Why is the integral from 0 to 1 of $\sin(2\pi nt) \sin(2\pi mt)$ equal to 0,5 if $m$ and $n$ are equal?

Question

I am interested in this result because I am studing Fourier Series. By the way, although I have studied Mathematical Analysis, my background is not so good. Could you please explain why the integral from $\int_0^1\sin(2\pi nt) \sin(2\pi mt)dt$ is $0,5$ in an easy to understand way. Thanks

Answer 1

We have that $$\sin{a}\sin{b}=\frac{1}{2}(\cos{(a-b)}-\cos{(a+b)}).$$ Applying to the integrand and integrating the cosines, we get sines that are between $0$ and $2\pi$ which is zero when $m$ is different from $n$. When $m$ equals $n$ $\cos{2\pi(m-n)t}=1$ and the integral is $\frac{1}{2}$

Answer 2

Use the first formula here to transform the product of sines: http://de.wikipedia.org/wiki/Formelsammlung_Trigonometrie#Produkte_der_Winkelfunktionen

$$\int_0^1 \sin(2\pi nt)\sin(2\pi mt) = \frac{1}{2} \int_0^1 \cos(2\pi (n-m)t) dt - \frac{1}{2} \int_0^1\cos(2\pi(n+m)t)dt$$

Therewith it should be easy to see or derive that the only way (assuming $n,m \in \mathbb{N}$) something nonzero will remain is when $n = m$. The result is then

$$\frac{1}{2} \int_0^1 \cos(0) dt = \frac{1}{2} \int_0^1 1 dt = \frac{1}{2}$$

Answer 3

When $n\neq m$, $\{1,\sin(2\pi t), \sin(4\pi t), \ldots, \sin(2\pi n t), \cos(2\pi t),\ldots, \cos(2\pi nt)\}$ are orthogonal functions for $t\in[0,1]$. That is, there inner product is zero when we define the inner product on the set as $$ \langle f,g\rangle = \int_0^1f(t)g(t)dt $$ and when $n=m$, we have $$ \int_0^1\sin^2(2\pi nt)dt = \frac{1}{2}\int_0^1dt - \frac{1}{2}\int_0^1\cos(4\pi n t)dt $$ The second integral is zero since $1$ and $\cos(2\pi nt)$ are orthogonal. If we define the inner product to be $$ \langle f,g\rangle = 2\int_0^1f(t)g(t)dt =\delta_{nm} = \begin{cases} 1,& \text{if }n=m\\ 0,& \text{otherwise} \end{cases} $$ and we take $1/\sqrt{2}$ instead of $1$ for our constant in the set, we have an orthonormal basis.