Question

@UsukiDoll

Answer 1

\[
\left[\frac{1}{x^2+a^2}\right](\xi) = \frac{\sqrt{2\pi}}{2a}e^{-a|\xi|}
\]

Answer 2

\[
\left[\frac{1}{x^2+b^2}\right](\xi) = \frac{\sqrt{2\pi}}{2b}e^{-b|\xi|}
\]

Answer 3

So \[
\sqrt{2\pi}\hat f(\xi) = \sqrt{2\pi}\frac{\sqrt{2\pi}}{2b}e^{-b|\xi|}= \frac{2\pi}{2b}e^{-b|\xi|}
\]

Answer 4

\[
\sqrt{2\pi}\hat f(\xi)\hat g(\xi) =\frac{2\pi}{2b}e^{-b|\xi|}\hat g(\xi) = \left[\frac{1}{x^2+a^2}\right](\xi)
\]

Answer 5

\[
\frac{2\pi}{2b}e^{-b|\xi|}\hat g(\xi) = \left[\frac{1}{x^2+a^2}\right](\xi) = \frac{\sqrt{2\pi}}{2a}e^{-a|\xi|}
\]Getting rid of the middle part:\[
\frac{2\pi}{2b}e^{-b|\xi|}\hat g(\xi) =\frac{\sqrt{2\pi}}{2a}e^{-a|\xi|}
\]

Answer 6

\[
\hat g(\xi) =\frac{\sqrt{2\pi}}{2a}e^{-a|\xi|}\frac{2b}{2\pi }e^{b|\xi|} = \frac{b}{a\sqrt{2\pi}}e^{-(a-b)|\xi|}
\]

Answer 7

They wanted it to be in the form: \[
e^{-c|\xi|}
\]So the let \(c=a-b\).

Answer 8

ahhh I see thanks :D sorry I was away eating lunch but this helps. Thank you!