Integration (:mod:`scipy.integrate`)
====================================
.. sectionauthor:: Travis E. Oliphant
.. currentmodule:: scipy.integrate
The :mod:`scipy.integrate` sub-package provides several integration
techniques including an ordinary differential equation integrator. An
overview of the module is provided by the help command:
.. literalinclude:: examples/4-1
General integration (:func:`quad`)
----------------------------------
The function :obj:`quad` is provided to integrate a function of one
variable between two points. The points can be :math:`\pm\infty`
(:math:`\pm` ``inf``) to indicate infinite limits. For example,
suppose you wish to integrate a bessel function ``jv(2.5,x)`` along
the interval :math:`[0,4.5].`
.. math::
:nowrap:
\[ I=\int_{0}^{4.5}J_{2.5}\left(x\right)\, dx.\]
This could be computed using :obj:`quad`:
>>> result = integrate.quad(lambda x: special.jv(2.5,x), 0, 4.5)
>>> print result
(1.1178179380783249, 7.8663172481899801e-09)
>>> I = sqrt(2/pi)*(18.0/27*sqrt(2)*cos(4.5)-4.0/27*sqrt(2)*sin(4.5)+
sqrt(2*pi)*special.fresnel(3/sqrt(pi))[0])
>>> print I
1.117817938088701
>>> print abs(result[0]-I)
1.03761443881e-11
The first argument to quad is a "callable" Python object (*i.e* a
function, method, or class instance). Notice the use of a lambda-
function in this case as the argument. The next two arguments are the
limits of integration. The return value is a tuple, with the first
element holding the estimated value of the integral and the second
element holding an upper bound on the error. Notice, that in this
case, the true value of this integral is
.. math::
:nowrap:
\[ I=\sqrt{\frac{2}{\pi}}\left(\frac{18}{27}\sqrt{2}\cos\left(4.5\right)-\frac{4}{27}\sqrt{2}\sin\left(4.5\right)+\sqrt{2\pi}\textrm{Si}\left(\frac{3}{\sqrt{\pi}}\right)\right),\]
where
.. math::
:nowrap:
\[ \textrm{Si}\left(x\right)=\int_{0}^{x}\sin\left(\frac{\pi}{2}t^{2}\right)\, dt.\]
is the Fresnel sine integral. Note that the numerically-computed
integral is within :math:`1.04\times10^{-11}` of the exact result --- well below the reported error bound.
Infinite inputs are also allowed in :obj:`quad` by using :math:`\pm`
``inf`` as one of the arguments. For example, suppose that a numerical
value for the exponential integral:
.. math::
:nowrap:
\[ E_{n}\left(x\right)=\int_{1}^{\infty}\frac{e^{-xt}}{t^{n}}\, dt.\]
is desired (and the fact that this integral can be computed as
``special.expn(n,x)`` is forgotten). The functionality of the function
:obj:`special.expn` can be replicated by defining a new function
:obj:`vec_expint` based on the routine :obj:`quad`:
>>> from scipy.integrate import quad
>>> def integrand(t,n,x):
... return exp(-x*t) / t**n
>>> def expint(n,x):
... return quad(integrand, 1, Inf, args=(n, x))[0]
>>> vec_expint = vectorize(expint)
>>> vec_expint(3,arange(1.0,4.0,0.5))
array([ 0.1097, 0.0567, 0.0301, 0.0163, 0.0089, 0.0049])
>>> special.expn(3,arange(1.0,4.0,0.5))
array([ 0.1097, 0.0567, 0.0301, 0.0163, 0.0089, 0.0049])
The function which is integrated can even use the quad argument
(though the error bound may underestimate the error due to possible
numerical error in the integrand from the use of :obj:`quad` ). The integral in this case is
.. math::
:nowrap:
\[ I_{n}=\int_{0}^{\infty}\int_{1}^{\infty}\frac{e^{-xt}}{t^{n}}\, dt\, dx=\frac{1}{n}.\]
>>> result = quad(lambda x: expint(3, x), 0, inf)
>>> print result
(0.33333333324560266, 2.8548934485373678e-09)
>>> I3 = 1.0/3.0
>>> print I3
0.333333333333
>>> print I3 - result[0]
8.77306560731e-11
This last example shows that multiple integration can be handled using
repeated calls to :func:`quad`. The mechanics of this for double and
triple integration have been wrapped up into the functions
:obj:`dblquad` and :obj:`tplquad`. The function, :obj:`dblquad`
performs double integration. Use the help function to be sure that the
arguments are defined in the correct order. In addition, the limits on
all inner integrals are actually functions which can be constant
functions. An example of using double integration to compute several
values of :math:`I_{n}` is shown below:
>>> from scipy.integrate import quad, dblquad
>>> def I(n):
... return dblquad(lambda t, x: exp(-x*t)/t**n, 0, Inf, lambda x: 1, lambda x: Inf)
>>> print I(4)
(0.25000000000435768, 1.0518245707751597e-09)
>>> print I(3)
(0.33333333325010883, 2.8604069919261191e-09)
>>> print I(2)
(0.49999999999857514, 1.8855523253868967e-09)
Gaussian quadrature (integrate.gauss_quadtol)
---------------------------------------------
A few functions are also provided in order to perform simple Gaussian
quadrature over a fixed interval. The first is :obj:`fixed_quad` which
performs fixed-order Gaussian quadrature. The second function is
:obj:`quadrature` which performs Gaussian quadrature of multiple
orders until the difference in the integral estimate is beneath some
tolerance supplied by the user. These functions both use the module
:mod:`special.orthogonal` which can calculate the roots and quadrature
weights of a large variety of orthogonal polynomials (the polynomials
themselves are available as special functions returning instances of
the polynomial class --- e.g. :obj:`special.legendre `).
