Integral $\int_{0}^{+\infty}\frac{t \sin(t)}{t^{2}+b^{2}}dt$

Question

I want to solve the integral $$\int_{0}^{+\infty}\frac{t \sin(t)}{t^{2}+b^{2}}dt$$ Which function and contour should I consider ?

Answer 1

The integrand, $\dfrac{t\sin t}{t^2+b^2}$ is an even function, so use

$$\int_0^\infty \frac{t\sin t}{t^2+b^2}\,dt = \frac{1}{2}\int_{-\infty}^\infty \frac{t\sin t}{t^2+b^2}\,dt.$$

Then replace $\sin t$ with $e^{it}$ to get

$$\int_0^\infty \frac{t\sin t}{t^2+b^2}\,dt = \frac{1}{2i} \int_{-\infty}^\infty \frac{t e^{it}}{t^2+b^2}\,dt.$$

The real part of $e^{it}$ produces an odd function that contributes $0$ to the overall integral, so we can take the imaginary part by dividing by $i$.

Then you can evaluate that with the usual semicircle in the upper half-plane.