What is the Cosine Fourier Transform of an Exponential Function?

[Jalo]

Homework Statement

Find the cosine Fourier transform of the function f(t)=e^-at

Homework Equations

The Attempt at a Solution

F(w)=(2/π)^0.5∫dt e^-atcos(wt)
The integral is from 0 to +∞

Using euler's formula I got the result

F(w)=(2/π)^0.5( e^it(w-a)/i(w-a) - e^-it(w+a)/i(w+a) )

I don't know what to do from here... I can't just substitute!
If anyone could point me in the right direction I'd appreciate!

 

[Simon Bridge]

You need a common denominator don't you?
If I understand you correctly - you converted the cosine in the integral into a sum of exponentials, then did the integration?

That would give you something like:
$$e^{-at}\cos(wt)=\frac{1}{2} e^{-at}\left ( e^{iwt}+e^{-iwt} \right ) = \frac{1}{2}\left (e^{(iw-a)t}+e^{-(iw+a)t}\right )$$ ...in the integrand - which does not look like it ends up looking like yours. Can you show the intermediate steps in your working?

Alternatively:
You could try using integration by parts instead (hint: twice).

 

[Jalo]

You need a common denominator don't you?
If I understand you correctly - you converted the cosine in the integral into a sum of exponentials, then did the integration?

That would give you something like:
$$e^{-at}\cos(wt)=\frac{1}{2} e^{-at}\left ( e^{iwt}+e^{-iwt} \right ) = \frac{1}{2}\left (e^{(iw-a)t}+e^{-(iw+a)t}\right )$$ ...in the integrand - which does not look like it ends up looking like yours. Can you show the intermediate steps in your working?

Alternatively:
You could try using integration by parts instead (hint: twice).

You're correct. I made a mistake! The actual result is:

[itex]{\frac{1}{2\sqrt{2π}}\left[\frac{exp(t(iw-a))}{iw-a}-\frac{exp(-t(iw+a))}{iw+a}\right]}[/itex]

The problem remains tho... I don't know what to do from here, since I'm integrating from 0 to ∞ !

 

[Simon Bridge]

Applying the limits - ##t=0## case is easy. It is ##t\rightarrow \infty## in the first term that gives you the problem?

To procede from here you need to convert the exponentials back into cosine and sine functions ... think in terms of a product of and exponential and trig functions.

I still think it is easier to just follow the hint in post #2.

 

[Jalo]

Applying the limits - ##t=0## case is easy. It is ##t\rightarrow \infty## in the first term that gives you the problem?

To procede from here you need to convert the exponentials back into cosine and sine functions ... think in terms of a product of and exponential and trig functions.

I still think it is easier to just follow the hint in post #2.

I tought about using your hint, but I'd end up with cosine and sine functions, and neither of them converge to a value as x goes to infinity! That's why I avoided using it!

 

[Dick]

I tought about using your hint, but I'd end up with cosine and sine functions, and neither of them converge to a value as x goes to infinity! That's why I avoided using it!

I think there is probably an unstated assumption that a>0. So your trig functions don't converge but they are bounded. e^(-at) goes to zero as t->infinity. What happens?

 

[Simon Bridge]

Thanks Dick - yes: that would be a problem if the trig functions were not multiplied by a decreasing exponential.