Use Fourier Transformï¼Œ$u(x,y) \rightarrow \hat{u} (k,y)$ $$-k^2\hat{u}+\hat{u}_{yy}=0\\\hat{u}(k,y)=A(k)e^{-|k|y}+B(k)e^{|k|y}$$

We discard second term because it is unbounded.$$\hat{u}(k,0) = A(k) = \hat{f}(k) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{1}{x^2 +1}e^{-ikx} dx = \frac{1}{2}e^{-|k|} \\u(x,y)=\frac{1}{2}\int_{-\infty}^{\infty} e^{-|k|y}e^{|-k|}e^{ikx} dk $$ Then the answer is same as Emily.

But for this one, from lecture notes, suppose we don't have a hint for $\hat{f}(k)$

We get IFT of $e^{-|k|y}$ is $ \frac{2y}{x^2+y^2}$ $$u(x,y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} f(x')\frac{2y}{((x-x')^2+y^2)}dx'\\= \frac{1}{\pi} \int_{-\infty}^{\infty} \frac{1}{(x'^2 +1)}\frac{y}{((x-x')^2+y^2)}dx'$$

This interval is supposed to be same as the correct answer. I am confused about this part.