How to calculate $\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2+ik_0x}e^{-ikx}dx$?

Question

$$A(k)=\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}u(x,0)e^{-ikx}dx$$ When $$u(x,0)=e^{-x^2+ik_0x}$$ We get $$A(k)=\frac{1}{\sqrt2}e^{-\frac{(k-k_0)^2}{4}}$$ How to get A(k)? I am stuck at here, $$ A(k) =\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2+ik_0x}e^{-ikx}dx =\frac{1}{\sqrt{2\pi}}\int^∞_{-∞}e^{-x^2}\int^∞_{-∞}e^{ik_0x-ikx}dx=\frac{1}{\sqrt{2\pi}}\sqrt{\pi}[\frac{e^{ik_0x-ikx}}{ik_0-ik}]^∞_{-∞}dx=\frac{1}{\sqrt{2}}[\frac{e^{ik_0x-ikx}}{ik_0-ik}]^∞_{-∞}dx$$

Answer 1

Let $a=k_{0}-k$ and \begin{align} I &= \int\limits_{-\infty}^{\infty} \mathrm{e}^{-x^{2}+iax} dx \\ &= \mathrm{e}^{-a^{2}/4} \int\limits_{-\infty}^{\infty} \mathrm{e}^{-(x-ia/2)^{2}} dx \end{align} we completed the square in the exponent.

\begin{align} I_{1} &= \int \mathrm{e}^{-(x-ia/2)^{2}} dx \\ &= \int \mathrm{e}^{-y^{2}} dy \\ &= \frac{\sqrt{\pi}}{2} \mathrm{erf}(y) \\ &= \frac{\sqrt{\pi}}{2} \mathrm{erf}\left(x-i\frac{a}{2} \right) \end{align}

Applying limits to $I_{1}$, we obtain \begin{align} \int\limits_{-\infty}^{\infty} \mathrm{e}^{-(x-ia/2)^{2}} dx &= \frac{\sqrt{\pi}}{2} \mathrm{erf}\left(x-i\frac{a}{2} \right) \Big|_{-\infty}^{\infty} \\ &= \frac{\sqrt{\pi}}{2} \Big[\lim_{x \to \infty} \mathrm{erf}\left(x-i\frac{a}{2} \right) - \lim_{x \to -\infty} \mathrm{erf}\left(x-i\frac{a}{2} \right) \Big] \\ &= \sqrt{\pi} \end{align}

Thus \begin{equation} I = \sqrt{\pi} \mathrm{e}^{-a^{2}/4} \end{equation} and \begin{equation} A(k) = \frac{1}{\sqrt{2}} \mathrm{e}^{-(k_{0}-k)^{2} /4} \end{equation}