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Could someone give a suggestion to calculate this limit please?

$$ \lim_{a \to \infty} \int\limits_0^{2010} \sqrt{x(1+\cos ax)} \ dx $$

I don't know methods for solving such limits with integrals. Thanks.

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"Solving" is the wrong word. One solves problems; one solves equations. One evaluates expressions, including limits,integrals, etc. –  Michael Hardy Sep 13 '11 at 21:52
    
Things like Lebesgue's convergence theorems won't work here because if "$\lim\limits_{a\to\infty}$" is pulled inside the integral, the limit doesn't exist. Somehow "squeezing" doesn't seem promising in this case. Is there some hope of actually evaluating the integral and then finding the limit? –  Michael Hardy Sep 13 '11 at 21:57
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This looks like something where you don't need a method -- you need a trick. The idea must be that when $a$ goes towards infinity, you can average over the cosine wave everywhere. I would expect the limit to be something like $$\int_0^{2010} \sqrt x\, dx \frac{1}{2\pi}\int_0^{2\pi} \sqrt{1+\cos t}\, dt$$but I don't have a rigorous proof of that ready. –  Henning Makholm Sep 13 '11 at 22:01
    
An initial guess: It will be half of $\int_0^{2010} \sqrt{x}\;dx$. –  Michael Hardy Sep 13 '11 at 22:01
    
Thank you! It is enough for my programm $\int_0^{2010} \sqrt x\, dx \frac{1}{2\pi}\int_0^{2\pi} \sqrt{1+\cos t}\, dt$ –  user16023 Sep 13 '11 at 22:11

1 Answer 1

up vote 8 down vote accepted

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?)

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You beat me to it. Nice answer :-) –  Srivatsan Sep 13 '11 at 23:32
    
Uh... yeah, you beat me to it, also... ! –  The Chaz 2.0 Sep 14 '11 at 0:26

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