Exponentials and trig functions

**leshields** · May 16th 2010, 07:27 AM

Hi, I'm having some trouble understanding the relationships between trig functions and exponential functions. I have a question with answers and just can't see how they simplify down to get to where they do

$C_{k}= 1/{2\pi} \int_{-\pi}^{\pi} te^{-ikt} dt = - {(-1)^k}/{ik}$
Now when i try to intergrate I get a bit of a crazy formula with lots of e's and I'm pretty confident it will simplify to what the answer is but I'm not sure how to.
The intergral I get is

$1/{2\pi} [ (-ik)^{-1} t e^{-ikt}]_{-\pi}^{\pi} + 1/{2ik\pi} [-e^{-ikt}/{ik}]_{-\pi}^{\pi}$

I can't really see how to simplify it to get required answer. At a guess the second part simplifies to zero Because $e^{ik\pi} - e^{-ik\pi} = 0$ ??? and the first part to $[e^{ik\pi} + e^{-ik\pi}]{-ik}^{-1}$ and if $e^{ik\pi} + e^{-ik\pi} = (-1)^k$ then im done, but this is making big guesses about how these exponentials work which I've no idea if its correct!! Sorry if this is really obvious and im being dumb! Thanks for any help - even if its just refferring me to a list of exponential identities!

**roninpro** · May 16th 2010, 11:32 AM

To be honest, I don't know much about calculus on complex numbers, but it seems to me that you can just write $e^{-ikt}=\cos(-kt)+i\sin(-kt)$ and then integrate the real and complex parts.

Have you tried this?

**Ackbeet** · June 4th 2010, 10:33 AM

As previously mentioned, the Euler formula for converting exponentials to trig functions is useful. Also, think about plugging in integer multiples of $\pi$ into trig functions:

$\cos(k\pi)=(-1)^{k}$ , and $\sin(k\pi)=0$ .

That might help simplify things a bit.

**Ackbeet** · June 4th 2010, 10:39 AM

Also, the following should help (if you convert to trig functions):

$\cos(-x)=\cos(x)$ , and $\sin(-x)=-\sin(x)$ .

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