# how to show that $\int_0^\infty \sin(x^2) dx$ converges [duplicate]

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

What test do I use to show that the following integral converges? $$\int_0^\infty \sin (x^2) \; dx$$

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.

• Should we be worried that when $M \to \infty \ \cos(M^2)$ is undefined?
– Alex
Jan 15, 2013 at 21:18
• @Alex we have $\cos (M^2)/M$ at the denominator.
– S L
Jan 15, 2013 at 21:21
• Not for this argument. For the first part (the evaluation), at the top we get $-\frac{1}{2M}\cos(M^2)$. Since $\cos$ wiggles between $-1$ and $1$, the $\frac{1}{2M}$ kills it. For the integral part, it is just Comparison Test, again using $|\cos(x^2)|\le 1$. Jan 15, 2013 at 21:23
• @AndréNicolas: of course! I confused it with something completely different: mathoverflow.net/questions/24579/convergence-of-a-series
– Alex
Jan 15, 2013 at 21:31
• @AndréNicolas This argument seems to generalize well to $\int_0^{\infty} \sin(x^{\alpha})dx$ and $\int_0^{\infty} \cos(x^{\alpha})dx$ with $|\alpha|\gt 1$. Can we say anything about $\int_0^{\infty} \cos(f(x))dx$ with $f(x)$ a polynomial? (This would include the Airy function $Ai(x)$, right?) Jan 16, 2013 at 3:52

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

• I should point out to the OP to pay special attention to the nice illustration of the integration contour used to evaluate the integral, which shows why the integral converges. Jan 15, 2013 at 21:02
• isn't there any easy method (something like comparison) just to show that is converges? I don't have to evaluate it. Also this is not complex analysis ... i guess there must be something nice and easy.
– S L
Jan 15, 2013 at 21:06
• looks like the the french version is same as the other answer. Nice +1 to everyone
– S L
Jan 15, 2013 at 21:16

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 $$\int_0^{\infty} \sin(x^2) dx = \text{Imag} \left( \int_0^{\infty} e^{ix^2} dx\right)=\dfrac{\sqrt{\pi}}{2\sqrt{2}}$$