I have a previous exam here for my course (Provided by the professor) that requires me to do a Fourier Transform of the following equation.

Here is the function: $f(x) = e^{-a^{2}x^{2}}cos{(bx)}$

He reminds us that $\mathfrak{F} [e^{-a^{2}x^{2}}] = \frac{1}{\sqrt{2a}}e^{-\frac{\xi^{2}}{4a}} $

His solution looks like this:
$(\mathbf{step 1})$
$\mathfrak{F}[e^{-a^{2}x^{2}}cos{(bx)}]=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}e^{-a^{2}x^{2}}cos(bx)cos(\xi x)dx$
(meaning he is using the cosine transform because it is an even function The hint is the $x^{2}$ to see it is even)

$(\mathbf{step 2})$
$\mathfrak{F}[e^{-a^{2}x^{2}}cos{(bx)}]=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}e^{-a^{2}x^{2}}\frac{1}{2}[cos((\xi+b)x)+cos((\xi-b) x)]dx$
(he used the identity: $cos(u)cos(v) = \frac{1}{2}[cos(u-v)+cos(u+v)$)

$(\mathbf{step 3})$
from here he simply states in one line:

Can someone fill the gap between step 2 and step 3? or at least fill in the note required to make this last step?

My thoughts were on the shift property of fourier transforms:

$\mathfrak{F}[f(x-a)]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x-a)e^{-i\xi x}dx = e^{ia\xi}F(\xi)$

If not this then some complicated form of converting the cosines into exponential and combining the first time with that. Then putting it into the form of the hint he gave.

Any thoughts from the math.stackexchange.com community?

Thank you,



I don't think going from step (2) to step (3) is obvious. You can do it with a trick. I'm renaming the constants so that they're not confused with yours. For $\alpha > 0$ and $\beta > 0$, define: $$ F(\beta)=\int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x)\,dx. $$ Notice that $$ F'(\beta)=-\int_{0}^{\infty}e^{-\alpha x^{2}}x\sin(\beta x)\,dx = \int_{0}^{\infty}\frac{d}{dx}\left(\frac{1}{2\alpha}e^{-\alpha x^{2}}\right)\sin(\beta x)\,dx $$ Now you can integrate by parts to obtain $$ F'(\beta)= \left.\frac{1}{2\alpha}e^{-\alpha x^{2}}\sin(\beta x)\right|_{x=0}^{\infty}-\frac{\beta}{2\alpha}\int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x)\,dx = -\frac{\beta}{2\alpha}F(\beta). $$ This ODE gives $F(\beta)=Ce^{-\beta^{2}/4\alpha}$, where $C$ is a constant to be determined. Setting $\beta=0$ gives the constant $C$: $$ \begin{align} C = F(0) & = \int_{0}^{\infty}e^{-\alpha x^{2}}\,dx = \frac{1}{2}\int_{-\infty}^{\infty}e^{-\alpha x^{2}}\,dx = \frac{1}{2}\left[\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-\alpha x^{2}-\alpha y^{2}}\,dxdy\right]^{1/2} \\ & = \frac{1}{2}\left[\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^{2}}r\,dr\,d\theta\right]^{1/2}=\frac{1}{2}\left[\frac{\pi}{\alpha}\int_{0}^{\infty}e^{-\alpha r^{2}}(2\alpha r)\,dr\right]^{1/2}=\frac{\sqrt{\pi}}{2\sqrt{\alpha}} \end{align} $$ Finally, $$ \int_{0}^{\infty}e^{-\alpha x^{2}}\cos(\beta x)\,dx = F(\beta) =\frac{\sqrt{\pi}}{2\sqrt{\alpha}}e^{-\frac{\beta^{2}}{4\alpha}}. $$ In your case $\alpha = a$, $\beta=\xi\pm b$, and you have to multiply by $\sqrt{2}/\sqrt{\pi}$. So it does work out, but I don't see anything obvious here.

  • $\begingroup$ Thanks @T.A.E. ! I appreciate your work on this. I'm rusty on my differentiation within integrals. Since youre differentiating I feel like I'd see a product rule come out of the cosine and exponent? Something like this happening:$\frac{\partial\text{}}{\partial x}\left(e^{-\alpha x^2}\cos(\beta x)\right)=-e^{\alpha\left(-x^2\right)}(2\alpha x(\cos(\beta x))+\beta(\sin(\beta x)))$ Maybe i'm not seeing a step you did at that part. $\endgroup$ – Dr. Dan Apr 19 '14 at 0:53
  • 1
    $\begingroup$ When you differentiate $F(\beta)$ with respect to $\beta$, the $\cos(\beta x)$ inside the integral becomes $-x\sin(\beta x)$. You can then take the extra $x$ and put it with $e^{-\alpha x^{2}}$ to write $-xe^{-\alpha x^{2}}=\frac{1}{2\alpha}\frac{d}{dx}e^{-\alpha x^{2}}$ $\endgroup$ – DisintegratingByParts Apr 19 '14 at 1:23
  • $\begingroup$ Oh okay thanks. I missed that you were differentiating w.r.t. $\beta$. I really like this way of looking at it. I also agree nothing is quite obvious with these Fourier transforms. Thank you for this. Seeing it through someone else's eyes really helps. $\endgroup$ – Dr. Dan Apr 19 '14 at 1:42


The professor told me that since the general Fourier Transform is:

$\mathfrak{F}[f(x)] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{i\xi x}dx=\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}f(x)cos(\xi x)dx$ works when it is even.

If you look at the form of (step 2). it can be tought of this way: the $\xi$ in the above cosine transform can be $\xi+b$ in the first term and $\xi-b$ in the 2nd term.

$\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}e^{-a^{2}x^{2}}\frac{1}{2}[cos((\xi+b)x)+cos((\xi-b) x)]dx$

knowing this we can use his hint: $\mathfrak{F} [e^{-a^{2}x^{2}}] = \frac{1}{\sqrt{2a}}e^{-\frac{\xi^{2}}{4a}} $

and the answer is: $\mathfrak{F}[e^{-a^{2}x^{2}}cos{(bx)}]=\frac{1}{2\sqrt{2}a}e^{-\frac{(\xi+b)^{2}}{4a^{2}}}+e^{-\frac{(\xi-b)^{2}}{4a^{2}}}$

The unfortunate part is that we are using $\xi$ twice now but it just needs to be recognized that this is a constant.


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