Why is $$\int_{0}^{\infty} \sin(t^2)\,dt=\sqrt{\frac{\pi}{8}}$$ I am kinda new to this kind of calculus and I only know that $$\int_{-1}^{1}\sqrt{1-x^2}dx=\frac{\pi}{2}$$

The problem in which I have to use it is: $$\lim_{x\to\infty}\frac{\int_{0}^{x} \sin(t^2)\,dt}{x^3}$$

  • 1
    $\begingroup$ The bottom one is a standard integral, it has an anti-derivative, and evaluates easily. The top one, on the other hand, does not have a closed anti-derivative and needs special analysis to find specific values. If you don't know anything about complex analysis, don't worry about that integral. $\endgroup$ – Kaynex Feb 16 '16 at 20:30
  • $\begingroup$ Well, I need to evaluate it because it's used in a limes in an other problem. If I can just prove that it is constant that would be fine with me too. $\endgroup$ – HeatTheIce Feb 16 '16 at 20:31
  • $\begingroup$ The top one is one of the two Fresnel integrals. wikiwand.com/en/Fresnel_integral The specific problem is mentioned later in the wikipedia page. $\endgroup$ – Dair Feb 16 '16 at 20:32
  • $\begingroup$ Do you know some complex analysis, in special complex integration? $\endgroup$ – DonAntonio Feb 16 '16 at 20:35
  • $\begingroup$ Yes, we talked about it in a lecture and the prof gave us some examples. But not that much because we will do it later. $\endgroup$ – HeatTheIce Feb 16 '16 at 20:35

You do not need to evaluate that integral in order to calculate the limit. Just note that $$ \Bigl|\int_0^x\sin (t^2)\,dt\Bigr|\leq\int_0^x|\sin(t^2)|\,dt $$ by the triangle inequality for integrals. Next, $|\sin(t^2)|\leq 1$ for all $0\leq t\leq x$, and hence $$ \Bigl|\int_0^x\sin (t^2)\,dt\Bigr|\leq\int_0^x1\,dt=x. $$ Thus, we find that, for all $x>0$, $$ -\frac{1}{x^2}=-\frac{x}{x^3}\leq\frac{\int_0^x\sin(t^2)\,dt}{x^3}\leq\frac{x}{x^3}=\frac{1}{x^2}. $$ Then, the squeeze theorem gives you that $$ \lim_{x\to+\infty}\frac{\int_0^x\sin(t^2)\,dt}{x^3}=0. $$


Hint. By setting, for any real number $t$ such that $t>0$, $$ I_t:=\int_0^\infty e^{-tx^2}\cos (x^2)\:dx,\quad J_t:=\int_0^\infty e^{-tx^2}\sin (x^2)\:dx. \tag1 $$ One may prove that $$ \begin{align} \left(I_t\right)^2-\left(J_t\right)^2 &=\left(\int_0^\infty e^{-tx^2}\cos (x^2)\:dx\right)^2-\left(\int_0^\infty e^{-tx^2}\sin (x^2)\:dx\right)^2 \\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\left(\cos (x^2)\cos (y^2)-\sin (x^2)\sin (y^2)\right)dxdy \\\\&=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\cos (x^2+y^2)\:dxdy \\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\cos(r^2)\:dr \\\\&=\frac{\pi}4\:\frac t{1+t^2}. \tag2 \end{align} $$ Similarly $$ \begin{align} 2\:I_tJ_t &=\int_0^\infty\!\int_0^\infty e^{-t(x^2+y^2)}\sin (x^2+y^2)\:dxdy \\\\&=\int_0^{\pi/2}d\theta\int_0^\infty r\:e^{-tr^2}\sin (r^2)\:dr \\\\&=\frac{\pi}4\:\frac1{1+t^2}. \tag3 \end{align} $$ From $(2)$ and $(3)$, one easily gets $$ I_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}+t}{1+t^2}},\quad J_t=\sqrt{\frac{\pi}8}\:\sqrt{\frac{\sqrt{1+t^2}-t}{1+t^2}} \tag4 $$ As $t\to 0$, this leads to

$$ \int_0^\infty \cos (x^2)\:dx=\int_0^\infty \sin (x^2)\:dx=\sqrt{\frac{\pi}8}. \tag5 $$


