Question: Show that satisfies $\mathit{u}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\frac{\mathbf{y}}{\mathbf{\pi }}{\mathbf{\int }}_{\mathbf{-}\mathbf{\infty }}^{\mathbf{\infty }}\frac{\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{d}\mathbf{t}}{{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{t}\mathbf{)}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}^{\mathbf{2}}}\mathbf{satisfies}\mathbf{}{\mathit{u}}_{\mathbf{x}\mathbf{x}}\mathbf{+}{\mathit{u}}_{\mathbf{y}\mathbf{y}}\mathbf{=}\mathbf{0}$.

Given that $u(\mathrm{x},\mathrm{y})=\frac{\mathrm{y}}{\mathrm{\pi }}{\int }_{-\infty }^{\infty }\frac{\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}}{{(\mathrm{x}-\mathrm{t})}^2+{\mathrm{y}}^2}$.

We know that$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{v}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{v}\mathbf{)}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\mathbf{(}\mathit{x}\mathbf{,}\mathit{u}\mathbf{)}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}$$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}{\mathbf{\int }}_{\mathbf{u}\left(x\right)}^{\mathbf{v}\left(x\right)}\mathit{f}\left(x,t\right)\mathit{d}\mathit{t}\mathbf{=}\mathit{f}\left(x,v\right)\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{f}\left(x,u\right)\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\mathbf{+}{\mathbf{\int }}_{\mathbf{u}}^{\mathbf{v}}\frac{\mathbf{\partial }\mathbf{f}}{\mathbf{\partial }\mathbf{x}}\mathit{d}\mathit{t}$

Using the formula

$$\begin{gathered} u_x=\frac{y}{\pi }\left[\frac{f\left(\infty \right)}{{\left(x-\infty \right)}^2.y^2}\frac{d}{dx}\left(\infty \right)-\frac{f\left(\infty \right)}{{\left(x+\infty \right)}^2+y^2}\frac{d}{dx}\left(-\infty \right)+{\int }_{-\infty }^{\infty }\frac{\partial }{\partial x}\left[\frac{f\left(t\right)}{{\left(x-t\right)}^2+y^2}\right]dt\right] \\ =\frac{y}{\pi }\left[0-0+{\int }_{-\infty }^{\infty }\frac{-2\left(x-t\right)f\left(t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^2}dt\right] \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{\left(x-t\right)f\left(t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^2}dt \end{gathered}$$

Solving further

localid="1659349480594" $$\begin{gathered} u_{xx}=\frac{-2y}{\pi }\left[\frac{\left(x-\infty \right)f\left(\infty \right)}{{\left(x-\infty \right)}^2+y^2}\frac{d}{dx}\left(\infty \right)-\frac{\left(x+\infty \right)f\left(\infty \right)}{{\left(x+\infty \right)}^2+y^2}\frac{d}{dx}\left(-\infty \right)+{\int }_{-\infty }^{\infty }\frac{\partial }{\partial x}\left[\frac{x-t}{{\left({\left(x-t\right)}^2+y^2\right)}^2}\right]f\left(t\right)dt\right] \\ =\frac{-2y}{\pi }\left[0-0+{\int }_{-\infty }^{\infty }\frac{{\left({\left(x-t\right)}^2+y^2\right)}^2\left(1\right)-\left(x-t\right)\left({\left(x-t\right)}^2+y^2\right)2\left(x-t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^4}f\left(t\right)dt\right] \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{\left({\left(x-t\right)}^2+y^2\right)\left[{\left(x-t\right)}^2+y^2-4{\left(x-t\right)}^2\right]}{{\left({\left(x-t\right)}^2+y^2\right)}^4}f\left(t\right)dt \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{y^2-3{\left(x-t\right)}^2}{{\left({\left(x-t\right)}^2+y^2\right)}^3}f\left(t\right)dt \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{y^2-3{\left(x-t\right)}^2}{{\left({\left(x-t\right)}^2+y^2\right)}^3}f\left(t\right)dt \end{gathered}$$

