Therefore D_n=\frac{2}{\pi(1-4n^2)}

[gabel]

$$D_n=\frac{1}{\pi}\int_0^\pi\sin{(t)}\cdot e^{-i2nt}dt=\frac{2}{\pi(1-4n^2)}$$

I have no idea on how they get from one side of the equation symbol to the other, can i get some tips and tricks ?

I have try ed writing sint as an exp function, but i don't feel it gets me anywhere close.

 

[SVXX]

and the formulae here -:

http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

will help you in solving the integral ;). Though I don't know why a simple by parts integration doesn't suit you..

 

[gabel]

Hmm, I am unsure on how to threat the variable n

 

[SVXX]

You needn't worry about 'n'. It isn't being integrated. Worry about 't', and 'n' will resolve itself after a little algebra. In simpler terms, n is a constant and t is the variable during integration.

 

[gabel]

Ok, this is where I am at now. But I am stuck once more.

$$D_k = \frac{2 cos^2(n\pi)+sin(2\pi n)}{1-4n^2}$$

 

[grey_earl]

$$D_n=\frac{1}{\pi}\int_0^\pi\sin{(t)}\cdot e^{-i2nt}dt=\frac{2}{\pi(1-4n^2)}$$

I think it's easier to write the sine as exponential, which gives you
$$D_n=\frac{1}{2 \pi i}\int_0^\pi (e^{i t} - e^{-i t}) e^{-i2nt}dt = \frac{1}{2 \pi i}\int_0^\pi (e^{-i(2n-1)t} - e^{-i(2n+1)t})dt$$
$$= \frac{1}{2 \pi i} \left[ \frac{1}{-i(2n-1)} ( e^{-i(2n-1)\pi} - 1 ) - \frac{1}{-i(2n+1)} ( e^{-i(2n+1)\pi} - 1 ) \right]$$
$$= \frac{1}{2 \pi i} \frac{1}{-i(2n-1)(2n+1)} \left[ (2n+1) e^{-i(2n-1)\pi} - (2n+1) - (2n-1) e^{-i(2n+1)\pi} + (2n-1) \right]$$
$$= \frac{1}{2 \pi (2n-1)(2n+1)} \left[ 2n e^{i\pi} - 2n e^{-i\pi} + e^{i\pi} + e^{-i\pi} - 2 \right]$$
$$= \frac{1}{2 \pi (2n-1)(2n+1)} \left[ 2n (-1) - 2n (-1) + (-1) + (-1) - 2 \right] = \frac{-4}{2 \pi (4n^2-1)} = \frac{2}{\pi (1-4n^2)}$$

 

[SVXX]

You're almost there, the penultimate step. Now cos(npi) = (-1)^n and sin(2npi) = 0. What would cos^2(npi) be? Hint : What does your answer require?

 

[klondike]

$$D_n=\frac{1}{\pi}\int_0^\pi\sin{(t)}\cdot e^{-i2nt}dt=\frac{2}{\pi(1-4n^2)}$$

I have no idea on how they get from one side of the equation symbol to the other, can i get some tips and tricks ?

I have try ed writing sint as an exp function, but i don't feel it gets me anywhere close.

Or it could be done by integrate by parts twice.(I know it's kinda retarded.)
$$\int_0^\pi\sin{t}e^{-j2nt}dt =$$
$$=-\cos{t}e^{-j2nt}-j2n \int_0^\pi\cos{t}\c e^{-j2nt}dt =$$
$$=-\cos{t}e^{-j2nt}-j2n \sin{t}e^{-j2nt}+4n^{2}\int_0^\pi\sin{t}\c e^{-j2nt}dt =$$

hence
$$\int_0^\pi\sin{t}\ce^{-j2nt}dt =\frac{e^{-j2nt}(-\cos{t}-j2n\sin{t})}{1-4n^{2}}|^{\pi}_{0}=\frac{2}{1-4n^{2}}$$