Fourier transformation of $\sin(ax)\exp(\frac{-x^2}{2})$

Question

I have been trying to calculate the Fourier transformation of $$\sin(ax)\exp\left(\frac{-x^2}{2}\right)$$ using $$\int_{\Bbb R} \cos(ax)\exp\left(\frac{-x^2}{2}\right)dx = \sqrt{2\pi}\exp\left(\frac{-a^2}{2}\right)$$, but haven't been able to succeed. I've mainly tried using integration by parts, but I don't feel like I'm getting somewhere.

Answer 1

Hint: $$\sin(ax) =\frac{1}{2i}(e^{aix}-e^{-aix})$$

Let $b= \pm a$ then using the parity argument of sine and cosine we get $$\mathcal{F}\left(e^{ibx}\exp\left(\frac{-x^2}{2}\right)\right)(y) = \frac{1}{2\pi} \int_{\Bbb R}e^{i(b-y)x}\exp\left(\frac{-x^2}{2}\right)dx \\= \frac{1}{2\pi} \int_{\Bbb R}\cos(\color{red}{b-y)}x)\exp\left(\frac{-x^2}{2}\right)dx =\color{red}{\frac{1}{\sqrt{2\pi}}\exp\left(\frac{-(b-y)^2}{2}\right)}$$

Answer 2

Hint: $e^{-\frac{x^2}{2}-ikx} = e^{-\frac{u^2}{2}-\frac{k^2}{2}}$ for $u = x+ik$. Then you can write $\sin(ax)$ as $\sin(au-b)$ and expand using the sum formula. You will have two term with $\sin(au)$ and $\cos(au)$. The $\sin$ term should be zero by symmetry.

You may have to integrate off the real axis.

Answer 3

This problem hinges on "complex" integrals having a zero part due to an odd integrand. We want to compute a Fourier transform, the imaginary number $iS(a,\,k):=\int \sin ax\exp-\frac{x^2}{2}i\sin kx dx$, from the real number $C(a,\,k):=\int \cos ax\exp-\frac{x^2}{2}\cos kx dx$. In particular $C\pm S=\int\cos(a\mp k)x\exp -\frac{x^2}{2}dx=\sqrt{2\pi}\exp-\frac{(a\mp k)^2}{2}$. Thus $iS=i\sqrt{8\pi}\exp-\frac{a^2+k^2}{2}\sinh ak$.