Compute the real integral $ \int_{0}^{\infty} \frac{x \sin(2x)}{x^2 + 1} \, dx. $

Question

Using the integration of the function $f$ with the prescription $f(z) = \frac{ze^{2iz}}{z^2 + 1}$ along the positively oriented boundary of the region $D = \{ z \in \mathbb{C} \mid |z| \leq R, \operatorname{Im}(z) \geq 0 \}$, where $R \in (1, \infty)$, compute the real integral

$$ \int_{0}^{\infty} \frac{x \sin(2x)}{x^2 + 1} \, dx. $$

Attempt: The boundary consists of the semicircle $|z| = R$ in the upper half-plane and the real interval $[-R, R]$. The function $f(z)$ has simple poles at $z = \pm i$, but only $z = i$ lies within the contour.

Using the residue theorem, the integral around the closed contour is $2\pi i$ times the residue at $z = i$: $$ \text{Res}\left(f(z), i\right) = \lim_{z \to i} (z - i) \frac{z e^{2iz}}{z^2 + 1} = \lim_{z \to i} \frac{z e^{2iz}}{z + i} = \frac{i e^{-2}}{2i} = -\frac{e^{-2}}{2}. $$

Am I on the right path? What can I do from here?

WolframAlpha indicates that the integral is $\frac{\pi}{2e^2}$. — Tuvasbien, Jul 12 '24 at 15:24
@SineoftheTime Rip, I even Approach0'd it but didn't find that one — FShrike, Jul 12 '24 at 15:34
Using a property of the Laplace Transform you can convert it to $$\int_0^\infty \dfrac{2\cos x}{x^2+4}, \mathrm dx$$ which is maybe easier to tackle with Residue Theorem. — Hug de Roda, Jul 12 '24 at 15:37

Claude Leibovici · Accepted Answer · 2024-07-12T16:06:17.383

If you start with $$\frac x {x^2+1}=\frac x {(x+i)(x-i)}=\frac{1}{2 (x+i)}+\frac{1}{2 (x-i)}$$ Using twice the obvious change of variable $$\int \frac {\sin(2x)}{x+a}\,dx=\cos (2 a) \,\text{Si}(2 a+2 x)-\sin (2 a)\, \text{Ci}(2 a+2 x)$$ If $a$ is an imaginary number $$\int_0^\infty \frac {\sin(2x)}{x+a}\,dx=\text{Ci}(2 a) \sin (2 a)+\frac{1}{2} (\pi -2 \text{Si}(2 a)) \cos (2 a)$$

Use it twice, play with the complex numbers to get the result.

Compute the real integral $ \int_{0}^{\infty} \frac{x \sin(2x)}{x^2 + 1} \, dx. $

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