Finding $ \lim_{a \to \infty} \int\limits_0^{2010} \sqrt{x(1+\cos ax)}\ \mathrm dx $ Could someone give a suggestion to calculate this limit please?
$$
\lim_{a \to \infty} \int\limits_0^{2010} \sqrt{x(1+\cos ax)}\ \mathrm dx
$$
I don't know methods for solving such limits with integrals. Thanks.
 A: This is one of those cases where you don't need a general method -- you need a trick (by which I mean a method that you don't expect to generalize to other problems).
The crucial idea is that when $a$ goes towards infinity, you can average over the cosine wave everywhere. Making this work rigorously is a bit involved, though:
For notational convenience, set $M=2010$ and $\nu = \frac{a}{2\pi}$, and let $L(\nu)=\int_0^M \sqrt{x(1+\cos 2\pi\nu x)}\,dx$. We're then seeking $\lim_{\nu\to\infty}L(\nu)$. It is clear that
$$\int_0^M \sqrt{\frac{\lfloor \nu x\rfloor}{\nu}}\sqrt{1+\cos 2\pi\nu x}\,dx
\le L(\nu) \le
\int_0^M \sqrt{\frac{\lceil\nu x\rceil}{\nu}}\sqrt{1+\cos 2\pi\nu x}\,dx$$
If for a moment we restrict our attention to $\nu$s that are multiples of $\frac{1}M$, we can integrate each wavelength (where $\lfloor \nu x \rfloor$ resp. $\lceil \nu x \rceil$ are constant) separately and combine to find
$$K \int_0^M \sqrt{\frac{\lfloor \nu x\rfloor}{\nu}}\,dx
\le L(\nu) \le
K \int_0^M \sqrt{\frac{\lceil\nu x\rceil}{\nu}}\,dx$$
where
$$K = \int_0^1 \sqrt{1+\cos 2\pi u}\,du = \frac{1}{2\pi} \int_0^{2\pi} \sqrt{1+\cos t}\,dt$$
However, the integrals surrounding $L(\nu)$ are now just Riemann sums for $\int_0^M \sqrt{x}\,dx$, and so both bounds for $L(\nu)$ approach
$$K\int_0^M\sqrt{x}\,dx = \frac{1}{2\pi} \int_0^{2\pi} \sqrt{1+\cos t}\,dt \int_0^M\sqrt{x}\,dx = \frac{4\sqrt 2}{2\pi}\cdot\frac{2}{3}\cdot M^{3/2} $$
which is therefore the only possible limit.
It remains only to see that the $\nu$s that are not multiples of $\frac{1}{M}$ cannot prevent the limit from existing. But even if the average over the last partial wavelength next to $x=M$ differs from $K$, the net error resulting from this can be at most $\frac 2\nu$, which goes towards zero too.
(After having done all this, we can see that the trick actually is a general method for finding $\lim_{\nu\to\infty} \int_a^b f(x)g(\nu x)\,dx$ when $g$ is a periodic function. But how often do you run into such problems?)
