_e(2t)}{(2t)^2 + 4} (2 \, dt)$

$I = \int_0^\infty \frac{\log_e(2) + \log_e(t)}{4(t^2 + 1)} (2 \, dt)$

$I = \frac{1}{2} \int_0^\infty \frac{\log_e(2) + \log_e(t)}{t^2 + 1} \, dt$

Splitting the integral into two parts:

$I = \frac{\log_e(2)}{2} \int_0^\infty \frac{1}{t^2 + 1} \, dt + \frac{1}{2} \int_0^\infty \frac{\log_e(t)}{t^2 + 1} \, dt$

Now, we evaluate each of these integrals individually:

1. First Integral: $\int_0^\infty \frac{1}{t^2 + 1} \, dt = \left[ \tan^{-1}(t) \right]_0^\infty = \frac{\pi}{2} - 0 = \frac{\pi}{2}$

2. Second Integral: Let $J = \int_0^\infty \frac{\log_e(t)}{t^2 + 1} \, dt$. Substitute $t = \frac{1}{u}$, so $dt = -\frac{1}{u^2} \, du$. The limits change as follows: when $t \to 0$, $u \to \infty$, and when $t \to \infty$, $u \to 0$.

$J = \int_\infty^0 \frac{\log_e(1/u)}{\left(\frac{1}{u}\right)^2 + 1} \left( -\frac{1}{u^2} \, du \right)$

$J = \int_0^\infty \frac{-\log_e(u)}{\frac{1 + u^2}{u^2}} \cdot \frac{1}{u^2} \, du = -\int_0^\infty \frac{\log_e(u)}{u^2 + 1} \, du$

$J = -J \implies 2J = 0 \implies J = 0$

Substituting the values of the two integrals back into the expression for $I$:

$I = \frac{\log_e(2)}{2} \cdot \left(\frac{\pi}{2}\right) + \frac{1}{2} \cdot (0)$

$I = \frac{\pi \log_e(2)}{4}$

Thus, the correct option is B.