The value of the integral int_0^infty frac{lo | vedclass

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Question

(B) Let $I = \int_0^\infty \frac{\log_e(x)}{x^2+4} dx$.Substitute $x = 2t$, so $dx = 2 dt$. As $x 	o 0, t 	o 0$ and as $x 	o \infty, t 	o \infty$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_0^\infty \frac{\log_e(x)}{x^2+4} dx$.Substitute $x = 2t$, so $dx = 2 dt$. As $x 	o 0, t 	o 0$ and as $x 	o \infty, t 	o \infty$.$I = \int_0^\infty \frac{\log_e(2t)}{4t^2+4} (2 dt) = \frac{1}{2} \int_0^\infty \frac{\log_e(2) + \log_e(t)}{t^2+1} dt$.$I = \frac{\log_e(2)}{2} \int_0^\infty \frac{1}{t^2+1} dt + \frac{1}{2} \int_0^\infty \frac{\log_e(t)}{t^2+1} dt$.For the second integral, let $t = 1/u$, then $dt = -1/u^2 du$. The limits change from $\infty$ to $0$.$\int_0^\infty \frac{\log_e(t)}{t^2+1} dt = \int_\infty^0 \frac{\log_e(1/u)}{(1/u)^2+1} (-1/u^2) du = \int_0^\infty \frac{-\log_e(u)}{1+u^2} du = -\int_0^\infty \frac{\log_e(u)}{u^2+1} du$.This implies $\int_0^\infty \frac{\log_e(t)}{t^2+1} dt = 0$.Thus, $I = \frac{\log_e(2)}{2} [\arctan(t)]_0^\infty = \frac{\log_e(2)}{2} (\frac{\pi}{2} - 0) = \frac{\pi \log_e(2)}{4}$.

The value of the integral $\int_0^\infty \frac{\log_e(x)}{x^2+4} dx$ is:

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