Integral $\int_{0}^{\infty} \frac{\sin{x}}{x^2+1} dx$

Question

How to evaluate :

$$\int_{0}^{\infty} \frac{\sin{x}}{x^2+1} dx$$

The integral from $-\infty$ to $\infty$ is quite easy, but how could we integrate this function from $0$ to $\infty$?

Answer 1

Using the fact that $x^2+1=(x-i)(x+i)$, we can rewrite the integral as $\displaystyle\frac1{2i}\bigg(\int_0^\infty\frac{\sin x}{x-i}dx$ $-\displaystyle\int_0^\infty\frac{\sin x}{x+i}\bigg)$, and then express each of these in terms of trigonometric integrals, by rewriting $\sin x$ as $\sin\Big((x\pm i)\mp i\Big)$, and then employing the angle addition formula for $\sin(a\pm b)$. Unlike

the related integral $\displaystyle\int_0^\infty\frac{\cos x}{x^2+1}dx=\frac\pi{2e}$ , this one does not seem to possess a closed form which

does not involve special functions.