Existence of Limit with Integrals.

[Bacle]

Hi, I saw a proof/argument done today that I think was wrong:

It is finding the limit as a->oo of the integral from 0 to b<oo:

Int_(0..b) Sqr[x(1 +cos(ax))]dx , where Sqr is the square root

Now, the argument given was that one could find a bound for the oscillation

of (1+cos(ax)).

The problem I have is that, no matter what the trick may be, cos(ax) will take

values of 1 , and of -1 (at 2kPi and 2(k+1)Pi respectively; k an integer), so that

the value of the limit will go from:

i) 0 , when cos(ax)=-1 , to:

ii) Int_
(0..b) Sqr[2x]dx =(Sqr2)x^(-1/2), which is an improper integral at x=0, but does not

go to zero.

So the limit does _not_ exist, right?

 

[jackmell]

No. The limits does exits. It's still just the limit of a (bounded) Riemann sum. First write it clearly what we're talking about:

$$\lim_{a\to\infty} \int_0^b \sqrt{x(1+\cos(ax))}dx,\quad b<\infty$$

Since b is finite and:

$$0\leq|1+\cos(ax)|\leq 2$$

then cannot we say:

$$0\leq \left|\int_0^b \sqrt{x(1+\cos(ax))}dx\right|\leq \int_0^b \sqrt{2x}dx$$

and therefore:

$$0\leq \lim_{a\to\infty} \int_0^b \sqrt{x(1+\cos(ax))}dx\leq \int_0^b \sqrt{2x}dx$$

 

[Bacle]

Well, yes, but I imagine the limit would exist if we can squeeze it from the right into
being 0, right? If we could, e.g., only say that the limit is betwwm 0 and 3, can we
then say the limit exists?

 

[Bacle]

I actually came up with numerical values near 80,000 for b near 2000; seems way too

far from 0.