I tried parts by integration but Im caught in an endless loop of ever growing in complexity integrals! I must be missing something.
If you're desperate, you could try working out a Taylor/Mclaurin series for it, and seeing if the integral of that is recognizable.
You can use a Maclaurin series to evaluate (or at least approximate) it...knowing that [tex]sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{(2n+1)}}{(2n+1)!}[/tex] you can replace x with 1/x and integrate to get: [tex]\int sin(\frac{1}{x})=\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{2(2n+1)!x^{2n}}[/tex]
Right, and the rest of the time they usually mean [tex]log_2[/tex] but anything other than [tex]log_e[/tex] gets a base.
It's typically for math/cs tpe situations and usually only applies to situations where hte base is not particuarly important.
yes, i did omit the modulus sign, however you should probably be told that log always means base e. This is completely standard in mathematics, and just one more thing they misteach at high school After all what other base would you possibly want?
This might help too: sin(1/x) is an odd function (meaning f(-x) = -f(x)). The definite integral of any odd function on the interval [-a,a] is 0.
one generally wouldn't integrate over a region where the function is not defined. (no choice at zero can make it continuous, interestingly enough, not that that's either here or there, and not that any choice would make the integral be anything but zero anyway, though 0 is the only choice that keeps it a genuine odd function.)
I should have been more careful when answering, but isn't the integral still well defined since {0} is a set of measure 0?
Do you mean to use Lebesgue integration? [tex]\lim_{x \rightarrow 0}[/tex] might also not exist and thus cause problems.
The function hasn't been defined at 0, that's all. It is true that any assignment of a value at zero will produce a function that is Riemann integrable (you don't need to use the machinery of Lesbegue integration on it). Only the setting the value at 0 to be 0 will provide an odd function. No assignment produces a continuous function. I wouldn't like to make any definitive statements about the propriety of integrating a function over region in which it contains points where it isn't defined, other than it seems something you shouldn't do.
Good point-- I had confused the Riemannian integral with the Lebesque. Time to shake the dust off my old real analysis books and review the basics. I think it will still work out: If I follow Matt's advice and extend the domain of [tex]\sin(\frac{1}{x})[/tex] to include 0 so that [tex]\sin(\frac{1}{x})=0[/tex], then I have a function which is bounded and continuous everywhere except 0. The Riemann Integral is defined for such a function, and since the function is odd and defined on all of [-a,a] its integral will be 0. Since the function is Riemann integrable on [-a,a] it will also be Lebesgue integrable on [-a,a] and the two integrals will agree. Now that I'm working with the Lebesgue integral my functions only need to be defined almost everywhere so the Lebesgue integral of [tex]\sin(\frac{1}{x})[/tex] from [-a,a] exists and is 0. Anybody see any problems with that? Analysis has never been my strong point. Edit: Matt's last post came up while I was in the midst of writing this so I didn't see it. It does sound like I'm responding directly to what he wrote though doesn't it?