$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} dx$ and $\int_{0}^{\infty} \frac {\ln(x)}{x^2+b^2} dx$

Question

Prove that $$\int_{0}^{\infty} \frac{\cos(x)}{1+x^2} dx = \frac {\pi}{2e}$$ My approach would be $$\lim_{n \to \infty} \int_{0}^{n} \frac{\cos(x)}{1+x^2} dx$$ and evaluate the limits of the sine and cosine integral functions, but I'm pretty sure there is an easier way.

The second integral is $$\int_{0}^{\infty} \frac {\ln(x)}{x^2+b^2} dx, b > 0$$ My approach; Let $f$ be an analytic function $$f(z)=\frac {\ln(z)}{z^2+b^2}$$ then the poles of $f$ would be at $$z=ib, z=-ib$$ Now I don't know what contour to draw and what would be inside it.

Answer 1

For the first, use the parity to write it as an integral over the entire real line, replace $\cos x$ by $e^{ix}$, and use the residue theorem to evaluate

$$\int_{-\infty}^\infty \frac{e^{iz}}{1+z^2}\,dz.$$

For the second integral, use a keyhole contour to integrate

$$\frac{(\log z)^2}{z^2+b^2}.$$

Since the values of $\log z$ differ by $2\pi i$ when approaching the positive half-axis from below and from above, after the cancelling of $\int_0^\infty \frac{(\log x)^2}{x^2+b^2}\,dx$, what remains is the desired integral (times a constant factor) and an integral of $\frac{C}{x^2+b^2}$ over the positive half-axis that poses no difficulty.

Answer 2

For the second one, We have to compute, $$\int_{0}^{\infty}\frac{\ln(x)}{x^2+b^2} dx $$

Let us compute the simpler case($b=1$) . i.e $$ I = \int_{0}^{\infty}\frac{\ln(x)}{x^2+1} dx $$ Substituting, $u= \frac{1}{x} \implies dx = \frac{-1}{u^2}du $

So our integral $$I = \int_{0}^{\infty}\frac{-\ln(u)}{u^2+1} du = -I \implies I=0 $$ Now, coming back to this problem, we substitute, $x=\frac{b}{u}$ in

$$ J = \int_{0}^{\infty}\frac{\ln(x)}{x^2+b^2} dx $$ Substitution will give us $$ J = \int_{0}^{\infty}\frac{\ln(b)}{x^2+b^2} dx - I $$ Since $I=0$ , we have $$ J = \ln(b)\int_{0}^{\infty}\frac{1}{x^2+b^2} dx = \frac{\ln(b)}{b} \times \frac{\pi}{2} \Box$$

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First one is extremely famous question, it is posted here many times. See here