Example 7.2.7: Integration by Parts and Limits
Suppose f:[a, b] R
is a continuously differentiable function. Show that:
- f(x) sin(nx) dx = 0
- f(x) cos(nx) dx = 0
We already mentioned that this will be an application of integration by parts. Let's focus on the first statement and let g'(x) = sin(nx), where g' is the function in the Integration by Parts theorem. Then:
f(x) sin(n x) dx = - 1/n cos(nx) f(x) + 1/n f'(x) cos(nx) dx
But |sin(nx)| 1 and |cos(nx)| 1, and since both f and f' are continuous functions on a closed, bounded interval there are constants K and L such that |f(x)| K and |f'(x)| L. Putting everything together we get:
| f(x) sin(n x) dx | K/n + L(b-a)/n
Thus,
f(x) sin(n x) dx = 0The proof of the second statement is left as an exercise.