# How find this $I=\int_{0}^{\infty}\frac{x\sin{(2x)}}{x^2+4}dx$

Find this integral $$I=\int_{0}^{\infty}\dfrac{x\sin{(2x)}}{x^2+4}dx$$

let $x=2t$, then $$I=\int_{0}^{\infty}\dfrac{t\sin{(4t)}}{(t^2+1)}dt$$ then $$I=1/2\int_{0}^{\infty}\sin{(4t)}d\ln{(t^2+1)}$$ then I can't.

This problem have without residue methods?

We have that $$F(y) = \int \frac{\sin(xy)}{x(x^2+1)}\mathrm{d}x = \frac{\pi}{2}(1-e^{-y})$$ A generalization of your integral is given as $$G(a,y) = \int \frac{\sin(xy)}{x(x^2+a^2)}\mathrm{d}x = a^{-2} F(ay) = \frac{\pi}{2a^2}(1-e^{-ay})$$ Differentiating twice yield $$\frac{\mathrm{d}^2 G}{\mathrm{d}y^2} = -\int_0^{\infty} \frac{x\sin(xy)}{x^2+a^2} = - \frac{\pi}{2} e^{-ay}$$ Hence $$\int_{0}^{\infty}\dfrac{x\sin{(2x)}}{x^2+4}\mathrm{d}x = -G''(2,2) = \frac{\pi}{2} e^{-4}$$ Where $G''(a,y)$ means differentiation with respect to $y$.