# Complex fourier series

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

2. 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$$^{-j2\pi*nt}$$

3. 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 im suppose to end up with $$\frac{1}{j2pi*n}$$ but it doesnt work out like it should.
Any help would be good thanks.

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HallsofIvy
Homework Helper
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"!)

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 shouldnt get the same answer as what I got using the Trig-Complex Relationship equation. Which is why im 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
Homework Helper
Once again, what is
$$e^{-2\pi i n}$$?

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]