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I have to solve the equation $$\int_{\mathbb R} \frac{f(t)}{1+(x-t)^2} dt =\frac{\sin x}{x}.$$ I tried change of variables to make the $\frac{1}{1+(x-t)^2}$ part resemble $e^{h(x)}$ so I can use the inverse Fourier transform. But I can't get it right.

Is it the right way to approach this problem?

Any help would be highly appreciated!

Thank you!

Ludolila
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  • Is anything given about $;f;$? Continuous, derivable...? The part you want to make ressemble $;e^{h};$ seems to me more like an arctangent... – DonAntonio Jan 28 '14 at 12:18
  • You seem to have $\left(f\ast \frac{1}{1+x^2}\right)(x)=\mathrm{sinc}(x)$. This does give you the fourier transform of $f$, assuming you have a reason to assume it exists (as per @DonAntonio's questions). Can you find the inverse transform? – Jonathan Y. Jan 28 '14 at 13:09
  • @DonAntonio: no, no additional information has been given. – Ludolila Jan 28 '14 at 13:53
  • @JonathanY. I think I understand. Do you mean that I should apply Fourier transform to both sides of the convolution to get: $\hat{f} \cdot \widehat{\frac{1}{1+x^2}}=\widehat{sinc (x)}$? Then I can calculate $\hat{f}$ and use inverse transform? – Ludolila Jan 28 '14 at 13:59
  • That's what I meant, but I thought that was your own suggestion originally. – Jonathan Y. Jan 28 '14 at 14:00
  • @JonathanY.: No, originally I was thinking to do something else. But your way is better! I solved it and got $f(x)=-\frac{1}{\pi (1+x^2)} (1+e\cdot(i+x)\cdot \sin (x))$. Does it look anything like the answer? Tried to check with wolfram, but it couldn't compute the integral... – Ludolila Jan 28 '14 at 14:58

1 Answers1

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I don't want to leave this unanswered, so I will answer my own question.

Note that the left side is in fact a convolution: $$\left( f \star \frac{1}{1+x^2} \right)(x)=\int_{\mathbb R}f(x)\cdot \frac{1}{1+(x-t)^2} dt $$

Thus $\left( f \star \frac{1}{1+x^2} \right)(x)=\frac{\sin x}{x}$ and therefore $\widehat{f}\cdot \widehat{\frac{1}{1+x^2}}=\widehat{\frac{\sin x}{x}}$.

We know (standard calculation) that $\widehat{\frac{1}{1+x^2}}(\omega)=\pi \cdot e^{-|\omega|}$ and $\widehat{\frac{\sin x}{x}}(\omega)= \left\{ \begin{array}{ll} \pi & |\omega| \leq 1 \\ 0 & |\omega|>1 \end{array} \right.$

Therefore $\widehat{f}(\omega) = \left\{ \begin{array}{ll} e^{|\omega|} &|\omega| \leq 1 \\ 0 & |\omega| > 1 \end{array} \right.$. Using inverse Fourier transform we get $$f(x)=\frac{-1}{\pi(1+x^2)}\left(1+e\cdot(i+x)\cdot \sin(x) \right) .$$

Ludolila
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