Can the Integral of Sin[x^2] Be Expressed in Exact Form?

[gamesguru]

I've attached my work, but basically, I'm trying to compute this:
\int_{0}^{\infty}2\sin({x^2})dx.
(The 2 is only there because when you expand it into an exponential function, it makes like easier.)
You can look at my work to see what I did. My question now is, is it possible to express that limit in exact form? If so, what is it? Why does the limit have this value?
http://img358.imageshack.us/img358/7304/file0004ch0.jpg
I also tried doing this with double integrals and polar coordinates to calculate the square of the integral but I ran into a problem deep into it of evaluating \sin{\infty} and \cos{\infty}.
If you're curious why I care about this, it's because this integral is important in the diffraction of light. While only the approximate answer is important in physics, I'm just curious about the mathematics behind it.
Thanks in advanced.

 

[flebbyman]

It has no elementary anti derivative so it can't be evaluated in a closed form, but you can only get a pretty good approximation, as you have done

 

[tiny-tim]

Hi gamesguru!

(1) 1.2533 = √(π/2).

(2) Do you know the trick for integrating \int_{0}^{\infty}e^{-x^2}dx ?

Try putting \int_{0}^{\infty}\sin{x^2}dx = imaginary part of \int_{0}^{\infty}e^{i x^2}dx.

(or perhaps \lim{\epsilon\to0}\int_{0}^{\infty}e^{(i\,-\,\epsilon) x^2}dx, for positive real epsilon)

 

[gamesguru]

@tiny-tim
I tried that method, read my post. I got to a point where I was evaluating cos[∞] or sin[∞]. Try it for yourself, it won't work.

 

[tiny-tim]

@tiny-tim
I tried that method, read my post.

ah … I must have stopped reading the post when I started looking at the .jpg, and then forgotten where to resume reading.

I got to a point where I was evaluating cos[∞] or sin[∞]. Try it for yourself, it won't work.

Yup … that's why I suggested the epsilon method.

(I haven't actually checked it myself … but the result is right, so I expect it works … over to you! )

 

[gamesguru]

I don't see how the epsilon method makes this more easy to integrate. I'd appreciate it if you should me how to do it, or at least get started.

 

[tiny-tim]

I don't see how the epsilon method makes this more easy to integrate.

Because you end up evaluating e^{-\epsilon x^2} e^{i x^2} from 0 to ∞ (which converges), instead of just e^{i x^2} (which doesn't).

 

[gamesguru]

Because you end up evaluating e^{-\epsilon x^2} e^{i x^2} from 0 to ∞ (which converges), instead of just e^{i x^2} (which doesn't).

In, e^{-\epsilon x^2} e^{i x^2} the first term goes to 1 (e^0=1), not 0. I still don't see how this helps...

 

[tiny-tim]

Yes, but e^-∞ = 0, which is what matters!

 

[gamesguru]

<br /> e^{-\epsilon x^2} e^{i x^2}=e^{-0x^2}e^{ix^2}=e^{ix^2}<br />
...what am i missing? I don't see a e^{-\infty}.

 

[tiny-tim]

ah … in my post #7, I was talking about x going from 0 to ∞.

I thought you had already found that that was the problem … that e^(ix^2), or sin(x^2), doesn't converge as x -> ∞.

But multiply it by e^(-(epsilon)x^2), where epsilon is fixed, and it converges to 0.

 

[gamesguru]

I'm not sure I follow you. I'd really like it if you just do it so I could see how it's done and possibly learn something. We're making no progress right now.

 

[K.J.Healey]

2*sqrt(Pi/8)

http://en.wikipedia.org/wiki/FresnelS