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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

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    If $m=n$ the integrand is positive on a nontrivial interval, so the integral of the product cannot be $0$ in this case. – coffeemath Jan 16 '15 at 18:47
  • my mistake, I editted the question –  Jan 16 '15 at 19:06
  • You are asking about the integral $\int\sin(2\pi nt)\sin(2\pi mt)\mathrm dt$ but only in the case where $n=m$... Why not simply ask about the integral of $\int\sin^2(2\pi nt)\mathrm d t$, which is clearly the same thing in that case?! – Mariano Suárez-Álvarez Jan 16 '15 at 19:08
  • I asked in the sin(2pint) sin(2pint) form because this is the form the teacher presented it –  Jan 16 '15 at 19:21
  • By the way, it seems I might have made mistakes with the values. Is the integral equal to 1 or 0,5? –  Jan 16 '15 at 19:22
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    It is equal to a 1/2 see my answer for why. – dustin Jan 16 '15 at 19:50

3 Answers3

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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 $\frac{1}{2}$

Daniel W. Farlow
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marwalix
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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

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

GDumphart
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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.

dustin
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