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

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


marked as duplicate by sdcvvc, Stefan Hansen, Davide Giraudo, Michael Greinecker, rschwieb Jan 17 '13 at 11:58

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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.

  • $\begingroup$ Should we be worried that when $M \to \infty \ \cos(M^2)$ is undefined? $\endgroup$ – Alex Jan 15 '13 at 21:18
  • $\begingroup$ @Alex we have $\cos (M^2)/M$ at the denominator. $\endgroup$ – Santosh Linkha Jan 15 '13 at 21:21
  • 2
    $\begingroup$ 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 $|\cos(x^2)|\le 1$. $\endgroup$ – André Nicolas Jan 15 '13 at 21:23
  • $\begingroup$ @AndréNicolas: of course! I confused it with something completely different: mathoverflow.net/questions/24579/convergence-of-a-series $\endgroup$ – Alex Jan 15 '13 at 21:31
  • $\begingroup$ @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?) $\endgroup$ – AndrewG Jan 16 '13 at 3:52

Lots of information here:


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:


  • $\begingroup$ 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. $\endgroup$ – Ron Gordon Jan 15 '13 at 21:02
  • $\begingroup$ 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. $\endgroup$ – Santosh Linkha Jan 15 '13 at 21:06
  • $\begingroup$ looks like the the french version is same as the other answer. Nice +1 to everyone $\endgroup$ – Santosh Linkha Jan 15 '13 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 $$\int_0^{\infty} \sin(x^2) dx = \text{Imag} \left( \int_0^{\infty} e^{ix^2} dx\right)=\dfrac{\sqrt{\pi}}{2\sqrt{2}}$$


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