Stuck on proving $\int_{-\infty}^\infty \cos(\frac{\pi}{a}x)\cos(\frac{3\pi}{a} x) \, \mathrm{d}x$ = 0 Can someone please help me to show how 
$$\int_{-\infty}^\infty \cos(\frac{\pi}{a}x)\cos(\frac{3\pi}{a} x) \, \mathrm{d}x = 0$$
Attempt:
Trig Identity yields
$$= \frac{1}{2} \int_{-\infty}^\infty \cos(\frac{4\pi}{a}x) + \cos(\frac{2\pi}{a} x) \, \mathrm{d}x$$ 
$$= \frac{a}{2} (\frac{\sin(\frac{4\pi}{a}x)}{4\pi} + \frac{\sin(\frac{2\pi}{a}x)}{2\pi}) $$ evaluated from $-\infty$ to $\infty$ 
What is a nontrivial way to show that the last expression is zero?
My course notes says something about stretching of the sine function, not good enough for me.
 A: The limit does not exist in the strictest sense.
However, in the theory of generalized functions (i.e., distribution theory), the limit as $\lim_{x\to \infty} \sin(ax) = 0$.

NOTE:
This is NOT a classical limit, but has a rigorous interpretation in the context of distribution theory.  Formal (non-rigorous) application of distribution theory is used pervasively by physicists and engineers to obtain results (usually correct) very quickly without the need to enforce rigor.  An example of such application is the formal use of "Dirac delta functions" to define Fourier integrals that do not converge classically.

A: Your first line after Trig Identity is quite helpful.  It should be clear that the integral over any interval of length $a$ is zero, as the first cosine goes through two complete cycles and the second goes through one.  In the usual sense the integral does not exist, as the value oscillates between $2$ and $-1$, as you would normally express it as $$\frac{1}{2} \int_{-\infty}^\infty \cos(\frac{4\pi}{a}x) + \cos(\frac{2\pi}{a} x) \mathrm{d}x=\frac 12 \lim_{b \to -\infty}\int_{b}^0 \cos(\frac{4\pi}{a}x) + \cos(\frac{2\pi}{a} x) \mathrm{d}x+\frac 12 \lim_{c \to \infty}\int_{0}^c \cos(\frac{4\pi}{a}x) + \cos(\frac{2\pi}{a} x) \mathrm{d}x$$ and neither limit exists.
