Fourier Series Integration by Parts Solution

[reece]

Homework Statement

Solve $$\frac{1}{1}$$ $$\int^{0}_{-1}$$ -t e$$^{-j2\pi*nt}$$dt

Homework Equations

So I use integration by parts
u = -t and dv = e$$^{-j2\pi*nt}$$ , du= -1 and v = $$\frac{1}{-j2\pi*n}$$e$$^{-j2\pi*nt}$$

The Attempt at a Solution

after integration by parts I get:
$$\frac{e^-j2\pi*nt}{j2pi*n}$$ + $$\frac{1}{(j2pi*n)^{2}}$$ [1 - e$$^{-j2\pi*nt}$$]

So basically I am suppose to end up with $$\frac{1}{j2pi*n}$$ but it doesn't work out like it should.
Any help would be good thanks.

 

[HallsofIvy]

First, you certainly do not get anything involving "t" when you have evaluated at t= 0 and -1! Did you mean
$$\frac{e^{2\pi i n}}{2\pi i n}+ \frac{1- e^{2\pi i n}}{(2\pi i n)^2}$$

What is $e^{-2\pi i nt}$?
(Sorry, but I just can't force my self to use "j" instead of "i"!)

 

[reece]

woops my bad, those t's arent spose to be in there. But the rest is right.

And yes it is what you typed out.

Im leading to think I shouldn't get the same answer as what I got using the Trig-Complex Relationship equation. Which is why I am getting something different.

The exponential part is what I integrate with to evaluate the complex Fourier series. See the equation. -t is the function.

 

[HallsofIvy]

Once again, what is
$$e^{-2\pi i n}$$?

 

[reece]

In trig FS, you evaluate each component a0, an, bn. Well the exponential replaces that I think.

im guessing you use Euler's Rules (i think) to evaluate it which says

sin theta = 1 / 2i [ e ^ i*theta - e ^ -i*theta]
cos theta = 1 /2 [ e ^ i*theta + e ^ -i*theta]

Hopefully that answered that.