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

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


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

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.

  • $\begingroup$ $\hat{f}(0)$ is just the Fourier transform of a Guassian evaluated at $-b$, which is what you want. Alternatively, you could have solved for $\hat{f}(-b)$ in solving your differential equation; all you need is the value at one point to solve. It works the same. $\endgroup$ Commented Jan 23, 2015 at 16:40
  • $\begingroup$ @T.A.E. I know what the result should be, but let's pretend for the moment that I didn't know. How would I then know to evaluate $\hat f(k)$ at $-b$? What I'm trying to do is to actually solve the integral, under the pretence that I have no idea what the final result is going to be. $\endgroup$ Commented Jan 23, 2015 at 16:51
  • $\begingroup$ It makes it Obviously easier to evaluate at $-b$, just from the form of $\hat{f}$. An ODE can be solved by specifying the value somewhere other than always at $0$. Otherwise, you've going to have a second Fourier transform to evaluate, and you end up back in the same place. $\endgroup$ Commented Jan 23, 2015 at 17:44
  • $\begingroup$ @T.A.E. Apologies, but I still don't see how evaluating $\hat f(-b)$ will make it easier? Ill get $$\hat f(k) = \hat f(-b) \exp \bigg(\frac{4\pi bk - k^{2}}{4\pi a}\bigg)$$ and $$\hat f(-b) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp(-\pi ab^{2} + 2\pi ib^{2} -ikb) dx$$ How is that easier to solve? $\endgroup$ Commented Jan 24, 2015 at 0:45
  • 1
    $\begingroup$ Actually, I was looking at a later expression. You should evaluate at $k=2\pi b$. Then your first expression following "We have that" $\hat{f}(2\pi b)$ leaves only the Gaussian integral to evaluate. That's the natural thing to do to get rid of the added term. $\endgroup$ Commented Jan 24, 2015 at 6:51

2 Answers 2


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.

  • 2
    $\begingroup$ Yeah I did that previously, but this time I specifically wanted to solve this problem by differentiating under the integral and using integration by parts. Thanks for the reply though (+1). $\endgroup$ Commented Jan 23, 2015 at 16:28

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 $b$.


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