Integral yielding auxiliary function for sine/cosine integrals

Question

The Abramowitz & Stegun section on exponential integrals and related functions includes the following for $\Re(z) \ge 0$:

$$ \int_0^\infty \frac{\sin t}{t+z} \mathrm{d}t = \int_0^\infty \frac{e^{-zt}}{t^2+1} \mathrm{d}t. $$

How can this be shown?

Answer 1

Start with the well-known and simple to derive $$\int_0^\infty \exp{(-t y)}\ \sin{\left(b \ y\right)} dy = \frac{b}{t^2+b^2} .$$ Set $b=1.$ Insert into right-hand side of your equation, Interchange $\int.$ Then $$\int_0^\infty \frac{e^{-zt}}{t^2+1} dt= \int_0^\infty dy \sin{y} \int_{0}^\infty e^{-t(z+y)}dt = \int_0^\infty \frac{\sin{y}}{y+z} dy$$ where in the last step another elementary integral has been used.