Integral Solutions for n,m Positive Integers

[yungman]

Homework Statement

$$\int ^b_0 cos(\frac{(n-m)\pi}{b}x) dx$$

$$\int ^b_0 cos(\frac{(n+m)\pi}{b}x) dx$$


n and m are positive integers.

The Attempt at a Solution

$$\int ^b_0 cos(\frac{(n-m)\pi}{b}x) dx = \frac{b\;sin[(n-m)\pi]}{(n-m)\pi}$$

Obviously answer is zero if n not equal to m. This is a sync function. I don't know how to derive the answer. From the graph, the answer should be b, but how do I derive it.


Also I want to verify:

$$\int ^b_0 cos(\frac{(n+m)\pi}{b}x) dx = \frac{sin[(n+m)\pi]}{(n+m)\pi} = \frac{b}{(n+m)\pi}$$


Thanks

 

[Hootenanny]

$$\int ^b_0 cos(\frac{(n-m)\pi}{b}x) dx = \frac{sin[(n-m)\pi]}{(n-m)\pi}$$

This isn't correct. Almost, but not quite. Surely your integral should be a function of b, no?

Obviously answer is zero if n not equal to m.

Indeed.

This is a sync function. I don't know how to derive the answer. From the graph, the answer should be 1, but how do I derive it.

Why do you think that be answer should be one?

Also I want to verify:

$$\int ^b_0 cos(\frac{(n+m)\pi}{b}x) dx = \frac{sin[(n+m)\pi]}{(n+m)\pi} = \frac{0}{(n+m)\pi}=0$$

You won't be able to verify this, because it isn't correct - see the above comments.

Thanks[/QUOTE]

 

[yungman]

This isn't correct. Almost, but not quite. Surely your integral should be a function of b, no?

Indeed.

Why do you think that be answer should be one?


You won't be able to verify this, because it isn't correct - see the above comments.

Thanks

[/QUOTE]

I have covered my tracks on my original post. Too bad I cannot change what you quote my answer to cover my track!!LOL!

Yes I was rushing to post late last night and forget the b. How does this look now? Also how do I derive the sync function?

Thanks

 

[Hootenanny]

I have covered my tracks on my original post. Too bad I cannot change what you quote my answer to cover my track!!LOL!

Yes I was rushing to post late last night and forget the b. How does this look now? Also how do I derive the sync function?

Thanks[/QUOTE]
The first one is now correct, the second one isn't - probably just another typo.

In order to compute the value of the integral when n=m, think about what happens in the limit as n-m approaches zero.

 

[yungman]

Yes, the second answer is wrong all together.

$$\frac {b \;sin(n+m)\pi}{(n+m)\pi}$$

where this is a sinc function equal to zero if n and m are integers.

Thanks

Alan