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Inverse Fourier Transform of:

$$\mathfrak{F}^{-1} \left \{ e^{-\frac{x^2}{2}}{\frac{sinx}{x}} \right \} $$

by using convolution theorem.

Since Fourier Transform convolution turns into multiplication - same property holds also for inverse. So, I am thinking that:

$$\mathfrak{F}^{-1} \left \{ e^{-\frac{x^2}{2}}{\frac{sinx}{x}} \right \}= \mathfrak{F}^{-1} \left \{ e^{-\frac{x^2}{2}} \right \} \mathfrak{F}^{-1} \left \{{\frac{sinx}{x}} \right \}$$

because of linearity.

Then first expression it is Gaussian and also its inverse is the same - so should I find only for second expression $\frac{sinx}{x}$ its Inverse Fourier Transform?

Melina
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  • On the right side of that equation, you should have the convolution of two functions. That's not what you have written now. – Ben Grossmann Feb 03 '16 at 15:19

1 Answers1

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No, you need to actually evaluate the convolution of the inverse transforms, which looks like

$$\begin{align}\sqrt{\frac{\pi}{2}} \int_{-1}^1 dk' \, e^{-(k-k')^2/2} &= \sqrt{ \pi} \int_{(k-1)/\sqrt{2}}^{(k+1)/\sqrt{2}} du \, e^{-u^2} \\ &= \frac{\pi}{2} \left [\operatorname{erf}{\left (\frac{k+1}{\sqrt{2}}\right )} - \operatorname{erf}{\left (\frac{k-1}{\sqrt{2}} \right )} \right ] \end{align}$$

Ron Gordon
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