asymptotic approximation of Fresnel integrals with complex argument

Question

It turns out that SciPy's Fresnel values are wrong for complex arguments and large enough absolute value. I'm trying to fix that.

The implementation is based on Zhang/Jin, Computation of special functions, which in turn is based on Abramowitz/Stegun, Handbook of Mathematical Functions. There we find for the Fresnel S integral (7.3.10) $$ S(z) = \frac{1}{2} - f(z) \cos\left(\frac{\pi}{2} z^2\right) - g(z) \sin\left(\frac{\pi}{2} z^2\right) $$ for all $z$ with the auxiliary functions (7.3.5), (7.3.6) $$ \begin{split} f(z) &= \left[\frac{1}{2} - S(z) \right] \cos\left(\frac{\pi}{2} z^2\right) - \left[\frac{1}{2} - C(z) \right] \sin\left(\frac{\pi}{2} z^2\right),\\ g(z) &= \left[\frac{1}{2} - C(z) \right] \cos\left(\frac{\pi}{2} z^2\right) + \left[\frac{1}{2} - S(z) \right] \sin\left(\frac{\pi}{2} z^2\right). \end{split} $$

Computation of $S$ for large values is done via the asymptotic expansion of $f$ (7.3.27) $$ \DeclareMathOperator\arg{arg} \pi z f(z)\sim 1 + \sum_{m=1}^\infty (-1)^m \frac{1\cdot 3\cdot\cdots \cdot (4m-1)}{(\pi z^2)^{2m}}, \quad z\to\infty, |\arg(z)|<\frac{\pi}{2}. $$ This is where I have problems understanding the approximation. Consider $f(iz)$; for the asymptotic $f_a$, we have $f_a(iz) = -if_a(z)$. However no such thing is true for $f$ itself. Numerical computation via the representations $$ \DeclareMathOperator\erf{erf} \begin{split} S(z) &= \frac{1 + i}{4} \left[\erf\left(\frac{1 + i}{2} \sqrt{\pi} z\right) - i \erf\left(\frac{1 - i}{2} \sqrt{\pi} z\right)\right]\\ C(z) &= \frac{1 - i}{4} \left[\erf\left(\frac{1 + i}{2} \sqrt{\pi} z\right) + i \erf\left(\frac{1 - i}{2} \sqrt{\pi} z\right)\right] \end{split} $$ shows that the approximation incorrect for everything off of the real axis, but the issue here could be numerical instability in the $\erf$ represenation too.

My current guess is that the above infinite sum is valid only in the area $|\arg{z}|<\pi/4$, which would already change how we compute the Fresnel integral values significantly.

To finish things off, here's a cplot of $f$ (for smaller $|z|$):

