Any clue how to solve this convolution integral?

Question

With other words: find a (closed) expression for $\;\overline{\mbox{sinc}}(x)$ . $$ \overline{\mbox{sinc}}(x) = \int_{-\infty}^{+\infty} \frac{\sin(\omega\xi)}{\omega\xi} e^{-\frac{1}{2}(x-\xi)^2/\sigma^2}\,d\xi $$

Answer 1

The convolution of two functions is equal to the inverse FT of the product of the respective FTs.

$$\int_{-\infty}^{\infty} dx \frac{\sin{\omega x}}{\omega x} e^{i k x} = \begin{cases}\pi/\omega & |k| \lt \omega \\ 0 & |k| \gt \omega \end{cases} $$

$$\int_{-\infty}^{\infty} dx e^{-x^2/(2 \sigma^2)} e^{i k x} = \sqrt{2 \pi} \sigma \, e^{-\sigma^2 k^2/2} $$

so that

$$\begin{align}\frac{\omega}{\pi} \int_{-\infty}^{\infty} dx' \frac{\sin{\omega x'}}{\omega x'} e^{-(x-x')^2/(2 \sigma^2)} &= \frac{1}{2 \pi} \sqrt{2 \pi} \sigma \int_{-\pi/\omega}^{\pi/\omega} dk \, e^{-\sigma^2 k^2/2} e^{-i k x}\\ &= \frac{\sigma}{\sqrt{2 \pi}} \int_{- \pi/\omega}^{\pi/\omega}dk \, e^{-(\sigma^2/2) (k^2 - i 2 k x/\sigma^2-x^2/\sigma^4)} e^{-x^2/(2 \sigma^2)}\\&= \frac{\sigma}{\sqrt{2 \pi}} \int_{- \pi/\omega}^{\pi/\omega}dk \, e^{-(\sigma^2/2) (k-i x/\sigma^2)^2}\\&=\frac{1}{\sqrt{\pi}}\int_{(\sigma/\sqrt{2})(-\pi/\omega + i x/\sigma)}^{(\sigma/\sqrt{2})(\pi/\omega + i x/\sigma)} du \, e^{-u^2} \\&=\frac12 \left [ \operatorname*{erf}{\left (\frac{\pi \sigma}{\sqrt{2} \omega}+i \frac{x}{\sqrt{2}} \right )}-\operatorname*{erf}{\left (-\frac{\pi \sigma}{\sqrt{2} \omega}+i \frac{x}{\sqrt{2}} \right )} \right ] \end{align}$$

Thus,

$$\begin{align}\int_{-\infty}^{\infty} dx' \frac{\sin{\omega x'}}{\omega x'} e^{-(x-x')^2/(2 \sigma^2)} &= \frac{\pi}{2 \omega} \left [ \operatorname*{erf}{\left (\frac{\pi \sigma}{\sqrt{2} \omega}+i \frac{x}{\sqrt{2}} \right )}-\operatorname*{erf}{\left (-\frac{\pi \sigma}{\sqrt{2} \omega}+i \frac{x}{\sqrt{2}} \right )} \right ]\\&= \frac{\pi}{\omega} \Re{\left [\operatorname*{erf}{\left (\frac{\pi \sigma}{\sqrt{2} \omega}+i \frac{x}{\sqrt{2}} \right )} \right]} \end{align}$$