Integrating using samples
-------------------------
There are three functions for computing integrals given only samples:
:obj:`trapz` , :obj:`simps`, and :obj:`romb` . The first two
functions use Newton-Coates formulas of order 1 and 2 respectively to
perform integration. These two functions can handle,
non-equally-spaced samples. The trapezoidal rule approximates the
function as a straight line between adjacent points, while Simpson's
rule approximates the function between three adjacent points as a
parabola.
If the samples are equally-spaced and the number of samples available
is :math:`2^{k}+1` for some integer :math:`k`, then Romberg
integration can be used to obtain high-precision estimates of the
integral using the available samples. Romberg integration uses the
trapezoid rule at step-sizes related by a power of two and then
performs Richardson extrapolation on these estimates to approximate
the integral with a higher-degree of accuracy. (A different interface
to Romberg integration useful when the function can be provided is
also available as :func:`romberg`).
Ordinary differential equations (:func:`odeint`)
------------------------------------------------
Integrating a set of ordinary differential equations (ODEs) given
initial conditions is another useful example. The function
:obj:`odeint` is available in SciPy for integrating a first-order
vector differential equation:
.. math::
:nowrap:
\[ \frac{d\mathbf{y}}{dt}=\mathbf{f}\left(\mathbf{y},t\right),\]
given initial conditions :math:`\mathbf{y}\left(0\right)=y_{0}`, where
:math:`\mathbf{y}` is a length :math:`N` vector and :math:`\mathbf{f}`
is a mapping from :math:`\mathcal{R}^{N}` to :math:`\mathcal{R}^{N}.`
A higher-order ordinary differential equation can always be reduced to
a differential equation of this type by introducing intermediate
derivatives into the :math:`\mathbf{y}` vector.
For example suppose it is desired to find the solution to the
following second-order differential equation:
.. math::
:nowrap:
\[ \frac{d^{2}w}{dz^{2}}-zw(z)=0\]
with initial conditions :math:`w\left(0\right)=\frac{1}{\sqrt[3]{3^{2}}\Gamma\left(\frac{2}{3}\right)}` and :math:`\left.\frac{dw}{dz}\right|_{z=0}=-\frac{1}{\sqrt[3]{3}\Gamma\left(\frac{1}{3}\right)}.` It is known that the solution to this differential equation with these
boundary conditions is the Airy function
.. math::
:nowrap:
\[ w=\textrm{Ai}\left(z\right),\]
which gives a means to check the integrator using :func:`special.airy `.
First, convert this ODE into standard form by setting
:math:`\mathbf{y}=\left[\frac{dw}{dz},w\right]` and :math:`t=z`. Thus,
the differential equation becomes
.. math::
:nowrap:
\[ \frac{d\mathbf{y}}{dt}=\left[\begin{array}{c} ty_{1}\\ y_{0}\end{array}\right]=\left[\begin{array}{cc} 0 & t\\ 1 & 0\end{array}\right]\left[\begin{array}{c} y_{0}\\ y_{1}\end{array}\right]=\left[\begin{array}{cc} 0 & t\\ 1 & 0\end{array}\right]\mathbf{y}.\]
In other words,
.. math::
:nowrap:
\[ \mathbf{f}\left(\mathbf{y},t\right)=\mathbf{A}\left(t\right)\mathbf{y}.\]
As an interesting reminder, if :math:`\mathbf{A}\left(t\right)`
commutes with :math:`\int_{0}^{t}\mathbf{A}\left(\tau\right)\, d\tau`
under matrix multiplication, then this linear differential equation
has an exact solution using the matrix exponential:
.. math::
:nowrap:
\[ \mathbf{y}\left(t\right)=\exp\left(\int_{0}^{t}\mathbf{A}\left(\tau\right)d\tau\right)\mathbf{y}\left(0\right),\]
However, in this case, :math:`\mathbf{A}\left(t\right)` and its integral do not commute.
There are many optional inputs and outputs available when using odeint
which can help tune the solver. These additional inputs and outputs
are not needed much of the time, however, and the three required input
arguments and the output solution suffice. The required inputs are the
function defining the derivative, *fprime*, the initial conditions
vector, *y0*, and the time points to obtain a solution, *t*, (with
the initial value point as the first element of this sequence). The
output to :obj:`odeint` is a matrix where each row contains the
solution vector at each requested time point (thus, the initial
conditions are given in the first output row).
The following example illustrates the use of odeint including the
usage of the *Dfun* option which allows the user to specify a gradient
(with respect to :math:`\mathbf{y}` ) of the function,
:math:`\mathbf{f}\left(\mathbf{y},t\right)`.
>>> from scipy.integrate import odeint
>>> from scipy.special import gamma, airy
>>> y1_0 = 1.0/3**(2.0/3.0)/gamma(2.0/3.0)
>>> y0_0 = -1.0/3**(1.0/3.0)/gamma(1.0/3.0)
>>> y0 = [y0_0, y1_0]
>>> def func(y, t):
... return [t*y[1],y[0]]
>>> def gradient(y,t):
... return [[0,t],[1,0]]
>>> x = arange(0,4.0, 0.01)
>>> t = x
>>> ychk = airy(x)[0]
>>> y = odeint(func, y0, t)
>>> y2 = odeint(func, y0, t, Dfun=gradient)
>>> print ychk[:36:6]
[ 0.355028 0.339511 0.324068 0.308763 0.293658 0.278806]
>>> print y[:36:6,1]
[ 0.355028 0.339511 0.324067 0.308763 0.293658 0.278806]
>>> print y2[:36:6,1]
[ 0.355028 0.339511 0.324067 0.308763 0.293658 0.278806]