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From (1), we have
Substituting this into (2) gives
Hence, for , we have
and
Alternatively, we can prove the same results by integrating by parts,
as follows.

(by (1)). Hence, . This argument also holds only for
, since otherwise becomes and so we get , so
that is not necessarily zero.
In the case when , we have

Since (1) is valid for as well as , we have
where
is the Kronecker delta.

** Next:** Mixed sine and cosine
** Up:** Orthogonality of trigonometric functions
** Previous:** Preliminary results
Alexander Frolkin
2001-02-17