Use Fourier transform to calculate double integral of harmonic function Let $$P_y(x)=\dfrac{1}{2\pi}\int_{-\infty}^\infty e^{-y|t|}e^{ixt}dt=\dfrac{1}{\pi}\dfrac{y}{x^2+y^2}.$$ 
Then $P_y(x)$ is harmonic in the upper half-plane $y>0$ and for $f\in L^1(\mathbb{R})$, $u(x,y)=P_y(x)\ast f(x)$ is also harmonic in the upper half-plane $y>0$.
Suppose $f$ belongs to the Schwartz class. How can we use the Fourier transform to calculate $$\int_0^\infty\int_\mathbb{R}|\nabla u(x,y)|^2ydxdy$$ in terms of $f$?
 A: Using Plancherel theorem for $y\not=0$
$$\int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 dx=\int_{\mathbb R} \left |\widehat{\frac{\partial u(\cdot,y)}{\partial x}}(\xi)\right |^2 d\xi=\int_{\mathbb R} |\xi|^2 \left |\widehat{u(\cdot,y)}(\xi)\right |^2 d\xi=2\pi\int_{\mathbb R} |\xi|^2  |\widehat{P_y} |^2 |\hat f|^2d\xi$$
From the definition of $P_y$ is clear that(using the inversion formula) $\widehat{P_y}=\frac{e^{-y|\xi|}}{\sqrt{2\pi}}$ so
$$\int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 dx=\int_{\mathbb R} |\xi|^2  e^{-2y|\xi|} |\hat f|^2d\xi$$
and hence
$$\int_0^\infty \int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 ydxdy=\int_0^\infty\int_{\mathbb R} |\xi|^2  e^{-2y|\xi|} |\hat f|^2d\xi y dy$$
Using Tonelli 
$$\int_0^\infty \int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 ydxdy=\int_{\mathbb R} |\xi|^2  \int_0^\infty e^{-2y|\xi|}ydy |\hat f|^2d\xi$$
and we have
$\int_0^\infty e^{-2y|\xi|}ydy=\frac{1}{4|\xi|^2}$ so
$$\int_0^\infty \int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial x}\right |^2 ydxdy=\frac 1 4\int_{\mathbb R} |\hat f|^2d\xi=\frac{\|f\|_{L^2(\mathbb R)}^2}{4}$$
on the other hand
$$\int_{\mathbb R} \left |\frac{\partial u(x,y)}{\partial y}\right |^2 dx=\int_{\mathbb R} \left |\widehat{\frac{\partial u(\cdot,y)}{\partial y}}(\xi)\right |^2 d\xi=2\pi \int_{\mathbb R} \left |\frac{\partial (\hat P_y \hat f)}{\partial y}(\xi)\right |^2 d\xi=2\pi \int_{\mathbb R} \left |\frac{\partial \hat P_y }{\partial y}(\xi)\right |^2 |\hat f|^2d\xi$$
and $\frac{\partial \hat P_y }{\partial y}=-\xi \hat P_y $ so where get the same integral.
Gathering all together
$$ \int_0^\infty\int_{\mathbb R} \left |\nabla u(x,y)\right |^2 ydxdy=\frac{\|f\|_{L^2(\mathbb R)}^2}{2}$$