Answer 1

I will consider the function $\operatorname{f}(z)$, the treatment of $\operatorname{g}(z)$ is analogous. By http://dlmf.nist.gov/7.12.ii, we have $$ \operatorname{f}(z) = \frac{1}{{\pi z}}\sum\limits_{m = 0}^{N - 1} {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}} + R_N^{(\operatorname{f})} (z) $$ where \begin{align*} R_N^{(\operatorname{f})} (z) & = \frac{{( - 1)^N }}{{\pi \sqrt 2 }}\int_0^{ + \infty } {\frac{{e^{ - \pi z^2 t/2} t^{2N - 1/2} }}{{1 + t^2 }}dt} \\ & = \frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}\frac{1}{{\Gamma \left( {2N + \frac{1}{2}} \right)}}\int_0^{ + \infty } {\frac{{e^{ - s} s^{2N - 1/2} }}{{1 + s^2 /(\pi z^2 /2)^2 }}ds} \\ & = \frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}\Pi _{2N + 1/2} (\pi z^2 /2), \end{align*} provided that $|\arg z|<\frac{\pi}{4}$ and $N\geq 0$. Here $\Pi_p(w)$ denotes one of Dingle's basic terminants: $$ \Pi _p (w) = \frac{1}{{\Gamma (p)}}\int_0^{ + \infty } {\frac{{e^{ - s} s^{p - 1} }}{{1 + (s/w)^2 }}ds} $$ for $|\arg w|<\frac{\pi}{2}$ and by analytic continuation elswhere. Using the expression for $R_N^{(\operatorname{f})} (z)$ in terms of this terminant, we can extend $R_N^{(\operatorname{f})} (z)$ to the universal covering of $\mathbb C \setminus \left\{ 0\right\}$. Now employing the estimates for the basic terminant established in https://doi.org/10.1007/s10440-017-0099-0, we obtain the bound \begin{align*} \left| {R_N^{(\operatorname{f})} (z)} \right| \le &\; \left| {\frac{1}{{\pi z}}( - 1)^N \left( {\frac{1}{2}} \right)_{2N} \frac{1}{{(\pi z^2 /2)^{2N} }}} \right| \\ & \times \begin{cases} 1 & \text{ if } \; \left|\arg z\right| \leq \frac{\pi}{8}, \\ \min\!\Big(\left|\csc ( 4\arg z)\right|,1 + \cfrac{1}{2}\chi(2N+1/2)\Big) & \text{ if } \; \frac{\pi}{8} < \left|\arg z\right| \leq \frac{\pi}{4}, \\ \cfrac{\sqrt {2\pi (2N + 1/2)} }{2\left| {\sin (2\arg z)} \right|^{2N+1/2} } + 1 + \cfrac{1}{2}\chi (2N +1/2) & \text{ if } \; \frac{\pi}{4} < \left|\arg z\right| < \frac{\pi}{2}. \end{cases} \end{align*} Here $\chi(p) =\sqrt{\pi}\Gamma(p/2+1)/\Gamma(p/2+1/2)$ for $p>0$. It is seen that the asymptotic expansion of $\operatorname{f}(z)$ is valid in every closed sub-sector of $|\arg z|<\frac{\pi}{2}$ in the sense of Poincaré. The Stokes lines are $\arg z =\pm \frac{\pi}{4}$ where terms are swiched on that remain exponentially small compared to any negative power of $z$ as long as we stay away from the rays $\arg z =\pm \frac{\pi}{2}$ (the anti-Stokes lines).

To obtain a better result and reveal the exponentially small terms, we can use the functional equation $$ \Pi _p (w) = \pm \pi i\frac{{e^{ \mp \frac{\pi }{2}ip} }}{{\Gamma (p)}}w^p e^{ \pm iw} + \Pi _p (we^{ \mp \pi i} ) $$ where $p>0$ and $w$ is any element of the universal covering of $\mathbb C \setminus \left\{ 0\right\}$ (the Riemann surface of the logarithm). With this functional equation, we find, after some algebra, $$ R_N^{(\operatorname{f})} (ze^{ \mp \frac{\pi }{2}i} ) = \pm iR_N^{(\operatorname{f})} (z) + \frac{{1 \mp i}}{2}e^{ \pm \frac{\pi }{2}iz^2 } . $$ This result is valid for all $N\geq 0$ and $z$ on the universal covering of $\mathbb C \setminus \left\{ 0\right\}$. You can see that if we omit the second term (which is exponentially small when $ 0 < \pm \arg z < \frac{\pi}{2}$) then, with $N=0$, we get the false result $$ \operatorname{f}(ze^{ \mp \frac{\pi }{2}i} ) = \pm i\operatorname{f}(z). $$ See also http://dlmf.nist.gov/7.4.

In summary, use $$ \operatorname{f}(z) \sim \frac{1}{{\pi z}}\sum\limits_{m = 0}^\infty {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}} $$ when $\left| {\arg z} \right| \le \frac{\pi }{4}$, and use $$ \operatorname{f}(z) \sim \frac{{1 \pm i}}{2}e^{ \pm \frac{\pi }{2}iz^2 } + \frac{1}{{\pi z}}\sum\limits_{m = 0}^\infty {( - 1)^m \left( {\frac{1}{2}} \right)_{2m} \frac{1}{{(\pi z^2 /2)^{2m} }}} $$ when $\frac{\pi }{4} < \pm \arg z \le \frac{{3\pi }}{4}$. Of couse by relaying on the symmetry relation http://dlmf.nist.gov/7.4.E8, we can assume $|\arg z|\leq \frac{\pi}{2}$.