/sci/ - Science & Math

>>14531403
Not quite.
If [math]U(x)[/math] and [math]\cos (\alpha x)[/math] have the same period ([math]2 \alpha \pi x[/math]) then we can break up
[math]\displaystyle \int_{- \infty}^{\infty} \cos (\alpha x) U(x) \ dx = \sum_{k \in \mathbb{Z}} \int_{k 2 \alpha \pi}^{(k + 1)2 \alpha \pi} \cos (\alpha x) U(x) \ dx = \sum_{k \in \mathbb{Z}} \int_{0}^{2 \alpha \pi} \cos (\alpha x) U(x) \ dx[/math]
, so the integral either zeroes or diverges.