python：积分未知的上限内含未知的fsolve

当然。

假设您要查找x，使得t *（1-x * t）的从t = 0到t = x的t上的积分为0。您可以通过定义两个函数来做到这一点。integrand(t, x)将评估t *（1-x * t），并将使用进行func(x)积分，将其作为积分的上限，并作为被积数的额外参数。这是一个演示：integrandquadx

import numpy as np
from scipy.integrate import quad
from scipy.optimize import fsolve


def integrand(t, x):
    return t*(1 - x*t)


def func(x):
    y, err = quad(integrand, 0, x, args=(x,))
    return y


# Use 1.0 as the initial guess.  Note that a bad initial guess
# might generate a warning and return the degenerate solution at x=0.
sol = fsolve(func, 1.0)

print "Solution:      ", sol[0]

# The exact solution that we want is sqrt(3/2)
print "Exact solution:", np.sqrt(1.5)

输出：

Solution:       1.22474487139
Exact solution: 1.22474487139