Question

Find the Fourier Cosine Transform.

Answer 1

\[\Large f(x) = e^{-x^2}\]

Answer 2

\[\Large F ( \lambda ) = \int\limits_0^\infty e^{-x^2} \cos (\lambda x)dx\]

Answer 3

The integration
..is scary

Answer 4

We can try making use of the fact that :
\[\Large \int\limits_0^\infty e^{-x^2}dx = \sqrt \pi/2\]

Answer 5

\[\Large \cos (\lambda x) \frac{\sqrt \pi}{2} - \int\limits_0^\infty \lambda \sin (\lambda x) \frac{\sqrt \pi}{2}dx \]

Answer 6

its an even function so the cosine should be the full transform itself

yes?

Answer 7

I don't get you ? Kindly elaborate a bit D:

Answer 8

i may be mixing my series and my transforms here - been a while - but my instinct is that the imaginary part should be zero if the function is even as the sin is odd and bits calcel out on either side of the y axis

Answer 9

BTW

this hits nail on head in terms of solution

Answer 10

but let me research that a little before i mislead you....

Answer 11

sure..though I think it can be easily integrated from where I left it ?

Answer 12

but the problem is how to substitute the limits in the end

Answer 13

Just take the real part of the full transform, which is fairly straightforward.

Answer 14

\[F(\lambda) = \int_{-\infty}^\infty dx e^{-x^2} e^{i\lambda x} \]
\[ = \int_{-\infty}^\infty dx e^{-(x^2-i\lambda x)} \]
complete the square:
\[ e^{(i\lambda/2)^2}\int_{-\infty}^\infty dx e^{-(x-i\lambda/2)^2} \]
\[ = \sqrt{\pi} \cdot e^{-\lambda^2/4} \]

The full transform turns out to be real, so the job is done. If you want to do the half-transform from \((0,\infty)\), you can just cut that in half.

Answer 15

Oh yes! Fourier transform is equivalent to the real part of fourier integral!

Answer 16

is this true in case of only even functions ?

Answer 17

No, that's true of every function. In the case of even functions, the cosine transform is equal to the full transform.

Answer 18

awesome! so I can see which integration is easy to perform and proceed accordingly :)
thanks!

Answer 19

@Jemurray3 I have a small doubt. What do you mean by "full transform" ? Sorry for such a basic question, just started with this topic recently. I thought we do :
\[F(\lambda) = \int\limits_{-\infty}^\infty dx e^{-x^2} e^{-i\lambda x}\] and then for inverse we take \[\Large F(s) = \int\limits_{-\infty}^\infty F(\lambda) e^{i\lambda s} ds\]
but you had e^(i lambda x) instead of e^(-i lambda x).

Answer 20

No problem. Two things - first, when I said full transform, I was referring to the fact that the Fourier Cosine Transform and the Fourier Sine Transform can be viewed as the real and imaginary parts of the standard Fourier Transform.

Second, this may seem like it doesn't make sense, but it doesn't actually matter whether you use a plus or a minus sign in your definition of the Fourier Transform, as long as you use the opposite sign for the inverse. Similarly, at some point you have to divide by \(2\pi\), but it doesn't actually matter whether you do it for the transform, the inverse, or you divide by a factor of \(\sqrt{2\pi}\) for each. The two most common definitions are:

\[ F(k) = \int_{-\infty}^\infty e^{-ikx}f(x) dx\]
\[ f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{ikx} F(k) dk \]

and

\[ F(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-ikx}f(x) dx\]
\[ f(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{ikx} F(k) dk \]

In your case, I didn't really worry about it because you might notice that \(\cos(\lambda x)\) is the real part of \(e^{i\lambda x}\) as well as \(e^{-i\lambda x}\), so it doesn't actually matter.

Answer 21

Sugoi! Thanks!! This is amazing :)