Good catch, I meant cos(t).The first thing I notice is that you have t in your integrand but are integrating over x.
Why?
Yeah, the duality is how it seems to be done often, but I thought I'd take a stab at the long route. Definitely showing to be tricky...I think if you split things out sufficiently, you will eventually get to the point of having a pair of coefficients each multiplying the Fourier transform of 1, which is what gives you the delta functions. Now, proving this last part is tricky. I think on of the standard ways is to do the opposite and show either that the Inverse Fourier Transform of a delta function is a constant, namely 1. Another way would be to show that the Fourier Transform of a delta function is equal to 1 and then invoke duality.
No, the cos(ω_o t) is fine, it's the dx that is the problem. It should be dt.Good catch, I meant cos(t).
Again, good catch. The math should be clear of variable typos now.No, the cos(ω_o t) is fine, it's the dx that is the problem. It should be dt.
You need the ω_o because it's what parks the detla functions.
No, w = 2∏f. N is just used as a placeholder for the integration limits because they are set to ±∞.I cannot see where 'n' is used in your integrand. Shouldn't ω be written with some dependence upon n?
My apologies for being "sloppy" (sarcasm). Writing this out in LaTeX makes it a lot harder to keep track of small typos. Whether the integration variable is stated as 'x' or 't', I figured it would be clear enough for those reading along. I have fixed all the cases of 'x' and 'dx' to their 't' counterparts. My apologies that this wasn't perfectly coded throughout this thread.That's not how you do something like that.
Your integrand has a variable over which you are integrating. You originally had this as x since your integrand ended with 'dx'. After integration, you end up with a result that still has the variable x in it. You then take the difference between this expression evaluated with x equal to the upper limit and this expression evaluated with x at the lower limit. That is how the limits enter into the picture.
You are being very, very sloppy with the notation and this is going to cause you all kinds of problems (probably already are).