2

I have to prove that integral

$I = \int_{0}^{+\infty}\sin(t^2)dt$ is convergent. Could you tell me if it's ok?

Let $t^2=u$ then $dt=\frac{du}{2\sqrt{u}}$

Now $$I = \int_{0}^{+\infty}\frac{\sin(u)du}{2\sqrt{u}}$$

Which is equal to $$\int_{0}^{1}\frac{\sin(u)du}{2\sqrt{u}} + \int_{1}^{+\infty}\frac{\sin(u)du}{2\sqrt{u}}$$

First of these is convergent because of the limit

$$\lim_{u\to 0}\frac{sin(u)}{2\sqrt{u}} = 0$$

Second is convergent from Dirichlet test.

Is it correct? Also how to find the value of this integral ($\sqrt{\frac{\pi}{8}}$) ?

shai horowitz
  • 2,086
janusz
  • 1,205
  • The first is correct. For the second, what is the Dirichlet test (I thought that was something for series)? About the value, when the limit of integration is finite it is given by Fresnel integral (see here). I am not sure there is an elementary way to derive the exact value of this integral. – Fimpellizzeri May 19 '16 at 18:20
  • @Fimpellizieri A lot of convergence tests work for series and integrals with only small changes in formulation. Then it's common to call both versions by the same name [whether the first discoverer of the test discovered both versions or not]. – Daniel Fischer May 19 '16 at 19:00

2 Answers2

2

To evaluate the integral, we analyze the closed-contour integral $I$ given by

$$I=\oint_C e^{iz^2}\,dz$$

where $C$ is comprised of (i) the line segment from $0$ to $R$, (ii) the circular arc from $R$ to $R(1+i)/\sqrt{2}$, and the line segment from $R(1+i)/\sqrt{2}$ to $0$.

Since $e^{iz^2}$ is analytic in and on $C$, Cauchy's Integral Theorem guarantees that $I=0$. Then, we have

$$\int_0^R e^{ix^2}\,dx+\int_0^{\pi/4}e^{iR^2e^{i2\phi}}iRe^{i\phi}\,d\phi-\frac{1+i}{\sqrt{2}}\int_0^R e^{-x^2}\,dx=0 \tag 1$$

Letting $R\to \infty$, the second integral on the left-hand side of $(1)$ approaches zero. Therefore, we find that

$$\begin{align} \int_0^\infty e^{ix^2}\,dx&=\frac{1+i}{\sqrt{2}}\int_0^\infty e^{-x^2}\,dx\\\\&=\frac{1+i}{\sqrt{2}}\frac{\sqrt{\pi}}{2} \tag 2 \end{align}$$

Finally, equating real and imaginary parts of $(2)$, we obtain

$$\int_0^\infty \sin(x^2)\,dx=\sqrt{\frac{\pi}{8}}$$

and

$$\int_0^\infty \cos(x^2)\,dx=\sqrt{\frac{\pi}{8}}$$

Mark Viola
  • 179,405
0

Putting $\sqrt{u}=t$ and $\dfrac{du}{2\sqrt{u}}=dt$,

$$I = \int_{0}^{+\infty}\frac{\sin(u)du}{2\sqrt{u}}\\=\int_0^{+\infty} \sin(t^2) dt\\=\int_0^{+\infty}\Im(e^{it^2})dt\\=\Im\left(\int_0^{+\infty}e^{it^2}dt\right)$$

Putting $it^2=-z$ and $t=\sqrt{-\frac1iz}=\sqrt{i}\sqrt{z}$, $$dt = \sqrt{i}\frac{1}{2\sqrt{z}}dz$$

Hence,

\begin{align*} I &=\Im\left(\int_0^{+\infty}e^{-z}\sqrt{i}\frac{1}{2\sqrt{z}}dz\right)\\ &=\Im\left(\frac{\sqrt{i}}{2}\int_0^{+\infty} z^{-\frac12}e^{-z} dz\right)\\ &=\Im\left(\frac{\sqrt{i}}{2} \Gamma\left(\frac12\right)\right)\\ &=\Im\left(\sqrt{i}\frac{\sqrt{\pi}}{2}\right)\\ &=\Im\left(\frac{1}{2\sqrt{2}}(1+i)\sqrt{\pi}\right) \tag{!!}\\ &=\fbox{$\sqrt{\frac{\pi}{8}}$} \end{align*}

Now I'm concerned about step $(!!)$ because I don't see why I should take that root of $i$ and not the other one. However since it yields the right answer, I think there must be some reason. Could anyone help me out?

Aritra Das
  • 3,528
  • This development lacks rigor. Upon enforcing the substitution, the limits of integration were not transformed appropriately. In addition, it is not true in general that $\sqrt{ab}=\sqrt{a}\sqrt{b}$ for complex $a$ and $b$. Caution needs to be taken there. I've posted a solution that shows one way forward. ;-)) – Mark Viola May 19 '16 at 18:34
  • @Dr.MV Moreover for some weird reason I have to take one particular square root of $i$ and not the other. This really is a ridiculous attempt. – Aritra Das May 19 '16 at 18:40
  • @Dr.MV Why do you say the limits weren't transformed properly? Isn't $\sqrt{-\infty}=\infty$ ? – Aritra Das May 19 '16 at 18:48
  • 1
    No, $\sqrt{-\infty}=\pm i \infty$. And in this transformation, one has $\sqrt{i\infty}=\pm\left(\frac{1+i}{\sqrt{2}}\right)\infty$. Then, one must decide on a suitable branch cut to define the square root as a single-valued function. – Mark Viola May 19 '16 at 18:56
  • @Dr.MV Aren't $\pm\infty$ one and the same thing in complex numbers? – Aritra Das May 19 '16 at 19:09
  • No they are not. – Mark Viola May 19 '16 at 19:12
  • @Dr.MV http://math.stackexchange.com/a/127550/229480 – Aritra Das May 19 '16 at 19:18
  • 1
    We are talking about two different things. In the case at hand, you have made a change of variables that augments the integration contour. It no longer is along the real axis upon enforcing the transformation. Rather, the new contour lies on a line that is parametrically described by $z=t\left(\frac{1+i}{\sqrt{2}}\right)$, where $t\in [0,\infty)$. The other concept - "the point at infinity - does not interfere here. The issue is the path. Now, as it turns out, because the integrand is analytic, the integral is path independent. – Mark Viola May 19 '16 at 19:35
  • @Dr.MV Thank you very much. – Aritra Das May 19 '16 at 19:39
  • 1
    You're welcome. My pleasure. -Mark – Mark Viola May 19 '16 at 19:43