Q:

Fourier Transform of Mixed Partial Derivative

My task at hand is to compute the Fourier transform of a mixed partial derivative given by:
$$\mathscr F\left(\frac{\partial^2 f}{\partial x\partial y} \right)(u,v)= \iint _{\Bbb R^2} e^{-i2\pi (ux+vy)}\frac{\partial^2 f}{\partial x\partial y}(x,y)  dxdy$$
Now, I have been advised that I simply need to use integration by parts to solve this.  However, my trouble comes in identifying the specific variables I need to isolate and apply with the integration by parts formula.  Any advice would be much appreciated.  Thank you.  

A:

Assume $f$, together with $\partial f/\partial x$ and $\partial f/\partial y$ vanish at infinity.  Then we integrate by parts with respect to $x$ first and the $y$ then.  We obtain
$$\mathscr F \left(\frac{\partial^2 f}{\partial x \partial y} \right)(u,v) = \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} e^{-2\pi i(ux+vy)} \frac{\partial^2 f}{\partial x \partial y}dx dy $$
$$ =- \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (-2\pi i u) e^{-2\pi i (ux+vy)}\frac{\partial f}{\partial y}dx dy = + \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} (-2\pi i u)(-2\pi i v) e^{-2\pi i (ux+vy)}f(x,y) dx dy $$
$$= - 4 \pi^2 uv  \mathscr F(f)(u,v).$$

