DLS
  • DLS
Find the fourier sine transform
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DLS
  • DLS
\[\Large f(x) = \frac{1}{x} e^{-ax}\]
DLS
  • DLS
We can either find.. \[\Large F_s(\lambda) = \int\limits_0^\infty \frac{1}{x}e^{-ax}\sin(\lambda x)dx\] or..take the full transform and consider only the imaginary part.. \[\Large \int\limits_0^\infty \frac{1}{x}e^{-ax}.e^{-isx}dx\]
DLS
  • DLS
^Correct me if I'm wrong^. However..I still find the integration a bit harsh in both the ways.

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DLS
  • DLS
@Jemurray3
anonymous
  • anonymous
Is there a particular reason you are only integrating from 0 to infinity rather than from -infinity to infinity?
DLS
  • DLS
In fourier transformation it should be from -infinity to +infinity (but in the last question it gave a different answer :/ ) and in the sine integral its correct if I consider my text book.
anonymous
  • anonymous
Actually, silly question, nevermind. It wouldn't converge otherwise. Okay.
DLS
  • DLS
yep..e^-x won't converge if we consider -infty.
anonymous
  • anonymous
There are a number of ways you can do it. I have a particular technique that I like, but it's a bit unusual. Is that fine with you?
DLS
  • DLS
I can only decide that after you show me XD
DLS
  • DLS
On a side note, both the above methods I mentioned are correct otherwise? Which one is easier to evaluate ?
anonymous
  • anonymous
\[F(s) = \int_{0}^\infty \frac{1}{x} e^{-ax}\sin(sx)dx\] Notice for the moment that \(F(0) = 0\). Next we differentiate with respect to \(s\). \[F'(s) = \int_0^\infty \frac{1}{x} e^{-ax} \cdot x \cdot \cos(sx)dx \] \[ = \int_0^\infty e^{-ax} \cos(sx)dx\] That integral is pretty easy - we can write it as the real part of this: \[ \int_0^\infty e^{-ax} e^{-isx} dx = \int_0^\infty e^{-(a+is)x} dx = \frac{1}{a+is}\] The real part of that is \(\frac{a}{a^2 + s^2}\), so we have \[F'(s) = \frac{a}{a^2+s^2} \] If you remember your inverse trig derivatives, that means \[F(s) = \tan^{-1}(\frac{s}{a}) + C\] Recalling that F(0) = 0, it follows that C = 0, and so the solution you're looking for is just \[F(s) = \tan^{-1}(\frac{s}{a}) \]
DLS
  • DLS
You never fail to impress me!! :D nice method :)
anonymous
  • anonymous
The really nice trick is when you invent a parameter to differentiate with respect to, then set it to 1 at the end of the calculation... but that's for another day.
DLS
  • DLS
:O yeah..definitely another day lol. I guess this trick is sufficient for most of the problems. :) thanks again!
IrishBoy123
  • IrishBoy123
.

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