I have a Fourier transform to complete with the definition of the Fourier Transform.
Let $\phi$ be defined as follows. $$\tag{1} \phi(x) = Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x} $$
I must complete the Fourier transform of the function.
The definition of a Fourier transform is as follows.
$$\tag{2} \hat f(k) = \frac{1}{\sqrt{2\pi}}\int f(x) e^{-ikx}dx $$
To compute the Fourier transform we must evaluate the following integral with $\phi$ substituted into (2).
$$\tag{3} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int \phi(x) e^{-ikx}dx $$
$$\tag{4} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x} e^{-ikx}dx $$
I have tried completing this integral with completion of squares. I cannot find a way to finish this integral. How can I solve this integral?
First, enforce the substitution $x-x_0\to x$ so that
$$\begin{align} \int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx&=e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-i(k-k_0)x}\,dx \end{align}$$
Then, enforce the substitution $x/\alpha\to x$ so that
$$e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-ikx}\,dx=\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx$$
Completing the square reveals
$$\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx$$
Enforcing the substitution $x-i(k-k_0)\alpha/2\to x$ yields
$$\begin{align}\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx&=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\\\\ &\times \int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx\end{align}$$
Applying Cauchy's Integral Theorem, we can deform the contour back onto the real line to obtain
$$\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx=\alpha e^{-((k-k_0)\alpha/2)^2}\underbrace{\int_{-\infty}^{\infty} e^{-x^2}\,dx}_{=\sqrt\pi}$$
Putting it all together, we find that
$$\int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx=\alpha e^{-i(k-k_0)x_0}\sqrt\pi e^{-((k-k_0)\alpha/2)^2}$$