Fourier transform $\frac{1}{t^2+6t+10}$

Question

I want to solve the equation $$\int^{\infty}_{-\infty}\frac{u(s)}{1+4(t-s)^2} ds= \frac{1}{t^2+6t+10}$$ using $$\mathcal{F}[\frac{1}{1+t^2}] = \pi e^{-|w|} $$

The left hand side is a convolution which I can solve using the hint. However I don't know how I Fourier Transform the right side.

When I fractionize the term, I get $$\frac{1}{t+(3+i)}*\frac{1}{t+(3-i)}$$ Now I don't know were to go with it. I can see that the term is an absolute square of either of the terms. My idea is to continue with the parseval identity, thought I still don't know exactly how to use it. Any help would be appreciated.

How does the Fourier transform interact with translations? If $g(x) = f(x-a)$, how are the Fourier transforms of $f$ and $g$ related? — Daniel Fischer, Nov 15 '15 at 16:08
That would be $G = e^{-aiw} * F$, thought I don't know if that's still the case if a is complex. — Shrubber, Nov 15 '15 at 16:40
For complex $a$, $g$ would generally not be defined. But $t^2 + 6t + 10 = (t+3)^2 +1$, try $a = -3$. — Daniel Fischer, Nov 15 '15 at 16:44

Jack D'Aurizio · Answer 1 · 2015-11-15T16:47:29.320

Let $v(s)=\frac{1}{1+4s^2}$. Your equation can be written as: $$ (u*v)(t) = \frac{1}{t^2+6t+10}\tag{1} $$ and by switching to Fourier transforms (defined through $\mathcal{F}(u)(\xi)=\widehat{u}(\xi)=\int_{-\infty}^{+\infty}u(x)e^{-2\pi i x \xi}\,dx$) we have: $$ \widehat{u}(\xi)\cdot \frac{\pi}{2} e^{-\pi|\xi|}= \widehat{u}(\xi) \cdot \widehat{v}(\xi) = \mathcal{F}\left(\frac{1}{(t+3)^2+1}\right)(\xi)=\pi e^{-2\pi|\xi|}\cdot e^{6\pi i\xi}\tag{2}$$ that implies: $$ \widehat{u}(\xi) = 2e^{-\pi|\xi|}\cdot e^{6\pi i\xi} \tag{3}$$ and by Fourier inversion: $$ \color{red}{u(t)} = \frac{2}{\pi}\cdot \frac{2}{1+(2\cdot(t+3))^2} = \color{red}{\frac{4}{\pi}\cdot\frac{1}{1+(2t+6)^2}}.\tag{4}$$

Fourier transform $\frac{1}{t^2+6t+10}$

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