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Mixed sine and cosine

Integrating by parts,
$\displaystyle L$ $\displaystyle :=$ $\displaystyle \int_0^\pi \sin mx \cos nx \d x$  
  $\displaystyle =$ $\displaystyle \frac{1}{n} \biggl[ \sin mx \sin nx \biggr]_0^\pi - \frac{m}{n}
\int_0^\pi \cos mx \sin nx  \d x$  
  $\displaystyle =$ $\displaystyle - \frac{m}{n} \biggl[ \frac{1}{m-n} \biggl( (-1)^{n-m} - 1 \biggr) +
L \biggr]$  

(by (3)). Hence,
$\displaystyle L \biggl( 1 + \frac{m}{n} \biggr)$ $\displaystyle =$ $\displaystyle \frac{m}{n(n-m)} \biggl( (-1)^{n-m}
- 1 \biggr)$  
$\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{\frac{m}{n-m} \biggl( (-1)^{n-m} - 1 \biggr)}{n+m}$  
  $\displaystyle =$ $\displaystyle \frac{m}{n^2 - m^2} \biggl( (-1)^{n-m} - 1 \biggr)$  

When $ n-m$ is odd, $ (-1)^{n-m} = -1$ and when $ n-m$ is even, $ (-1)^{n-m} = 1$. Thus, we can express this as

$\displaystyle \int_0^\pi \sin mx \cos nx  \d x = \begin{cases}
0, & n-m \text{ even} \\
2m/(m^2 - n^2), & n-m \text{ odd}
\end{cases}$



Alexander Frolkin 2001-02-17