Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have seen the Fresnel integral

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

evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, however, does anyone know a real analysis method to evaluate this integral?

share|improve this question
1  
See mathdl.maa.org/images/cms_upload/Chen-CMJ0926332.pdf. –  KCd Aug 29 '12 at 1:56
add comment

5 Answers

up vote 11 down vote accepted

Let $u=x^2$, then $$ \int_0^\infty \sin(u) \frac{\mathrm{d} u}{2 \sqrt{u}} $$ The real analysis way of evaluating this integral is to consider a parametric family: $$\begin{eqnarray} I(\epsilon) &=& \int_0^\infty \frac{\sin(u)}{2 \sqrt{u}} \mathrm{e}^{-\epsilon u} \mathrm{d} u = \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\int_0^\infty u^{2n+\frac{1}{2}} \mathrm{e}^{-\epsilon u} \mathrm{d} u \\ &=& \frac{1}{2} \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} \Gamma\left(2n+\frac{3}{2}\right) \epsilon^{-\frac{3}{2}-2n} \\ &=& \frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(2n+\frac{3}{2}\right)}{\Gamma\left(2n+2\right)} \\ &\stackrel{\Gamma-\text{duplication}}{=}&\frac{1}{2 \epsilon^{3/2}} \sum_{n=0}^\infty \left(-\frac{1}{\epsilon^2}\right)^n\frac{\Gamma\left(n+\frac{3}{4}\right)\Gamma\left(n+\frac{5}{4}\right)}{\sqrt{2} n! \Gamma\left(n+\frac{3}{2}\right)} \\ &=& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} {}_2F_1\left(\frac{3}{4}, \frac{5}{4}; \frac{3}{2}; -\frac{1}{\epsilon^2}\right) \\ &\stackrel{\text{Euler integral}}{=}& \frac{1}{(2 \epsilon)^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{1}{\operatorname{B}\left(\frac{5}{4}, \frac{3}{2}-\frac{5}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{3}{2}-\frac{5}{4} -1} \left(1+\frac{x}{\epsilon^2}\right)^{-3/4} \mathrm{d} x \\ &=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma\left(\frac{3}{2}\right)} \frac{\Gamma\left(\frac{3}{2}\right)}{\Gamma\left(\frac{5}{4}\right) \Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{5}{4}-1} (1-x)^{\frac{1}{4} -1} \left(\epsilon^2+x\right)^{-3/4} \mathrm{d} x \end{eqnarray} $$ Now we are ready to compute $\lim_{\epsilon \to 0} I(\epsilon)$: $$\begin{eqnarray} \lim_{\epsilon \to 0} I(\epsilon) &=& \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \int_0^1 x^{\frac{1}{2}-1} \left(1-x\right)^{\frac{1}{4}-1} \mathrm{d} x = \frac{1}{2^{3/2}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right)} \\ &=& \frac{1}{2^{3/2}} \Gamma\left(\frac{1}{2}\right) = \frac{1}{2} \sqrt{\frac{\pi}{2}} \end{eqnarray} $$

share|improve this answer
    
It seems that you have assumed $\epsilon > 1$ in the intermediate steps, and identified $I(\epsilon)$ with the last line for all $\epsilon > 0$ by analytic continuation argument. Is it right? –  sos440 Aug 28 '12 at 4:28
3  
The original integral $I(\epsilon)$ is defined for $\epsilon>0$ and is a continuous function of $\epsilon$. The convergence of the series requires $\epsilon>1$, but the Euler integral now converges for $\epsilon>0$. By the principle of analytic continuation $I(\epsilon)$ equals to the Euler integral. –  Sasha Aug 28 '12 at 4:35
    
Thanks, now I clearly understood (except the step using hypergeometric series, which I barely know.) –  sos440 Aug 28 '12 at 4:41
add comment

Aside from some trigonometric substitutions and identities, we will need the following identity, which can be shown using integration by parts twice: $$ \int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}\tag{1} $$ We will also use the standard arctangent integral: $$ \int_0^\infty\frac{\mathrm{d}t}{a^2+t^2}=\frac\pi{2a}\tag{2} $$ Now $$ \begin{align} &\left(\int_0^\infty\color{#C00000}{\sin}(x^2) e^{-\lambda x^2}\,\mathrm{d}x\right)^2\\ &=\int_0^\infty\int_0^\infty \color{#C00000}{\sin}(x^2)\color{#C00000}{\sin}(y^2) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.1}\\ &=\frac12\int_0^\infty\int_0^\infty \left(\cos(x^2-y^2) \color{#FF0000}{-}\cos(x^2+y^2)\right) e^{-\lambda(x^2+y^2)}\,\mathrm{d}y\,\mathrm{d}x\tag{3.2}\\ &=\frac12\int_0^{\pi/2}\int_0^\infty \left(\cos(r^2\cos(2\phi)) \color{#FF0000}{-}\cos(r^2)\right)e^{-\lambda r^2} \,r\,\mathrm{d}r\,\mathrm{d}\phi\tag{3.3}\\ &=\frac14\int_0^{\pi/2}\int_0^\infty \left(\cos(s\cos(2\phi)) \color{#FF0000}{-}\cos(s)\right) e^{-\lambda s} \,\mathrm{d}s\,\mathrm{d}\phi\tag{3.4}\\ &=\frac14\int_0^{\pi/2}\left(\frac{\lambda}{\cos^2(2\phi)+\lambda^2} \color{#FF0000}{-}\frac{\lambda}{1+\lambda^2}\right)\,\mathrm{d}\phi\tag{3.5}\\ &=\frac12\int_0^{\pi/4} \frac{\lambda}{\cos^2(2\phi)+\lambda^2}\,\mathrm{d}\phi \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.6}\\ &=\frac14\int_0^{\pi/4} \frac{\lambda\,\mathrm{d}\tan(2\phi)} {1+\lambda^2+\lambda^2\tan^2(2\phi)} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.7}\\ &=\frac14\int_0^\infty\frac{\mathrm{d}t}{1+\lambda^2+t^2} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.8}\\ &=\frac{\pi/8}{\sqrt{1+\lambda^2}} \color{#FF0000}{-}\frac{\lambda\pi/8}{1+\lambda^2}\tag{3.9} \end{align} $$

