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$$ f(t)=\frac{\sin(at)}{t} $$

Since the term is parameterized, it's easy to see that if I take the first derivative with respect to 'a', then the function becomes considerably easier. I do this to the Fourier Transform and obtain: $$ \frac{\partial }{\partial a}\Im (f(t))=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{\infty }\cos(at))e^{itx}dt $$

However, this is an integral of an even function times an odd function, which equals 0 and raises my suspicion. I've tried implementing Euler's cosine form and got nowhere.

Also I'm using the imaginary symbol as the Fourier transform. Why? It looks cool.

Damien L
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Gerg
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1 Answers1

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The differentiating inside the integral trick requires several conditions be checked first. If you notice, the integral on the right is not even defined.

Check out http://ocw.mit.edu/courses/mathematics/18-304-undergraduate-seminar-in-discrete-mathematics-spring-2006/projects/integratnfeynman.pdf

and The Integral that Stumped Feynman?

Also you have to be a bit careful with how you're defining everything. You should call the Fourier transform $\hat{f}(x)$ rather than $f(t)$ since it is a different function in the variable $x$.

muzzlator
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  • Edited for definition – Gerg Mar 20 '13 at 08:22
  • I'll take a look at the link – Gerg Mar 20 '13 at 08:22
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    That first link also has a version of the integral you are interested in. They choose a different parameter to differentiate with respect to so that the function is dominated by $|e^{-bt}|$ – muzzlator Mar 20 '13 at 08:31
  • I see it now. Took the derivative to the wrong one. Sorry, for the late reply, I'm working on about 10 other things for students tomorrow. – Gerg Mar 20 '13 at 09:46
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    @muzzlator : the integral on the right is very well defined in the [best integration theory](http://en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral). – Damien L Mar 20 '13 at 10:39