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I'm trying to do an exercise in Folland (8.45), and I'm having to calculate $W_t=\left[\frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}\right]^\vee,$ where $\vee$ is the inverse Fourier transform. We can make free use of the identity: $$ \chi_{[-a,a]}^\vee(x) = \frac{\sin(2\pi a x)}{\pi x },$$

I must have some elementary confusions.

Here's my attempt:

$$\begin{eqnarray*} W_t&=&\int_\mathbb{R} \frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}\cdot e^{-2\pi i |\xi|x}\ dx \\ &=&\frac12 \int_\mathbb{R} \chi_{[-t,t]}^\vee(|\xi|)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=&\frac12 \int_\mathbb{R} \left(\int_\mathbb{R} \chi_{[-t,t]}(x)\cdot e^{-2\pi i |\xi| y}\ dy\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac12 \int_\mathbb{R} \left(\int_{-t}^t e^{-2\pi i |\xi| y}\ dy\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac12 \int_\mathbb{R} \left(\left(\frac{1}{2\pi i|\xi|}\right)\left(e^{2\pi i t}-e^{-2\pi i t}\right)\right)\cdot e^{-2\pi i |\xi|x}\ dx \\ &=& \frac{(e^{2\pi i t}-e^{-2\pi i t})}{4\pi i}\int_\mathbb{R} \frac{e^{-2\pi i |\xi|x}}{|\xi|}\ dx, \end{eqnarray*} $$

which doesn't seem to converge...what am I doing wrong?

If it matters, the bigger question is to show (using the above identity) that $$u(x,t):=f*\partial_t W_t(x)+g*W_t(x)=\frac12[f(x+t)+f(x-t)]+\frac12\int_{x-t}^{x+t}g(s)\ ds,$$ subject to the initial conditions $$(\partial_t^2-\Delta)u=0,\ \ u(x,0)=f(x), \ \ \partial_t u(x,0)=g(x).$$

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Ah, a lack of convergence. Sounds like distributions are at hand.. – anon Apr 21 '12 at 23:13
up vote 3 down vote accepted

There's a mistake in the penultimate line of the equation array – the $|\xi|$ is missing in the exponents. If you correct that, you'll find that the whole calculation is one big detour and you could have directly obtained the last line by using

$$\sin(2\pi t|\xi|)=\frac1{2\mathrm i}\left(\mathrm e^{2\pi\mathrm it|\xi|}-\mathrm e^{-2\pi\mathrm it|\xi|}\right)$$

at the very beginning. This in turn shows that the main problem lies at the very beginning, where you misapplied the definition of the inverse Fourier transform. You should be integrating over $\xi$, and by integrating over $x$ instead you introduced the divergent integral that you got back out in the end.

By the way, all the absolute value signs are unnecessary, since $\displaystyle\frac{\sin (2\pi t|\xi|)}{2\pi|\xi|}=\frac{\sin (2\pi t\xi)}{2\pi\xi}$.

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Ok, but then aren't we still left with the seemingly (to me) intractable $\int_\mathbb{R} \frac{1-e^{-4\pi i t \xi}}{\xi}\ d\xi$? – user21725 Apr 22 '12 at 3:36
By the way that's a nice observation regarding the use of the sine identity. Folland seems to think these detours are useful as learning experiences, though. :) – user21725 Apr 22 '12 at 3:41
@Eric: Your question was "what am I doing wrong?", not "how can I do it right?" :-) There's no need to calculate anything; all you need is the identity in your second equation. Fourier transform it to get the Fourier transform of the $\operatorname{sinc}$ function; then you just have to take the complex conjugate and perhaps take into account some normalization factors to get its inverse Fourier transform. – joriki Apr 22 '12 at 7:51

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