# Simple Integration Orthogonal Sin

1. Nov 21, 2013

### Omega0

Hi,

take the function

I(m,n) = Integral from 0 to 1 of sin(m*pi*x)*sin(n*pi*x) over dx

depending from n and m, being +-1, +-2, and so on.
If I use sin(x)sin(y)=1/2(cos(x-y)-cos(x+y)) I get sin^2(n*pi*x)=1/2-1/2cos(2n*pi*x)

or I(n,n)=1/2

because the integral of cosine over full periods is zero.

Well, the integral for n not equal to m is zero (easy to see, sin is zero for n*pi).

Now let's integrate via partial integration without knowledge of formulars between
periodic functions.

After two integrations I get

I(m,n) = (m/n)^2 I(m,n)

which means that if the integration is correct

1. m=+-n or
2. I(m,n)=0

You can choose m and n independently from each other. This means
that I(m,n)=0 is proven for m not equal +-n.

This is interesting from a point that for example J. Farlow in "Partial Differential Equations
for Scientists and Engineers" does not mention the negative sign. He completely skips this
calculation.

This is an indirect prove. Is it correct?

Thanks!

2. Nov 22, 2013

### HallsofIvy

Yes, that is correct.