I have to solve the following exercise. Using the Fourier transform to solve the integral equation $$ \frac{1}{1+x^2}=\int_{-\infty}^{\infty}f(y)\frac{1}{x-y}dy. $$

Some hints?

  • $\begingroup$ Do you mean verify that you can find a function f so that this equality is true? $\endgroup$ – Paul Jan 24 '18 at 12:35
  • $\begingroup$ Yes, I have to find $f(y)$. $\endgroup$ – Jeji Jan 24 '18 at 12:37
  • $\begingroup$ You should add a precise definition: due to the singularity at $y=x$, the RHS doesn't make sense as an ordinary integral. It may be meant as a Cauchy principal value, but in that case, you should add (or link to) the correct definition. Maybe you don't like doing any research for your exercises, but why do you expect others to be eager to do that for you? $\endgroup$ – user436658 Jan 24 '18 at 12:52
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    $\begingroup$ Fourier transform both sides. The right hand side is then a transform of a convolution. Use your FT table and get an algebraic equation for F(w), the transform of f. $\endgroup$ – Paul Jan 24 '18 at 12:52

Take $h(x)={1\over x}$ therefore $$f(x)*h(x)=\frac{1}{1+x^2}$$where $*$ denotes convolution operator. Taking Fourier transform from both sides we get:$$F(\omega)H(\omega)=\pi e^{-|\omega|}\to F(\omega)=isgn(\omega)e^{-|\omega|}=ie^{-\omega}u(\omega)-ie^{\omega}u(-\omega)\to \\f(x)=-\frac{1}{\pi}\frac{x}{1+x^2}$$

  • $\begingroup$ The student asked for a hint. $\endgroup$ – Paul Jan 24 '18 at 13:55

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