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?


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 $y$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.