Find the fourier sine transform

Question

dls · Answer

$$\Large f(x) = \frac{1}{x} e^{-ax}$$

dls · Answer

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 · Answer

^Correct me if I'm wrong^.
However..I still find the integration a bit harsh in both the ways.

dls · Answer

@Jemurray3

anonymous · Answer

Is there a particular reason you are only integrating from 0 to infinity rather than from -infinity to infinity?

dls · Answer

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 · Answer

Actually, silly question, nevermind.  It wouldn't converge otherwise.  Okay.

dls · Answer

yep..e^-x won't converge if we consider -infty.

anonymous · Answer

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 · Answer

I can only decide that after you show me XD

dls · Answer

On a side note, both the above methods I mentioned are correct otherwise? Which one is easier to evaluate ?

anonymous · Answer

$$F(s) = \int_{0}^\infty \frac{1}{x} e^{-ax}\sin(sx)dx$$
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 · Answer

You never fail to impress me!! :D
nice method :)

anonymous · Answer

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 · Answer

:O yeah..definitely another day lol. I guess this trick is sufficient for most of the problems. :)
thanks again!

irishboy123 · Answer

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