Solving further

localid="1659350725416" $$\begin{gathered} u\left(x,y\right)=\frac{y}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)}{{\left(x-t\right)}^2+y^2} \\ =\frac{y}{\pi }\left[\frac{f\left(\infty \right)}{{\left(x-\infty \right)}^2+y^2}\frac{d}{dy}\left(\infty \right)-\frac{f\left(-\infty \right)}{{\left(x-\infty \right)}^2+y^2}\frac{d}{dy}\left(-\infty \right)+{\int }_{-\infty }^{\infty }\frac{\partial }{\partial y}\left[\frac{1}{{\left(x-t\right)}^2+y^2}\right]f\left(t\right)dt\right] \\ +\frac{y}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)}{{\left(x-t\right)}^2+y^2} \\ {u}_y=\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{yf\left(t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^2}dt+\frac{1}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)dt}{{\left(x-t\right)}^2+y^2} \end{gathered}$$

Solving for$u_{yy}$

$$\begin{gathered} u_{yy}=\frac{-2y}{\pi }\left[\frac{yf\left(\infty \right)}{{\left({\left(x-\infty \right)}^2+y^2\right)}^2}\frac{d}{dy}\left(\infty \right)-\frac{yf\left(-\infty \right)}{\left({\left(x+\infty \right)}^2+y^2\right)}\frac{d}{dy}\left(-\infty \right)+{\int }_{-\infty }^{\infty }\frac{\partial }{\partial x}\left[\frac{yf\left(t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^2}\right]dt+\left(\frac{-2}{\pi }\right){\int }_{-\infty }^{\infty }\right] \\ =\frac{1}{\pi }\left[\frac{f\left(\infty \right)}{{\left(x-\infty \right)}^2+y^2}\frac{d}{dy}\left(\infty \right)+{\int }_{-\infty }^{\infty }\frac{\partial }{\partial y}\left(\frac{f\left(t\right)}{{\left(x-t\right)}^2+y^2}\right)dt\right] \\ =\frac{-2y}{\pi }\left[0-0+\left(\frac{\left({\left(x-t\right)}^2+y^2\right)\left(1\right)y^2\left({\left(x-t\right)}^2+y^2\right)2y}{{\left({\left(x-t\right)}^2+y^2\right)}^4}\right)\right]+\left[\frac{yf\left(t\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^2}\right]dt+\left(\frac{-2}{\pi }\right){\int }_{-\infty }^{\infty } \\ =\frac{1}{\pi }\left[0-0{\int }_{-\infty }^{\infty }\frac{-2y}{{\left({\left(x-t\right)}^2+y^2\right)}^2}f\left(t\right)dt\right] \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{\left({\left(x-t\right)}^2-3y^{2}f\left(t\right)\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^3}dt-\frac{4y}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)dt}{{\left({\left(x-t\right)}^2+y^2\right)}^2} \end{gathered}$$

$$\begin{gathered} u_{xx}+u_{yy}=\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{y^2-3(x-t)2}{{\left({\left(x-t\right)}^2+y^2\right)}^3}f\left(t\right)dt \\ -\frac{2y}{\pi }{\int }_{-\infty }^{\infty }\left[\frac{\left({\left(x-t\right)}^2+3y^2\right)}{{\left({\left(x-t\right)}^2+y^2\right)}^3}f\left(t\right)dt-\frac{4y}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)dt}{{\left({\left(x-t\right)}^2+y^2\right)}^2}\right] \\ =\frac{-2y}{\pi }{\int }_{-\infty }^{\infty }\frac{-2\left[{\left(x-t\right)}^2+y^2\right]}{{\left({\left(x-t\right)}^2+y^2\right)}^3} \\ =-\frac{4y}{\pi }{\int }_{-\infty }^{\infty }\frac{f\left(t\right)dt}{{\left({\left(x-t\right)}^2+y^2\right)}^2} \\ =0 \end{gathered}$$

Hence$u_{xx}+u_{yy}=0.$ .