Integral Equation $\int_{-\infty}^\infty \frac{f(y)}{1+(x-y)^2}\mathrm dy =0 \quad \forall x$ I want to calculate $f(y)$ such that 
$$\int_{-\infty}^\infty \frac{f(y)}{1+(x-y)^2}\mathrm dy =0 \quad \forall x$$ 
Can we prove that the solution to this problem is $f(y)=0?$
Thank you so much
 A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$\widetilde{\vphantom{\large A}\cdots}$ ( tilde's ) are Fourier Transforms.
\begin{align}
0
&=
\int_{-\infty}^{\infty}\dd x\,\expo{-\ic kx}
\int_{-\infty}^\infty {{\rm f}\pars{y} \over 1 + \pars{x - y}^{2}}\,\dd y
=
\int_{-\infty}^{\infty}\dd x\,{\rm f}\pars{y}\expo{-\ic ky}
\int_{-\infty}^\infty {\expo{-\ic kx} \over \pars{x - \ic}\pars{x + \ic}}\,\dd x
\\[3mm]&=
\tilde{\rm f}\pars{k}\pi\expo{-\verts{k}}
\quad\imp\quad
\tilde{\rm f}\pars{k} = 0
\quad\imp\quad
{\rm f}\pars{x} = 0
\end{align}
