$\int_0^\pi\int_0^\infty e^{-xy}\sin kx~dy~dx=$ ?

Question

$$\int_0^\pi\int_0^\infty e^{-xy}\sin kx~dy~dx=~?$$ My computation is

$$\int_0^\infty e^{-xy}\sin kx~dx=\frac{1}{k+y^2}$$

so

$$\int_0^\pi\frac{1}{k+y^2}dy=\frac{\sqrt{k}}{k}\arctan\pi$$

$$\int_0^\pi\int_0^\infty e^{-xy}\sin kx~dy~dx=\frac{\sqrt{k}}{k}\arctan\pi$$

Is my result wrong?

Answer 1

The improper integral is not right; a simple dimensional analysis should tell you so without any further computation. The result is as follows:

$$\begin{align}\int_0^{\infty} dx \, e^{-x y} \sin{k x} &= \Im{\left [\int_0^{\infty} dx \, e^{-(y-i k) x} \right ]} \\ &= \Im{\left [\frac{1}{y-i k} \right ]}\\ &=\frac{k}{k^2+y^2}\end{align}$$

$$\int_0^{\pi} dy \, \frac{k}{k^2+y^2} = \arctan{\frac{\pi}{k}}$$