How the heck do I integrate sin(1/x) ?

1. Feb 18, 2004

Matt Jacques

I tried parts by integration but Im caught in an endless loop of ever growing in complexity integrals! I must be missing something.

2. Feb 18, 2004

matt grime

is that an indefinite or definite integral?

3. Feb 18, 2004

HallsofIvy

Staff Emeritus
Do you have any reason to believe its anti-derivative is an elementary function?

4. Feb 18, 2004

PrudensOptimus

easy man

-cosintegral[1/x] + xSin[1/x]

5. Feb 19, 2004

Matt Jacques

I got that, too. No way to further simplify?

6. Feb 19, 2004

matt grime

apart from that the integral of 1/x is log(x) you mean?

7. Feb 19, 2004

NateTG

If you're desperate, you could try working out a Taylor/Mclaurin series for it, and seeing if the integral of that is recognizable.

8. Feb 20, 2004

Spectre5

You can use a Maclaurin series to evaluate (or at least approximate) it...knowing that

$$sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{(2n+1)}}{(2n+1)!}$$

you can replace x with 1/x and integrate to get:

$$\int sin(\frac{1}{x})=\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{2(2n+1)!x^{2n}}$$

Last edited: Feb 20, 2004
9. Feb 20, 2004

PrudensOptimus

Wrong. &int;1/x dx = ln |x| + C.

&int;1/(x(ln 10)) dx = log |x| + C.

10. Feb 20, 2004

master_coda

When a mathematician says "log" they are generally talking about the natural logarithm.

11. Feb 20, 2004

NateTG

Right, and the rest of the time they usually mean $$log_2$$
but anything other than $$log_e$$ gets a base.

12. Feb 20, 2004

master_coda

I don't see too many mathematicians refer to $\log_2$ as $\log$.

13. Feb 21, 2004

NateTG

It's typically for math/cs tpe situations and usually only applies to situations where hte base is not particuarly important.

14. Feb 21, 2004

matt grime

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?

15. Feb 24, 2004

curiousbystander

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.

16. Feb 24, 2004

matt grime

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.)

17. Feb 24, 2004

curiousbystander

I should have been more careful when answering, but isn't the integral still well defined since {0} is a set of measure 0?

18. Feb 24, 2004

NateTG

Do you mean to use Lebesgue integration?

$$\lim_{x \rightarrow 0}$$ might also not exist and thus cause problems.

19. Feb 24, 2004

matt grime

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.

20. Feb 24, 2004

curiousbystander

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 $$\sin(\frac{1}{x})$$ to include 0 so that $$\sin(\frac{1}{x})=0$$, 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 $$\sin(\frac{1}{x})$$ 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?

Last edited: Feb 24, 2004