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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?

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This is a duplicate of this Functions defined by integrals (problem 10.23 from Apostol's Mathematical Analysis)

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 $y$.

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