Why ∫02πsin2xdx≠∫02πsin⁡xsin⁡nxdx→n=1 By orthogonality, we know that ∫02πsin⁡mxsin⁡nxdx=π if m=n and 0 otherwise. Nevertheless, when I am calculating it as follows, I get 0. ∫02πsin⁡xsin⁡nxdx=12[sin⁡[(1−n)x]1−n−sin⁡[(1+n)x]1+n]02π But the above quantity is 0 even if n=1. Shouldn't it be π?

Question

Why ∫02πsin2xdx≠∫02πsin⁡xsin⁡nxdx→n=1 By orthogonality, we know that ∫02πsin⁡mxsin⁡nxdx=π if m=n and 0 otherwise. Nevertheless, when I am calculating it as follows, I get 0. ∫02πsin⁡xsin⁡nxdx=12[sin⁡[(1−n)x]1−n−sin⁡[(1+n)x]1+n]02π But the above quantity is 0 even if n=1. Shouldn&#039;t it be π?

Matias Lang · Accepted Answer

Your work so far is correct, but things like these are often a sort of &quot;by cases&quot; thing: in this case, the cases are n=1 and n≠1. When n≠1, you can show that expression is 0, but its

Karli Kaiser · Accepted Answer

In the limit of n→0 you have  limn→0xsin⁡(1−n)x(1−n)x=x  Therefore  limn→0∫02πsin⁡xsin⁡(nx)dx=12[x−sin⁡2x2]02π=π

Why \int_0^{2\pi} \sin^2 xdx \neq \int_0^{2\pi} \sin x \sin nx dx \to n=1By orthogonality

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