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UPDATE: The problem reduces to this for me:

I ran into an issue with a change of variable integral:

I have $\int\limits_{-\infty}^{\infty}e^{-\frac{1}{2}(x+\frac{iy}{2})^{2}}dx$.

I do the change of variable $u = \frac{x + \frac{iy}{2}}{\sqrt{2}}du$.

This changes the integral to $\sqrt{2}\int\limits_{*}^{*}e^{-u^{2}}du$. If my new bounds are the same as my old bounds, then I'm done since the integral is known to be $\sqrt{\pi}$. But since I have done a complex translation I am not so sure. Can I say that $\infty + i = \infty$ ?

Original Problem: I need to find the Fourier transform of $g(x) = (x+1)^2e^{-x^2/2}$.

This is defined by:

$$\widehat{g}(y) = \int\limits_{-\infty}^{\infty}g(x)\,e^{-ixy}\,dx.$$

This works out as:

$$ \begin{eqnarray*} \widehat{g}(y) &=& \int\limits_{-\infty}^{\infty}(x+1)^2\,e^{-x^2/2}\,e^{-iyx}\,dx\\ &=& \int\limits_{-\infty}^{\infty}(x+1)^2e^{-x^2/2 - ixy}\,dx. \end{eqnarray*}$$

But I don't know how to evaluate this. Is using software appropriate for something like this?

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    Using software help in a problem of this kind is probably not appropriate. The next step that seems to suggest itself here is to complete the square in $-x^2/2-iyx$ and try to get it into the form $-\frac12(x+\alpha)^2+\beta$.2011-11-13
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    Another way would be to calculate first and second derivative of $f(x) = e^{-x^2 / 2}$ and express $g(x)$ as a combination of $f$, $f'$ and $f''$, remember the Fourier transform of $f$ and how Fourier transforms interact with differentiation.2011-11-13
  • 0
    These are both very good ideas. I'll try them. Thank you both so much!2011-11-13
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    OK I went with the second approach, which boiled down to evaluating $\int\limits_{-\infty}^{\infty}e^{-u^2}du$. I used the fact that this is equal to $\sqrt{\pi}$. (I don't know how to show this part but I don't know if it would normally be done?2011-11-13
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    @Kyle: this is **the** classic: evaluate $I = \int_{-\infty}^{\infty} e^{-u^2} \,du$ by computing $$I^2 = \int_{-\infty}^{\infty} e^{-u^2-v^2}\,du\,dv = \pi $$ using polar coordinates...2011-11-13
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    That completely reminds me how we did this exact thing in class. I feel so thick, haha. Thanks for the reminder. So obviously it is fair game then since we proved it in class.2011-11-13
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    To give another hint regarding t.b.'s approach, set $f(x)=e^{-x^2/2}$. Then $g(x)=x^2f(x)+2xf(x)+f(x)$. A basic property of the Fourier transform is that $\widehat{x^nf}(y)={d^n\over dy^n}\widehat f(y)$, up to some normalization (with your definition, I believe it should be multiplied by $i^n$). From here, it should be pretty clear how to find $\widehat g(y)$.2011-11-13
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    Well the result we have from class is that $\widehat{f'}(y) = iy\widehat{f}(y)$. This was sufficient to get me to the point mentioned in the top of the question as an update. I've almost got it I think.2011-11-13
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    The change of variables $u = 2^{-1/2} (x + iy/2)$ is a complex change of variables. You swap integrating along one line with integrating along another. The bounds are thus the same for $x$.2011-11-14

1 Answers 1

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If I understand the comments correctly, the remaining problem seems to be to find the Fourier transform of $$\color{blue}{f(x) = e^{-x^2/2}}.$$ You can do this by completing the squares and substitution, as outlined in the comments, but I prefer this argument:

Let's compute formally first, deferring the justification of the interchange of differentiation and integration in $(!!)$ to the end of the answer: $$ \begin{align*} \frac{d}{dk} \widehat{f}(k) &= \frac{d}{dk} \int_{-\infty}^{\infty} e^{-x^2/2}e^{-ikx}\,dx \\ &\!\color{red}{\stackrel{(!!)}{=}} \int_{-\infty}^\infty e^{-x^2/2}\, (-ix)\, e^{-ikx}\,dx \tag{!!} \\ &= \int_{-\infty}^\infty i \left( \frac{d}{dx}e^{-x^2/2} \right) \, e^{-ikx}\, dx \\ &= -\int_{-\infty}^\infty i e^{-x^2/2}(-ik)e^{-ikx}\,dx && \text{(integration by parts)}\\ &= -k \,\hat{\!f}(k). \end{align*} $$ This gives us the ordinary differential equation $\hat{\!f}'(k) = -k \,\hat{f}(k)$ with the initial condition $$ \hat{\!f}(0) = \int_{-\infty}^{\infty} e^{-x^2/2}\,dx = \sqrt{2\pi}, $$ which shows that $$\color{blue}{\hat{f}(k) = \sqrt{2\pi}\,e^{-k^2/2}}.$$


In order for you to be able to check your own work, here's my solution of the exercise (N. S. suggested a slightly different approach, but neither is simpler or more efficient, they come down to the same):

We have $f'(x) = -x\,f(x)$ and $f''(x) = (x^2-1)f(x)$ so that $$g(x) = (x+1)^2 f(x) = f''(x) -2f'(x) + 2f(x),$$ hence $$ \begin{align*} \hat{g}(k) &= \widehat{f''}(k) - 2\widehat{f'}(k) + 2\widehat{f}(k) \\ &= (-k^2+2ik+2) \hat{f}(k) \end{align*} $$ which gives us $$\color{red}{\hat{g}(k) = \sqrt{2\pi}\,(-k^2+2ik+2)\,e^{-k^2/2}},$$ where you can write $-k^2+2ik+2 = 1+(1+ik)^2$ if you prefer.


It remains to justify $(!!)$: Compute $$ \begin{align*} \frac{d}{dk}\hat{\!f}(k) & = \lim_{h\to 0} \frac{1}{h} \left(\hat{\!f}(k+h)-\hat{\!f}(k)\right) \\ &= \lim_{h\to0} \int_{-\infty}^{\infty} e^{-x^2/2} \frac{e^{-ikx}}{h}\underbrace{\left(e^{-ihx}-1\right)}_{\large 2i e^{ihx/2}\sin{\frac{hx}{2}}}\,dx. \end{align*} $$ Using that $|\sin{\frac{hx}{2}}| \leq \dfrac{|hx|}{2}$ we see that the absolute value of the integrand is bounded above by $|x|e^{-x^2/2}$ which is clearly integrable, so we conclude by the dominated convergence theorem.