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Preliminary results

Using Euler's formula,

$\displaystyle e^{ix} = \cos x + i \sin x
$

we can write
$\displaystyle e^{imx} e^{inx}$ $\displaystyle =$ $\displaystyle ( \cos mx + i \sin mx ) ( \cos nx + i \sin nx )$  
  $\displaystyle =$ $\displaystyle \cos mx \cos nx + i \sin nx \cos mx + i \sin mx \cos nx - \sin
nx \sin mx$  

from which,

\begin{displaymath}
\begin{split}
I := \int_0^\pi e^{imx} e^{inx}  \d x = & \in...
...l( \sin mx \cos nx + \cos mx \sin nx \biggr)  \d x
\end{split}\end{displaymath}

We now evaluate $ I$,
$\displaystyle I$ $\displaystyle =$ $\displaystyle \int_0^\pi e^{i(m+n)x}  \d x$  
  $\displaystyle =$ $\displaystyle \frac{1}{i(m+n)} \biggl[ e^{i(m+n)x} \biggr]_0^\pi$  
  $\displaystyle =$ $\displaystyle \frac{-i}{m+n} \biggl( (-1)^{m+n} - 1 \biggr)$  

Equating real and imaginary parts gives

$\displaystyle \int_0^\pi \cos mx \cos nx \d x = \int_0^\pi \sin mx \sin nx \d x$ (1)

We now consider
$\displaystyle e^{-imx} e^{inx}$ $\displaystyle =$ $\displaystyle ( \cos mx - i \sin mx ) ( \cos nx + i \sin nx )$  
  $\displaystyle =$ $\displaystyle \cos mx \cos nx + i \cos mx \sin nx - i \sin mx \cos nx + \sin
mx \sin nx$  

and so

\begin{displaymath}
\begin{split}
J := \int_0^\pi e^{-imx} e^{inx} \d x &= \int...
...l( \cos mx \sin nx - \sin mx \cos nx \biggr)  \d x
\end{split}\end{displaymath}

Evaluate $ J$,
$\displaystyle J$ $\displaystyle =$ $\displaystyle \int_0^\pi e^{i(n-m)x}  \d x$  
  $\displaystyle =$ $\displaystyle \frac{1}{i(n-m)} \biggl[ e^{i(n-m)x} \biggr]_0^\pi$  
  $\displaystyle =$ $\displaystyle \frac{i}{m-n} \biggl( (-1)^{n-m} - 1 \biggr)$  

and compare real and imaginary parts:

$\displaystyle \int_0^\pi \cos mx \cos nx \d x + \int_0^\pi \sin mx \sin nx \d x = 0$ (2)

$\displaystyle \int_0^\pi \cos mx \sin nx  \d x = \frac{1}{m-n}\biggl((-1)^{n-m} - 1\biggr) + \int_0^\pi \sin mx \cos nx  \d x$ (3)

Clearly, (2) and (3) are only valid for $ n\neq m$ since the derivation involves expressions containing $ \frac{1}{m-n}$.
next up previous
Next: Pair of sines or Up: Orthogonality of trigonometric functions Previous: Orthogonality of trigonometric functions
Alexander Frolkin 2001-02-17