how to show that $\int_0^\infty \sin(x^2) dx$ converges 
Possible Duplicate:
Prove: $\int_{0}^{\infty} \sin (x^2) dx$ converges. 

What test do I use to show that the following integral converges?
$$ \int_0^\infty \sin (x^2) \; dx$$
 A: Lots of information here: 
http://en.wikipedia.org/wiki/Fresnel_integral
See especially the section Evaluation.

@rlgordonma & @experimentX
I just see the french like their Fresnel so much, their wikipedia page actually has a section on convergence as well as derivations of the final value:
http://fr.wikipedia.org/wiki/Int%C3%A9grale_de_Fresnel
A: Consider the triangle $\Delta$ with vertices at $(0,0), (T,0), (T,T)$ in the complex plane. Since $\exp(iz^2)$ is entire, we have 
$$\int_{\Delta} \exp(iz^2) dz = 0$$
Further, the integral on the side perpendicular to the $X$ axis, as $T \to \infty$ is 0, since $$\lim_{T \to \infty} \left \vert \int_{T}^{T+iT} \exp(iz^2) dz \right \vert \leq  \lim_{T \to \infty} \int_{T}^{T+iT} \left \vert \exp(iz^2) \right \vert \vert dz \vert = \lim_{T \to \infty} \int_0^T \exp(-2Tx) dx\\ = \lim_{T \to \infty} \dfrac{1-\exp(-2T^2)}{2T} = 0$$
Hence, the integral along the $X$ axis equals the integral along the hypotenuse i.e. $$\int_{0}^T \exp(iz^2) dz = \int_{0}^{T+iT} \exp(iz^2) dz$$
Setting $z= (1+i)w$, we get that
$$\int_{0}^{T+iT} \exp(iz^2) dz = \int_0^T \exp(i(1+i)^2 w^2) (1+i) dw = (1+i) \int_0^T \exp(-2w^2) dw$$
Hence, $$\lim_{T \to \infty}\int_{0}^{T} \exp(iz^2) dz = (1+i) \dfrac{\sqrt{\pi}}{2\sqrt{2}}$$
Now, note that $$\int_0^{\infty} \sin(x^2) dx = \text{Imag} \left( \int_0^{\infty} e^{ix^2} dx\right)=\dfrac{\sqrt{\pi}}{2\sqrt{2}}$$
A: We deal with the integral from (say) $1$ to $\infty$.
In principle we should look at $\int_1^M \sin(x^2)\,dx$, then let $M\to\infty$.  
Use integration by parts. Let $f(x)=\frac{1}{x}$ and $g'(x)=x\sin(x^2)$. Then $f'(x)=-\frac{1}{x^2}$ and we can take $g(x)=-\frac{1}{2}\cos x^2$. 
We end up with
$$\int_1^M\sin(x^2)\,dx=\left. -\frac{1}{2x}\cos(x^2)\right|_1^M -\int_1^M \frac{1}{2x^2}\cos(x^2)\,dx.$$
Now let $M\to \infty$. Note that the remaining integral behaves nicely as $M\to\infty$, since $\int_1^\infty \frac{dx}{x^2}$ converges, and $|\cos(x^2)|$ is bounded. 
