Can scipy.integrate.fixed_quad compute integral with functional boundaries?

Question

I would like to numerically integrate function over a triangle similarly as

import scipy.integrate as integrate
inside = lambda x: integrate.quad(lambda x,y: 1, 0, x, args=(x))[0]
outside = integrate.quad(inside, 0, 1)[0]
print(outside)
0.5

but using scipy.integrate.fixed_quad function (which has integration order n as parameter). However, when I write

inside = lambda x: integrate.fixed_quad(lambda x,y: 1, 0, x, args=(x), n=5)
print(inside(5))

Traceback (most recent call last): File "", line 1, in File "", line 1, in File "/Users/username/anaconda/lib/python3.5/site-packages/scipy/integrate/ quadrature.py", line 82, in fixed_quad return (b-a)/2.0 * np.sum(w*func(y, *args), axis=0), None

TypeError: () argument after * must be an iterable, not int

I don't know what I'm doing wrong as I'm following the documentation on scipy.integrate.fixed_quad.

Cleb Cleb · Accepted Answer · 2017-02-16T10:02:56

The problem is your definition of args, args=(x). It should be passed as a tuple, so you need to add an additional comma to make it a tuple:

inside = lambda x: integrate.fixed_quad(lambda x,y: 1, 0, x, args=(x,), n=5)

Then

inside(5)

yields

(5.0, None)

The line

integrate.quad(lambda x,y: 1, 0, x, args=(x))[0]

works as in quad it is checked whether args is a tuple; if not it is converted (directly taken from the source code):

if not isinstance(args, tuple):
        args = (args,)

In fixed_quad that is not the case and that's why you received the error in one but not both cases.

Can scipy.integrate.fixed_quad compute integral with functional boundaries?

1 Answers