# Toronto Math Forum

## APM346-2018S => APM346––Home Assignments => Web Bonus Problems => Topic started by: Victor Ivrii on February 16, 2018, 08:27:43 AM

Title: Web Bonus problem -- Reading Week
Post by: Victor Ivrii on February 16, 2018, 08:27:43 AM
$\renewcommand{\Re}{\operatorname{Re}}$
Find Fourier transform of $e^{-\frac{1}{2}a x^2}$, where $a\in \mathbb{C}$ with $\Re a>0$.

To do this

(a) Check that $u=e^{-\frac{1}{2}a x^2}$ satisfies $u'(x) = -ax u(x)$; then using Fourier Transform properties conclude that $\hat{u}$ satisfies $i \omega \hat{u}(\omega)= -i a\hat{u}'(\omega)$ and solving this equation find $\hat{u}$ up to a constant factor $C$.

(b) Then calculate $I:=\hat{u}(0)=\int e^{-\frac{1}{2}a x^2}\,dx$ in the same way as we did it for real $a>0$ but justify the correct "sign" (since $I$ is recovered up to a factor $\pm 1$), comparing with the case of real $a>0$.
Title: Re: Web Bonus problem -- Reading Week
Post by: Adam Gao on February 17, 2018, 01:28:15 PM
Part (a):

To check that $u = e^{\frac{1}{2}ax^2}$ satisfies $u'(x) = -ax u(x)$, differentiate $u$. I am not sure if we can just treat $u$ like an exponential or if we have to expand $u = e^{\frac{1}{2}\mathrm{Re}(a)x^2}e^{\frac{1}{2}\mathrm{Imaginary}(a)x^2}$ using euler's formula first.

Next, use $u'(x) = -ax u(x)$ to show $i\omega{u}(\omega) = -ia\hat{u}'(\omega)$.

First, use fourier transform property $g(x) = xf(x) \rightarrow \hat{g}(k) = i\hat{f}'(k)$.
Using $u'(x) = g(x)$ and $-au(x) = f(x)$, as well as the definition of a fourier transform, we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$.

Now use fourier transform property $g'(x) = f(x) \rightarrow \hat{g}(k) = ik\hat{f}(k)$. Using $-axu(x) = f(x)$ and $u'(x) = g'(x)$ we get $\int_{\infty}^{\infty} -axu(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Replacing $-axu(x)$ with $u'(x)$ we get $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = i\omega \hat{u} (\omega)$. Apply this equality to the previously shown equation $\int_{\infty}^{\infty} u'(x)e^{-i\omega x} = -ia\hat{u}'(\omega)$ and get $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega)$.

Finally, find $\hat{u}$. Consider that $i\omega \hat{u} (\omega) = -ia\hat{u} ' (\omega) \rightarrow -\frac{\omega}{a} \hat{u} (\omega) = \hat{u} ' (\omega)$. This is an ODE with an exponential solution, $\hat{u} (\omega) = Ce^{-\frac{\omega ^2}{2a}}$.
Title: Re: Web Bonus problem -- Reading Week
Post by: Jingxuan Zhang on February 17, 2018, 03:10:50 PM
I find the second part perplexing: what are we to compare $I$ with? and shouldn't $I=\hat{u}(0)$ be immediate once the formula for $\hat{u}$ is determined by Adam?
Title: Re: Web Bonus problem -- Reading Week
Post by: Jingxuan Zhang on February 17, 2018, 03:26:11 PM
Part (b): normalize the integrand we have
$$I=\hat{u}(0)=\int e^{-\frac{1}{2}a x^2}\,dx=\sqrt{\frac{2\pi}{a}}\cdot\underbrace{\frac{1}{\sqrt{2\pi}}\int e^{-z^2/2}\,dz}_{1}=\sqrt{\frac{2\pi}{a}}$$
But here really $\sqrt{a}$ is ill-defined: $a\in\mathbb{C}$ might result in this root having multiple value. The sign therefore depends on the branch chosen for $a$. In particular the principal value of $I$ will have a positive sign if $\Im(a)\geq0$ and a negative sign otherwise.

EDIT: I consulted Ahlfors' text on complex variable, which I have completely forgotten now. Write $a=\alpha+i\beta$ then
$$\sqrt{\frac{2\pi}{\alpha}}=\pm\frac{\sqrt{2\pi}}{\alpha^2+\beta^2}\left(\sqrt{\frac{\alpha+\sqrt{\alpha^2+\beta^2}}{2}}-i\frac{\beta}{|\beta|}\sqrt{\frac{-\alpha+\sqrt{\alpha^2+\beta^2}}{2}}\right)$$

If $\alpha>0,\beta=0$ then of course this is just the simple formula on LS.
Title: Re: Web Bonus problem -- Reading Week
Post by: Victor Ivrii on March 24, 2018, 07:59:26 AM
$\renewcommand{\Re}{\operatorname{Re}}\renewcommand{\Im}{\operatorname{Im}}$
Root does not have multiple values, or rather, it may have but then it does not change the result. So, let us calculate
$$\int_{-\infty}^{\infty}e^{-\frac{1}{2}az^2}\,dz$$
where $\Re a>0$ (we need it, otherwise integral diverges).

Let us make $z=\alpha w$ with $\alpha=1/\sqrt{a}$ where $\sqrt{a}$ is defined uniquely in the right complex half-plane, with $\Re \alpha>0$...