scipy.special.fresnel#

scipy.special.fresnel(z, out=None) = <ufunc 'fresnel'>#

Fresnel integrals.

The Fresnel integrals are defined as

\[\begin{split}S(z) &= \int_0^z \sin(\pi t^2 /2) dt, \\ C(z) &= \int_0^z \cos(\pi t^2 /2) dt.\end{split}\]

See [dlmf] for details.

Parameters:

zarray_like: Real or complex valued argument
out2-tuple of ndarrays, optional: Optional output arrays for the function results

Returns:

S, C2-tuple of scalar or ndarray: Values of the Fresnel integrals

See also

fresnel_zeros: zeros of the Fresnel integrals

References

[dlmf]

NIST Digital Library of Mathematical Functions https://dlmf.nist.gov/7.2#iii

Examples

>>> import numpy as np
>>> from scipy.special import fresnel, erf

First, we verify the following limits:

\[\lim_{z \to \infty} S(z) = \lim_{z \to \infty} C(z) = 1/2.\]

>>> S, C = fresnel([0.1, 1, 10, 100, np.inf])
>>> S
array([0.00052359, 0.43825915, 0.46816998, 0.4968169 , 0.5       ])
>>> C
array([0.09999753, 0.7798934 , 0.49989869, 0.4999999 , 0.5       ])

Next, we verify the following relation to the error function:

\[C(z) + i S(z) = \frac{1 + i}{2} \operatorname{erf}\left(\frac{\sqrt{\pi}(1 - i)}{2} z\right).\]

>>> z = np.linspace(-10, 10)
>>> S, C = fresnel(z)
>>> RHS = (1 + 1j)/2 * erf((np.sqrt(np.pi)*(1 - 1j))/2*z)
>>> np.allclose(C + 1j*S, RHS)
True

Finally, we plot \(C(z)\) against \(S(z)\) to get the Euler or Cornu spiral.

>>> import matplotlib.pyplot as plt
>>> z = np.linspace(-10, 10, num=1000)
>>> S, C = fresnel(z)
>>> fig, ax = plt.subplots()
>>> ax.plot(C, S)
>>> ax.set_aspect('equal')
>>> ax.set_xlabel('$C(z)$')
>>> ax.set_ylabel('$S(z)$')
>>> plt.show()

../../_images/scipy-special-fresnel-1.png