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?

Thanks for your help.

Solutions Collecting From Web of "Fourier Transform of $f(x) = \exp(-\pi ax^{2} + 2\pi ibx)$"

Hint to solve the integral:

$$ \exp \left(ax^2 +bx\right) = \exp \left(\left(\sqrt{a}x+\frac{b}{2\sqrt{a}}\right)^2 -\frac{b^2}{4a} \right),$$ 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

$$ F(k) = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + 2\pi ibx}e^{-ikx}dx = \int_{-\infty}^{\infty} e^{-\pi ax^{2} + (2\pi b-k)ix}dx. $$

To evaluate the last integral complete the square and then use Gaussian integrals.