I've tried looking at
$$\mathcal{F}\left(\sin(t)e^{-x^2}\right) = \dfrac{1}{\sqrt{\tau}}\int_{-\infty}^\infty \sin(x)e^{-x^2}e^{-iwx}\,\mathrm{d}x$$ but this seems like a dead end, as I don't know any easy ways to integrate $e^{-x^2}$ to make use of partial integration

I thought there might be some trick involving the derivatives of $f(x)$, as there is with $e^{-x^2}$, but I can't see any obvious solutions using that either, as $f(x ) = \dfrac{\cos(x)e^{-x^2}-f'(x)}{2x}$.

  • 1
    $\begingroup$ Hint: Only $\sin(wx)$ contributes, for parity reasons. Next, $\sin(x)\sin(wx)$ is a difference of cosines. Next, every cosine is the real part of a complex exponential. Finally, $$\int_{-\infty}^\infty e^{isx}e^{-x^2}dx=e^{-s^2/4}\int_{-\infty}^\infty e^{-(x-is/2)^2}dx=\ldots$$ $\endgroup$ – Did Sep 19 '16 at 12:29
  • $\begingroup$ The result should not be very simple (it does not appear for example in Gradshteyn & Ryzhik) $\endgroup$ – Jean Marie Sep 19 '16 at 12:51
  • $\begingroup$ @JeanMarie Please... $\endgroup$ – Did Sep 19 '16 at 12:57

Hints (although already indicated by Did): You may write $\sin x = \frac{e^{ix}-e^{-ix}}{2i}$ and then use $$e^{-z^2}\int_{-\infty}^{\infty} e^{-x^2 - 2zx} dx =\int_{-\infty}^{\infty} e^{-(x+z)^2} dx = \sqrt{\pi}$$ valid for any complex number $z$.

  • $\begingroup$ It is impossible: the result cannot be a constant. The mistake (very understandable) comes from the fact that you are not allowed to do a change of variable in this way for what has to be considered as a complex integral. $\endgroup$ – Jean Marie Sep 19 '16 at 12:53
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    $\begingroup$ @JeanMarie "the result cannot be a constant" Sorry but where do you see that this states the result is constant? The change of variable the OP recommends is valid (and ultra standard in this context, if I may add). $\endgroup$ – Did Sep 19 '16 at 12:55
  • $\begingroup$ @Did You are right, my remark is false. But, do you agree with me that it means that, by multiplying LHS and RHS by $e^{z^2}$ the result is "something like" (up to constants etc...) $\sqrt{\pi}e^{z^2}$, which sounds me a rather strange result... $\endgroup$ – Jean Marie Sep 19 '16 at 13:28
  • $\begingroup$ @JeanMarie Why "strange"? Note that the identity in this post will be used for some purely imaginary numbers $z$, hence $z^2$ real, $z^2<0$. $\endgroup$ – Did Sep 19 '16 at 13:57
  • $\begingroup$ @Did finally, after many doubts and stupid remarks (:, I managed to find my way to a solution (see the last lines aftr my answer). $\endgroup$ – Jean Marie Sep 21 '16 at 13:53

Let us denote by FT the Fourier Transform $f \longrightarrow F$ defined by

$$F(\xi):=\int_{-\infty}^{+\infty}f(t)e^{-2 i \pi \xi t}dt$$

Let us write the initial function as the product:

$$f(t)g(t) \ \ \text{where} \ \ f(t):=\frac{\sin(t)}{t} \ \ \text{and} \ \ g(t):=te^{-t^2}$$

Let us take the convention $a=\frac{1}{2 \pi}.$ The Fourier Tranforms of $f$ and $g$ are:

  • $F(\xi):=FT(f)(\xi)=\pi C_{[-a,a]} \ (\xi)$ where $C_I$ means "characteristic function of interval $I$" (classical FT of "sinc" function) and

  • $G(\xi):=FT(g)(\xi)=-i \pi \xi e^{-(\pi \xi)^2} \ $ using a result given in (Fourier transform of $te^{-t^2}$?)

Using the convolution theorem, one gets:

$H(\xi):=(F\star G)(\xi)=\pi C_{[-a,a]}(\xi) \ \star \ -i\pi \xi e^{-(\pi \xi)^2}$

Let us differentiate $H$ by convolving $F'$ with $G$ under the form:

$H'(\xi):=(F'\star G)(\xi)=\pi \left(\delta(\xi+a)-\delta(\xi-a)\right) \star -i\pi \xi e^{-(\pi \xi)^2}$

Otherwise said :

$H'(\xi):=(F'\star G)(\xi)=-i \pi^2 \left( (\xi+a) e^{-\pi^2 (\xi+a)^2}-(\xi-a) e^{-\pi^2 (\xi-a)^2} \right)$

finally giving, by taking a primitive (with the help of Mathematica), the following Fourier transform:

$$H(\xi)= i \frac{\sqrt{\pi}}{2} (1-e^{2 \pi \xi})e^{-(\frac{1}{2}+\pi \xi)^2}=-i\sqrt{\pi}e^{-1/4}e^{-(\pi \xi)^2}\sinh(\pi \xi)$$

There is an interesting remark with the second expression of $H(\xi)$ : the Fourier transform of $\sin(t)e^{-t^2}$ bears (up to some constants) a striking similitude with the original function, a $\sinh$ taking the place of the $\sin$ function.


Refering to my other answer, since $$\mathcal{F}\{\sin(x)\}=\frac{1}{2i}\mathcal{F}\{e^{ix}-e^{-ix}\}=\pi i[\delta(\omega+1)-\delta(\omega-1)]$$ the FT of $e^{-x^2}\sin(x)$ is $$\pi i [F(\omega+1)-F(w-1)]$$ where $$F(\omega)=\mathcal{F}\{e^{-x^2}\}$$


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