$(3.1)$ change the square of the integral into a double integral

$(3.2)$ use $2\color{#C00000}{\sin}(A)\color{#C00000}{\sin}(B)=\cos(A-B)\color{#FF0000}{-}\cos(A+B)$

$(3.3)$ convert to polar coordinates

$(3.4)$ substitute $s=r^2$

$(3.5)$ apply $(1)$

$(3.6)$ pull out the constant and apply symmetry to reduce the domain of integration

$(3.7)$ multiply numerator and denominator by $\sec^2(2\phi)$

$(3.8)$ substitute $t=\lambda\tan(2\phi)$

$(3.9)$ apply $(2)$

Finally, take the square root of both sides of $(3)$ and let $\lambda\to0^+$ to get $$ \int_0^\infty\color{#C00000}{\sin}(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{4} $$

Addendum

I just noticed that the same proof works for $$ \int_0^\infty\cos(x^2)\,\mathrm{d}x=\sqrt{\frac\pi8}\tag{5} $$ if each red $\color{#C00000}{\sin}$ is changed to $\cos$ and each red $\color{#FF0000}{-}$ sign is changed to $+$.

share|improve this answer
    
@Sasha: this answer should be better than my last :-) –  robjohn Sep 3 '12 at 3:27
    
I love this method! It is a novel application of the polar coordinate transform for me. –  sos440 Sep 4 '12 at 6:49
    
This is nice that only standard "freshmen calculus" knowledge is used. –  i707107 Oct 25 '13 at 1:51
    
Actually, justification of $\lambda\rightarrow 0$ would be something nontrivial. –  i707107 Nov 6 '13 at 9:23
    
I mean in the integral, explaining why the limit of LHS is the value at 0. –  i707107 Nov 6 '13 at 9:29
show 4 more comments

First, use the substitution $t = x^2$ to obtain

$$ I := \int_{0}^{\infty}\sin\left(x^2\right)\;dx=\int_{0}^{\infty}\frac{\sin t}{2\sqrt{t}}\;dt. $$

This integral converges conditionally, thus we make an integration by parts to obtain an absolutely convergent integral as follows:

$$I = \left[\frac{1-\cos t}{2\sqrt{t}}\right]_{0}^{\infty}-\int_{0}^{\infty}\left(\frac{d}{dt}\frac{1}{2\sqrt{t}}\right)(1 - \cos t)\;dt =\int_{0}^{\infty}\frac{1 - \cos t}{4t^{3/2}}\;dt.$$

Now, from gamma integral,

$$ \begin{align*} I &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\frac{\Gamma(3/2)}{t^{3/2}}\right)(1 - \cos t)\;dt \\ &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\left(\int_{0}^{\infty}u^{1/2}e^{-tu}\;du\right)(1 - \cos t)\;dt\\ &=\frac{1}{4\Gamma(3/2)}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt \\ &=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dudt, \end{align*} $$

where in the last line we have used the fact that $\Gamma(3/2) = \frac{1}{2}\sqrt{\pi}$, which is a direct consequence of the Gaussian integral. (Of course, this integral can be evaluated by a famous real analysis technique.) By Tonelli's theorem we can change the order of integration, and with the substitution $u = v^2$ we obtain

$$\begin{align*}I &=\frac{1}{2\sqrt{\pi}}\int_{0}^{\infty}\int_{0}^{\infty}u^{1/2}e^{-tu}(1 - \cos t)\;dtdu \\ &= \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty}\frac{u^{1/2}}{u(1+u^2)}\;du = \frac{1}{2\sqrt{\pi}} \int_{0}^{\infty} \frac{2}{1+v^4} \; dv. \end{align*}$$

To evaluate the last integral, we use the following decomposition

$$ \frac{2}{1+v^4} = \frac{1+v^{-2}}{(v-v^{-1})^2-2} - \frac{1-v^{-2}}{(v+v^{-1})^2-2}. $$

