fourier series analysis, show that for every integer n, using euler's formulas relating trigonometric and exponential functions

Question

Show that for every integer $n$,

$$\int_0^{\pi} \cos nt~\sin t~\mathrm{d}t = \begin{cases} \dfrac{2}{1-n^2} & \text{if } n \text{ is even} \\[10pt] 0 &\text{if } n \text{ is odd} \end{cases}$$

by using Euler's formulas relating trigonometric and exponential functions:

$$\cos x=\frac{1}{2}(\mathrm{e}^{ix}+\mathrm{e}^{-ix}), \ \ \ \sin x=\frac{1}{2i}(\mathrm{e}^{ix}-\mathrm{e}^{-ix})$$

(Here $i$ is the imaginary unit; remember that $\mathrm{e}^{\pi i} = -1$)

I am not sure how to go about doing this problem, if anyone could help it would be much appreciated, thank you for your help in advance.

Answer 1

Consider if the function is even or odd for $n$ even or odd (with respect to the mid point $\pi /2$), this will get your odd case dealt with. To deal with the even case, use the double angle formula repeatedly.

Answer 2

If you expand this integral you should get: $$\int_0^\pi \cos(nt)\sin(t)dt = \frac{1}{4i}\int_0^\pi \left(e^{int}+e^{-int}\right)\left(e^{it}-e^{-it}\right)dt$$ $$= \frac{1}{4i}\int_0^\pi e^{i(n+1)t} - e^{i(n-1)t} + e^{-i(n-1)t} - e^{-i(n+1)t}dt$$

Note that this is exactly the sum of sines: $$= i\int_0^\pi \sin((n+1)t)-\sin((n-1)t)dt$$ Now perform the integration and note that $\cos(n\pi)=(-1)^n$.