# Fourier Transform of $f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$

I was trying to take the FT of

$$f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$$

This is just the shifting rule applied to the FT of

$$g(x) = \exp(-\pi ax^{2})$$

which is given by

$$\hat g(k) = \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{-k^{2}}{4\pi a} \bigg)$$

Hence, we should get

$$\hat f(k) = \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{- (k – 2\pi b)^{2}}{4\pi a} \bigg)$$

However, when trying to derive this result I got stuck and if someone could help me I would really appreciate it.

We have that

$$\hat f(k) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx$$

Differentiating wrt to $k$ and integrating by parts, we find

\begin{align} \hat f'(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) (-ix) \exp(-ikx) dx \\ &= \frac{i}{\sqrt{2\pi}} \int_{-\infty}^{\infty} -x \exp(-\pi ax^{2}) \exp(i(2 \pi b – k)x) dx \\ \end{align}

with

\begin{align} u &= \exp(i(2 \pi b – k)x) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \frac{\exp(-\pi ax^{2})}{2\pi a} \\ u’ &= i(2 \pi b – k)\exp(i(2 \pi b – k)x) \ \ \ \ \ v’ = -x \exp(-\pi ax^{2}) \end{align}

Hence we get

\begin{align} \hat f'(k) = \frac{i}{\sqrt{2\pi}} \bigg[\frac{\exp(-\pi ax^{2}) \exp(i(2 \pi b – k)x)}{2\pi a}\Biggr \rvert_{-\infty}^{\infty} \\ – \bigg(\frac{i(2 \pi b – k)}{2\pi a} \bigg) \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ \end{align}

noting that the evaluated expression equals $0$. Therefore

\begin{align} \hat f'(k) &= \frac{i}{\sqrt{2\pi}} \bigg[- \bigg(\frac{i(2 \pi b – k)}{2\pi a} \bigg) \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ &= \frac{2\pi b – k}{2\pi a} \bigg[\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) \exp(-ikx) dx \bigg] \\ &= \frac{2\pi b – k}{2\pi a} \hat f(k) \end{align}

Solving, we find

\begin{align} \hat f(k) &= \hat f(0) \exp \bigg(\frac{4\pi bk – k^{2}}{4\pi a} \bigg) \\ &= \hat f(0) \exp \bigg(\frac{- (k – 2\pi b)^{2} + 4\pi^{2}b^{2}}{4\pi a} \bigg) \end{align}

The only way I can see to get the correct result is if

\begin{align} \hat f(0) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ax^{2} + 2\pi ibx) dx \\ &= \frac{1}{\sqrt{2\pi a}} \exp \bigg(\frac{-4\pi^{2}b^{2}}{4\pi a} \bigg) \end{align}

and the only way I thought I may be able to solve the integral for $\hat f(0)$ is to change to polar coordinates.

So, does anyone have suggestions on how to solve the $\hat f(0)$ integral?

#### Solutions Collecting From Web of "Fourier Transform of $f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$"
$$\exp \left(ax^2 +bx\right) = \exp \left(\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2 -\frac{b^2}{4a} \right),$$ for given constants $a$ and $b$.
If your goal to find the Fourier transform of the function $f(x) = e^{-\pi ax^{2} + 2\pi ibx}$ then here is an easier approach
$$F(k) = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + 2\pi ibx}e^{-ikx}dx = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + (2\pi b-k)ix}dx.$$