Thus with the substitution $z = v - v^{-1}$ and $w = v + v^{-1}$, the integral becmes

$$\begin{align*}I &= \frac{1}{2\sqrt{\pi}} \left( \int_{0}^{\infty} \frac{d(v-v^{-1})}{(v-v^{-1})^2+2} - \int_{0}^{\infty} \frac{d(v+v^{-1})}{(v+v^{-1})^2-2} \right) \\ &= \frac{1}{2\sqrt{\pi}} \left( \int_{-\infty}^{\infty} \frac{dz}{z^2+2} - \int_{\infty}^{\infty} \frac{dw}{w^2-2} \right) = \frac{1}{2\sqrt{\pi}} \left( \frac{\pi}{\sqrt{2}} - 0 \right) = \sqrt{\frac{\pi}{8}}. \end{align*}$$

Only a slight modification of this argument immediately yields

$$ \int_{0}^{\infty} \cos\left(x^2\right) \; dx = \sqrt{\frac{\pi}{8}}. $$

share|improve this answer
    
@Sasha, Thank you. At first, I mistakenly posted a calculation for $\int_{0}^{\infty}\cos\left(x^2\right)\;dx$. When making amends in a hurry, I forgot fixing that. –  sos440 Aug 28 '12 at 4:02
    
This way of evaluating $\int_0^\infty \frac{dv}{1+v^4}$ is new to me. Another, maybe shorter way, would be use $v=\left(\frac{u}{1-u}\right)^{1/4}$: $$\int_0^\infty \frac{dv}{1+v^4} = \int_0^1 \frac{1}{4} (1-u)^{3/4-1} u^{1/4-1} \mathrm{d} u = \frac{1}{4} \operatorname{B}\left(\frac{1}{4}, \frac{3}{4}\right) = \frac{1}{4} \sqrt{2} \pi $$ –  Sasha Aug 28 '12 at 4:15
    
@Sasha, I also strongly agree that the use of beta function saves an enormous amount of effort. I also enjoy exploiting it. Here, however, I decided to try some sort of minimalism in a burden of background knowledge, to make a partial excuse for an alleged claim that real-analysis-approach is much harder. –  sos440 Aug 28 '12 at 4:24
add comment

Here is paper that addresses this exact question by H. Flanders, "On Fresnel integrals", American Mathematical Monthly, vol. 89, no. 4, 1982, pp. 264-266.

Here is another paper on the topic that does not require a subscription.

share|improve this answer
add comment

Integrate $\displaystyle{J \equiv \int_{0}^{\infty}{\rm e}^{{\rm i}x^{2}}\,{\rm d}x}$. $\quad\Im J = ?$.

\begin{eqnarray*} J^{2} & = & \int_{0}^{\infty}{\rm e}^{{\rm i}x^{2}}\,{\rm d}x \int_{0}^{\infty}{\rm e}^{{\rm i}y^{2}}\,{\rm d}y = {\pi \over 2}\int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho^{2}}\rho\,{\rm d}\rho = {\pi \over 4}\int_{0}^{\infty}{\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho \\[3mm] & = & {\pi \over 4}\int_{-\infty}^{\infty} \left(% {\rm i}\int_{-\infty}^{\infty}{{\rm d}k \over 2\pi}\, {{\rm e}^{-{\rm i}k\rho} \over k + {\rm i}0^{+}} \right) {\rm e}^{{\rm i}\,\rho}\,{\rm d}\rho = {\rm i}\,{\pi \over 4}\int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}} \int_{-\infty}^{\infty}{{\rm d}\rho \over 2\pi}\,{\rm e}^{{\rm i}\left(1 - k\right)\rho} \\[3mm] & = & {\rm i}\,{\pi \over 4}\int_{-\infty}^{\infty}{{\rm d}k \over k + {\rm i}0^{+}} \delta\left(1 - k\right) = {\rm i}\,{\pi \over 4}{1 \over 1 + {\rm i}0^{+}} = {\rm i}\,{\pi \over 4}\left\lbrack 1 - {\rm i}\pi\,\delta\left(1\right)\right\rbrack = {\rm i}\,{\pi \over 4} = {\rm e}^{{\rm i}\pi/2}\,{\pi \over 4} \\[1cm]&&\mbox{} \end{eqnarray*}

$$ J = \sqrt{\,{\rm e}^{{\rm i}\pi/2}\,{\pi \over 4}\,} = {\rm e}^{{\rm i}\pi/4}\,\sqrt{\,\pi \over 4\,} $$

$$ \int_{0}^{\infty}\sin\left(x^{2}\right)\,{\rm d}x = \Im J = \overbrace{\quad\sin\left(\pi \over 4\right)\quad}^{=\ 1/\sqrt{\,2\,}} \sqrt{\,\pi \over 4\,} = \sqrt{\,\pi \over 8\,} $$

share|improve this answer
    
A DownVote when somebody does not understand a neat procedure is better than an UpVote. –  Felix Marin Nov 4 '13 at 4:24
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.