Show that the solution to the following Neumann problem \begin{cases} u_{xx} + u_{yy} =0 \hspace{0.5cm} -\infty <x<\infty, y>0\\ u_{y}(x,0) = f(x), \hspace{0.5cm} -\infty <x<\infty\\ u(x,y)\hspace{0.3cm} \text{and}\hspace{0.3cm}u_{y}(x,y) \to 0 \hspace{0.3cm}\text{as}\hspace{0.3cm} (x^2 + y^) \to \infty \end{cases} where the function $f$ is piecewise smooth and absolutely integrable in $(-\infty , \infty)$ , can be written as $$u(x,y) = c + \frac{1}{2\pi}\int_{-\infty}^{\infty}{f(\xi)\ln[y^2 + (x-\xi)^2]} d\xi$$ where $c$ is an arbitrary constant.
My attempt:
Applying Fourier transform on the variable $x$ to the PDE $u_{xx}+u_{yy} =0$ and using the property of linearity, we have $$\mathcal{F}[u_{xx}] + \mathcal{F}[u_{yy}] =0$$ it follows that $$\frac{d^2}{dy^2}\hat{u} - \xi^2 \hat{u} =0$$ We then have the problem of ODE \begin{cases} \frac{d^2}{dy^2}\hat{u} - \xi^2 \hat{u} =0\\ \frac{d}{dy}\hat{u}(\xi,0) = \hat{f}(\xi) \end{cases}
The general solution to this problem is $$\hat{u}(\xi,y) = A(\xi)e^{\xi y} + B(\xi)e^{-\xi y}$$
Since $u$ is bounded when $y \to \infty$, its transform must be bounded when $y \to \infty$. This implies that $A(\xi) = 0$ for $\xi> 0$. If $\xi <0$ then $B(\xi) = 0$. So $$ \hat{u}(\xi,y) = Ke^{-|\xi| y} \hspace{0.3cm} \text{where $K$ is a constant}$$ Then $$\frac{d}{dy}\hat{u}(\xi,y) = -K|\xi|e^{-|\xi|y}$$
$$\Rightarrow \frac{d}{dy}\hat{u}(\xi,0) = -K|\xi| = \hat{f}(\xi) \Rightarrow K = -\frac{\hat{f}(\xi)}{|\xi|}$$
$$\therefore \hat{u}(\xi,y) = -\frac{\hat{f}(\xi)}{|\xi|}e^{-|\xi|y} = -\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\frac{f(x)e^{-|\xi|y}e^{-i\xi x}}{|\xi|}}dx$$
Now, applying inverse transform to both sides, we have
\begin{align} u(x,y) &= \frac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}{\left[- \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{\frac{f(\tau)e^{-|\xi|y}e^{-i\xi \tau}}{|\xi|}}d\tau \right]e^{i\xi x}}d\xi\\ &= \frac{1}{2\pi}\int_{-\infty}^{\infty}{f(\tau)}\int_{-\infty}^{\infty}{\frac{-e^{\left(\xi \left[i(x-\tau)\right]-|\xi|y\right)}}{|\xi|}}d\xi d\tau \end{align}
However, I have not been able to obtain the expression mentioned in the statement. How can I get that? Have I made a mistake? Any help is appreciated