Inverse fourier transform using tables

**math2011** · September 29th 2011, 04:58 AM

By using tables find the inverse Fourier transform of $\displaystyle e^{-\frac{k^2}{4} + \frac{2}{3} k}$ .

The definition of inverse Fourier transform of $F(k)$ in my lecture notes is $f(x) = \mathcal{F}^{-1} \{ F(k) \} = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx} F(k) dk$ .

I guess this would involve convolution as $\mathcal{F}^{-1} \{ e^{-\frac{k^2}{4}} \} = \sqrt{2} e^{-x^2}$ . What is $\mathcal{F}^{-1} \{ e^{\frac{2}{3} k} \}$ ? How do I get this from the Fourier transforms table?

The given solution is $\sqrt{2} e^{\frac{4}{9}} e^{-i \frac{4}{3} x - x^2}$ .

**CaptainBlack** · September 29th 2011, 06:09 AM

Originally Posted by math2011

By using tables find the inverse Fourier transform of $\displaystyle e^{-\frac{k^2}{4} + \frac{2}{3} k}$ .

The definition of inverse Fourier transform of $F(k)$ in my lecture notes is $f(x) = \mathcal{F}^{-1} \{ F(k) \} = \frac{1}{\sqrt{2\pi}} \int^\infty_{-\infty} e^{-ikx} F(k) dk$ .

I guess this would involve convolution as $\mathcal{F}^{-1} \{ e^{-\frac{k^2}{4}} \} = \sqrt{2} e^{-x^2}$ . What is $\mathcal{F}^{-1} \{ e^{\frac{2}{3} k} \}$ ? How do I get this from the Fourier transforms table?

The given solution is $\sqrt{2} e^{\frac{4}{9}} e^{-i \frac{4}{3} x - x^2}$ .

Complete the square of the exponent then use the appropriate translation theorem and the inverse transform of $e^{-w^2}$

CB

**math2011** · September 30th 2011, 05:22 AM

Thank you. That worked.

$e^{-\frac{k^2}{4} + \frac{2}{3} k} = e^{-(x - \frac{4}{3})^2 + \frac{4}{9}}$
$\mathcal{F}^{-1} \{e^{-(x - \frac{4}{3})^2 + \frac{4}{9}} \} = \sqrt{2} e^{\frac{4}{9}} e^{-x^2} e^{-\frac{4}{3} i x}$

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