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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?

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  • $\begingroup$ You were correct to pursue completing the square. You will need to evaluate a Gaussian integral thereafter. And to be rigorous, you need to deform the contour back to the real line by appealing to Cauchy's Integral Theorem. And that is about it. $\endgroup$ – Mark Viola Nov 20 '17 at 2:00
  • $\begingroup$ After completing the square, $\hat\phi$ will become the following integral. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2}}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ $\endgroup$ – Jeremy Nov 20 '17 at 2:33
  • $\begingroup$ There is an error in the previous comment. $\hat\phi$ will be defined as follows. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2})^2}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ $\endgroup$ – Jeremy Nov 20 '17 at 2:39
  • $\begingroup$ After evaluating the Gaussian integral $\hat\phi$ is defined as follows. $$\hat\phi(k) = \frac{1}{\sqrt{2}}\int e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ $\endgroup$ – Jeremy Nov 20 '17 at 2:41
  • $\begingroup$ I substituted $u = x-\frac{(2x_0 + ik_0a^2)}{2}$ in. After squaring both sides and changing the integral to polar coordinates, I found that the integral evaluated to $\sqrt{\pi}$. This equation I have arrived at does not match the provided solution. I may have made an error in evaluating the Gaussian integral. Where did I go wrong? $\endgroup$ – Jeremy Nov 20 '17 at 2:46
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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

$$\int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx=\alpha e^{-i(k-k_0)x_0}\sqrt\pi e^{-((k-k_0)\alpha/2)^2}$$

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  • $\begingroup$ I think the same techniques you've applied in your solution can be applied to the original problem. I think you have misread the problem statement. You have two exponentials with $(x-x_0)$ in both. Please note in the original problem statement that one exponential contains $(k-k_0)$ $\endgroup$ – Jeremy Nov 20 '17 at 3:45
  • $\begingroup$ Jeremy, I edited on 20 November in response to your posted comment herein. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to both up vote and accept an answer as you see fit. Happy holidays! -Mark $\endgroup$ – Mark Viola Dec 18 '17 at 18:40
  • $\begingroup$ Please feel free to up vote and accept an answer as you see fit. And Happy Holidays! $\endgroup$ – Mark Viola Dec 24 '17 at 17:29
  • $\begingroup$ Your solution is incorrect by a factor of $\frac1{\sqrt{2}}$. $\endgroup$ – Jeremy Dec 29 '17 at 22:48
  • $\begingroup$ @Jeremy Why do you say that? $\endgroup$ – Mark Viola Dec 29 '17 at 22:50

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