I'm not sure what the top integral is useful for computing the final limit, which can be done with a straightforward application of l'Hôpital (in the form “whatever/$\infty$”), via the fundamental theorem of calculus: $$ \lim_{x\to\infty}\frac{\displaystyle\int_0^x\sin(t^2)\,dt}{x^3}= \lim_{x\to\infty}\frac{\sin(x^2)}{3x^2}=0 $$

However, it's not difficult to show that the integral converges. Let's do $$ \int\sin(t^2)\,dt= \frac{1}{2}\int\frac{\sin u}{\sqrt{u}}\,du= -\frac{1}{2}\frac{\cos u}{\sqrt{u}} -\frac{1}{4}\int\frac{\cos u}{\sqrt{u^3}}\,du $$ The first step is $u=\sqrt{t}$, the second is integration by parts. Now $$ \int_1^{\infty}\frac{\cos u}{\sqrt{u^3}}\,du $$ converges absolutely, because $$ \frac{|\cos u|}{\sqrt{u^3}}\le\frac{1}{\sqrt{u^3}} $$ and $$ \int_1^{\infty}\frac{1}{\sqrt{u^3}}\,du $$ is convergent. Set $$ k=\int_1^{\infty}\frac{\cos u}{\sqrt{u^3}}\,du $$ so we have $$ \int_1^{\infty}\sin(t^2)\,dt= -\frac{k}{4} -\frac{1}{2}\lim_{v\to\infty}\left[\frac{\cos u}{\sqrt{u}}\right]_1^v $$ which is finite, say $K$. Then your integral is $$ \int_0^{\infty}\sin(t^2)\,dt=K+\int_0^1\sin(t^2)\,dt $$

The fact that this integral is $\sqrt{\pi/8}$ is much more complicated to establish and requires higher level methods.


I thought it might be instructive to present a way forward that uses contour integration in the complex plane. To that end, let $I$ be the integral given by

$$\begin{align} I&=\int_0^\infty e^{it^2}\,dt\\\\ &= \lim_{R\to \infty}\int_{0}^R e^{it^2}\,dt \end{align}$$

Next, we analyze the closed-contour integral $J$ given by

$$J=\oint_{C_1+C_2+C_3} e^{iz^2}\,dz$$


(i) $C_1$ is the straight line segment from $(0,0)$ to $(R,0)$;

(ii) $C_2$ is the circular arc of radius $R$ from $(R,0)$ to $(R/\sqrt 2,R/\sqrt 2)$;

(iii) $C_3$ is the straight line segment from $(R/\sqrt 2,R/\sqrt 2)$ to $(0,0)$.

Therefore, we can write

$$\begin{align} J&=\int_0^R e^{ix^2}\,dx+\int_0^{\pi/4} e^{i R^2e^{i2\theta}}\,iRe^{i\theta}\,d\theta+\int_{R}^{0} e^{i(1+i)^2t^2/2}\,\frac{1+i}{\sqrt 2}\,dt\\\\ &\int_0^R e^{ix^2}\,dx+\int_0^{\pi/4} e^{i R^2e^{i2\theta}}\,iRe^{i\theta}\,d\theta-\frac{1+i}{\sqrt 2}\int_{0}^{R}e^{-t^2}\,dt \tag 1 \end{align}$$

Since $e^{iz^2}$ is analytic in and on $C_1+C_2+C_3$, Cauchy's Integral Theorem guarantees that $J=0$.

And as $R\to \infty$, the first integral on the right-hand side of $(1)$ approaches $I$, the second integral approaches $0$, and the third approaches $(1+i)\sqrt{\pi/8}$ (see Gaussian Integral).

Thus, we conclude that

$$I=\int_0^\infty \cos(t^2)\,dt+i\int_0^\infty \sin(t^2)\,dt=(1+i)\sqrt{\pi/8}$$

whereupon equating real and imaginary parts reveals

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \cos(t^2)\,dt=\int_0^\infty \sin(t^2)\,dt=\sqrt{\pi/8}}$$

as was to be shown!


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