I was asked to determine whether the integrals $\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ and $\int_{0}^{\infty}dx\int_{0}^{\infty}e^{-xy}\sin(x)dy$ converge, and if they do, calculate them.
I'm not sure what this notation means. Is $\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ supposed to mean $\int_{0}^{\infty}\int_{0}^{\infty}e^{-xy}\sin(x)dxdy$? Or does this literally mean the product of the integrals: $\int_{0}^{\infty} dy\cdot \int_{0}^{\infty}e^{-xy}\sin(x)dx$? where the $y$ in the second integral is just some unknown parameter now, not related to the first integral.
Anyone seen this notation before?
$\int_{0}^{\infty}dy\int_{0}^{\infty}e^{-xy}\sin(x)dx$ means $\int_{0}^{\infty}\left(\int_{0}^{\infty}e^{-xy}\sin(x)dx\right)dy$. It is very common to put the $dx$ (or $dy$) at the end of an integral (thus signifying the end of the integrand). While still common, it is less common to do so in repeated integrals.
Probably, the purpose of this is to make it clearer which integral sign (with bounds) belongs to which variable. In this case both $x$ and $y$ go from $0$ to $\infty$, but that will not always be so, and then it is important to knwo which variable has which bounds. Sticking the $d$ right next to the $\int$ it belongs to makes that easier.