# Integration of Sin[x^2]?

1. ### gamesguru

85
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?

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.

Last edited: Mar 29, 2008
2. ### flebbyman

28
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

3. ### tiny-tim

26,054
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)

4. ### gamesguru

85
@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.

5. ### tiny-tim

26,054
ah … I must have stopped reading the post when I started looking at the .jpg, and then forgotten where to resume reading.
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! )

6. ### gamesguru

85
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.

7. ### tiny-tim

26,054
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).

8. ### gamesguru

85
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...

9. ### tiny-tim

26,054
Yes, but e^-∞ = 0, which is what matters!

10. ### gamesguru

85
$$e^{-\epsilon x^2} e^{i x^2}=e^{-0x^2}e^{ix^2}=e^{ix^2}$$
...what am i missing? I don't see a $e^{-\infty}$.

11. ### tiny-tim

26,054
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.

12. ### gamesguru

85
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.

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