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!