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

The Fresnel integrals are defined as follows: \[\begin{eqnarray} S(t)&=& \int_0^t dr \sin r^2,\nonumber\\ C(t)&=& \int_0^t dr \cos r^2 \tag{1}. \end{eqnarray}\] 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: \[\begin{eqnarray} I&=& \oint_C dz e^{- z^2} \tag{2}, \end{eqnarray}\] where the contour \(C\) is illustrated in Fig. 1.

Let’s first evaluate the integral on \(\gamma_0\) in the limit \(R\to\infty\): \[\begin{eqnarray} I_{\gamma_0}&=& \lim_{R\to\infty} \int_0^R dr e^{- r^2}=\frac{\sqrt{\pi}}{2}, \tag{3} \end{eqnarray}\] where the details of the derivation can be found here. Now consider the (absolute value of the) integral on \(\gamma_R\) in the limit \(R\to\infty\): \[\begin{eqnarray} \left| I_{\gamma_R}\right|&=& \left| \lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{i\theta}e^{- R^2 (\cos^2\theta-\sin^2\theta+2i \cos\theta\sin\theta)}\right| = \left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{i\theta+i\sin(2\theta)}e^{- R^2 \cos(2\theta)}\right|\nonumber\\ &\leq&\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \cos(2\theta)} \right|. \tag{4} \end{eqnarray}\] 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 Fig. 2.

We can now go back to Eq. (6) make use of the bound:

\[\begin{eqnarray} \left| I_{\gamma_R}\right| &\leq&\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \cos(2\theta)} \right| \leq\left|\lim_{R\to\infty} R \int_0^{\frac{\pi}{4}} d\theta e^{- R^2 \left(1-\frac{4\theta}{\pi}\right)} \right| =\left|\lim_{R\to\infty} R e^{- R^2} \int_0^{\frac{\pi}{4}} d\theta e^{R^2\frac{4\theta}{\pi}} \right|\nonumber\\ &\leq& \left|\lim_{R\to\infty} R \frac{\pi}{4 R^2} \left(1- e^{- R^2}\right) \right| =\left|\lim_{R\to\infty} \frac{\pi}{4 R} \left(1- e^{- R^2}\right) \right|=0. \tag{5} \end{eqnarray}\]

Finally, let’s look at the integral on \(\gamma_1\) in the limit \(R\to\infty\): \[\begin{eqnarray} I_{\gamma_1}&=& \lim_{R\to\infty} R \int_R^{0} dr e^{\frac{i\pi}{4}}e^{- r^2 \frac{i\pi}{2}} =-\frac{1+i}{\sqrt{2}}\lim_{R\to\infty} \int_0^R dr \left( \cos r^2 -i\sin r^2\right) =-\frac{1+i}{\sqrt{2}} \int_0^\infty dr \left( \cos r^2 -i\sin r^2\right)\nonumber\\ &=&-\frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 +i\left[\int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right] \right). \tag{6} \end{eqnarray}\] As we have computed individual pieces of the integral Eq.(2), we can assemble them and state that they need to add to \(0\) since \(e^{-z^2}\) is analytic everywhere. Therefore we have: \[\begin{eqnarray} \oint_C dz e^{- z^2}=0&=&I_{\gamma_0}+I_{\gamma_R}+I_{\gamma_1}\nonumber\\ &=&\frac{\sqrt{\pi}}{2}+0-\frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 +i\left[\int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right] \right) \tag{7}. \end{eqnarray}\] Matching the real and imaginary parts, we get:

\[\begin{eqnarray} \frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 +\int_0^\infty dr \sin r^2 \right)&=&\frac{\sqrt{\pi}}{2},\nonumber\\ \frac{1}{\sqrt{2}}\left( \int_0^\infty dr \cos r^2 -\int_0^\infty dr \sin r^2 \right)&=&0, \tag{8} \end{eqnarray}\] from which we get \[\begin{eqnarray} \int_0^\infty dr \cos r^2 =\int_0^\infty dr \sin r^2 &=&\frac{1}{2}\sqrt{\frac{\pi}{2}}\simeq 0.626 \tag{9}. \end{eqnarray}\]

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