\(\text{Integral of the month: } \int dr \cos r^2\)

Abstract

Fresnel integrals are a pair of integrals that are used to calculate the diffraction patterns of light waves. This post explores the mathematics of Fresnel integrals, a fundamental wave phenomenon that limits the resolution of optical instruments like telescopes. We first derive the results in the large \(r\) limit using the residue theorem. Then we will see how to evaluate the integrals numerically.

Keywords

diffraction, fresnel, fraunhofer

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The Fresnel integrals are defined as follows: \[\begin{eqnarray} S(t)&=& \int_0^t dr \sin r^2,\nonumber\\ C(t)&=& \int_0^t dr \cos r^2 \label{eq:cst}. \end{eqnarray}\] For a general value of \(t\), the integrals need to be evaluated numerically. However, the asymptotic values \(C(t)\) and \(S(t)\) can be calculated via the closed contour integral below: \[\begin{eqnarray} I&=& \oint_C dz e^{- z^2} \label{eq:cint}. \end{eqnarray}\] where the contour \(C\) is illustrated in Figure 1.

Figure 1: The contour to evaluate the integral. The return path, \(\gamma_1\), is chosen such that the integrand reduces to the regular Gaussian.

Let’s first evaluate the integral on \(\gamma_0\) in the limit \(R\to\infty\): \[\begin{eqnarray} I_{\gamma_0}&=& \lim_{R\to\infty} \int_0^R dr e^{- r^2}=\frac{\sqrt{\pi}}{2}, \label{eq:gamma0} \end{eqnarray}\] where the details of the derivation can be found here. Now consider the (absolute value of the) integral on \(\gamma_R\) in the limit \(R\to\infty\): \[\begin{eqnarray} \left| I_{\gamma_R}\right|&=& \left| \lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{i\theta}e^{- R^2 (\cos^2\theta-\sin^2\theta+2i \cos\theta\sin\theta)}\right| = \left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{i\theta+i\sin(2\theta)}e^{- R^2 \cos(2\theta)}\right|\nonumber\\ &\leq&\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \cos(2\theta)} \right|. \label{eq:gammaR} \end{eqnarray}\] Let’s try to put a bound on \(\cos(2\theta)\) in the range \(0\leq\theta\leq\pi/4\). At \(\cos(2\theta)\vert_{\theta=0}=1\) and \(\cos(2\theta)\vert_{\theta=\pi/4}=0\). We can draw a line that connects these two points: \(1-\frac{4\theta}{\pi}\). Since \(\frac{d^2}{d\theta^2}\cos(2\theta)=-4 \cos(2\theta)<0\) for \(0<\theta<\pi/4\), we know that \(\cos(2\theta)<1-\frac{4\theta}{\pi}\) in this range. This observation is illustrated in Figure 2.

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Figure 2: \(\cos 2\theta\) and a bound on it with the line \(1-\frac{4\theta}{\pi}\)

We can now go back to Eq. \(\ref{eq:gamma1}\) and make use of the bound:

\[\begin{eqnarray} \left| I_{\gamma_R}\right| &\leq&\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \cos(2\theta)} \right| \leq\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \left(1-\frac{4\theta}{\pi}\right)} \right|\nonumber\\ &=&\left|\lim_{R\to\infty} R e^{- R^2} \int_0^{\frac{\pi}{4}} d\theta e^{R^2\frac{4\theta}{\pi}} \right| \leq \left|\lim_{R\to\infty} R \frac{\pi}{4 R^2} \left(1- e^{- R^2}\right) \right|\nonumber\\ &=&\left|\lim_{R\to\infty} \frac{\pi}{4 R} \left(1- e^{- R^2}\right) \right|=0. \label{eq:gammaR2} \end{eqnarray}\]

Finally, let’s look at the integral on \(\gamma_1\) in the limit \(R\to\infty\): \[\begin{eqnarray} I_{\gamma_1}&=& \lim_{R\to\infty} R \int_R^{0} dr e^{\frac{i\pi}{4}}e^{- r^2 \frac{i\pi}{2}} =-\frac{1+i}{\sqrt{2}}\lim_{R\to\infty} \int_0^R dr \left( \cos r^2 -i\sin r^2\right)\nonumber\\ &=& -\frac{1+i}{\sqrt{2}} \int_0^\infty dr \left( \cos r^2 -i\sin r^2\right)\nonumber\\ &=&-\frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 +i\left[\int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right] \right). \label{eq:gamma1} \end{eqnarray}\] As we have computed individual pieces of the integral Eq. \(\ref{eq:cint}\), we can assemble them and state that they need to add to \(0\) since \(e^{-z^2}\) is analytic everywhere. Therefore we have: \[\begin{eqnarray} I&=&\oint_C dz e^{- z^2}=0=I_{\gamma_0}+I_{\gamma_R}+I_{\gamma_1}\nonumber\\ &=&\frac{\sqrt{\pi}}{2}+0-\frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 +i\left[\int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right] \right) \label{eq:cintf}. \end{eqnarray}\] Matching the real and imaginary parts, we get:

\[\begin{eqnarray} \frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 \right)&=&\frac{\sqrt{\pi}}{2},\nonumber\\ \frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right)&=&0, \label{eq:cintparts} \end{eqnarray}\] from which we get \[\begin{eqnarray} \int_0^\infty dr \cos r^2 =\int_0^\infty dr \sin r^2 &=&\frac{1}{2}\sqrt{\frac{\pi}{2}}\simeq 0.626 \label{eq:cintpartsf}. \end{eqnarray}\]

Now we can conclude with the plots of the Fresnel integrals in Figure 3 .

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Figure 3: Left: Fresnel integrals as a function of their argument, Right: parametric plot of the integrals forming the Euler spiral.