Solve this integral equation using Fourier transform

$$\int_{-\infty}^{\infty} \frac{f(t)}{(x-t^2)+a^2} dt= \frac{\sqrt{2} \pi}{x^2 + b^2}$$ for $b> a > 0 $

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1 Answer 1


I think that the solution in this way is impossible, so I modify the question to become

$$\int_{-\infty}^{\infty} \frac{f(t)}{(x-t)^2+a^2} dt= \frac{\sqrt{2} \pi}{x^2 + b^2}$$ for $b> a > 0 $

My Solution:-

Applying Fourier transform , we get

$$\sqrt{2\pi}F(k).T[\frac{1}{(x-t)^2+a^2}]=T[\frac{\sqrt{2} \pi}{x^2+b^2}]$$ $$\sqrt{2\pi}F(k).\sqrt{\frac {\pi}{2}}\frac{e^{-a |k|}}{a}=\sqrt{2} \pi\sqrt{\frac {\pi}{2}}\frac{e^{-b |k|}}{b}$$ $$F(k)=\sqrt{\pi}\frac{a}{b} e^{- |k|(b-a)}$$ Applying the inverse transform , we get $$f(x)=T^{-1}[\sqrt{\pi}\frac{a}{b} e^{- |k|(b-a)}]$$ $$f(x)=\sqrt{\pi}\frac{a}{b} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{- |k|(b-a)}e^{ikx} dk $$ $$f(x)=\frac{a}{b\sqrt{2}} \int_{-\infty}^{\infty} e^{- |k|(b-a)}e^{ikx} dk $$ $$f(x)=\frac{a}{b\sqrt{2}} [\int_{-\infty}^{0} e^{k(b-a)}e^{ikx} dk +\int_{0}^{\infty} e^{-k(b-a)}e^{ikx} dk]$$ $$f(x)=\frac{a}{b\sqrt{2}} [\int_{-\infty}^{0} e^{k[(b-a)+ix]} dk +\int_{0}^{\infty} e^{-k[(b-a)-ix]} dk]$$ $$f(x)=\frac{a}{b\sqrt{2}} [\int_{0}^{\infty} e^{-k[(b-a)+ix]} dk +\int_{0}^{\infty} e^{-k[(b-a)-ix]} dk]$$ $$f(x)=\frac{a}{b\sqrt{2}} [\frac{1}{(b-a)+ix}+\frac{1}{(b-a)-ix}]$$ $$f(x)=\frac{a}{b\sqrt{2}} [\frac{2(b-a)}{(b-a)^2+x^2}]$$ $$f(x)=\frac{a\sqrt{2}}{b} [\frac{(b-a)}{(b-a)^2+x^2}]$$

Is this true solution?

Is there a simplification of the final answer


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