本文整理汇总了Python中sympy.integrals.meijerint.meijerint_definite函数的典型用法代码示例。如果您正苦于以下问题：Python meijerint_definite函数的具体用法？Python meijerint_definite怎么用？Python meijerint_definite使用的例子？那么恭喜您, 这里精选的函数代码示例或许可以为您提供帮助。
在下文中一共展示了meijerint_definite函数的4个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于我们的系统推荐出更棒的Python代码示例。
示例1: try_meijerg
def try_meijerg(function, xab):
ret = None
if len(xab) == 3 and meijerg is not False:
x, a, b = xab
try:
res = meijerint_definite(function, x, a, b)
except NotImplementedError:
from sympy.integrals.meijerint import _debug
_debug("NotImplementedError from meijerint_definite")
res = None
if res is not None:
f, cond = res
if conds == "piecewise":
ret = Piecewise((f, cond), (self.func(function, (x, a, b)), True))
elif conds == "separate":
if len(self.limits) != 1:
raise ValueError("conds=separate not supported in " "multiple integrals")
ret = f, cond
else:
ret = f
return ret
开发者ID:brajeshvit，项目名称:virtual，代码行数:22，代码来源:integrals.py
示例2: test_meijerint
def test_meijerint():
from sympy import symbols, expand, arg
s, t, mu = symbols("s t mu", real=True)
assert integrate(
meijerg([], [], [0], [], s * t) * meijerg([], [], [mu / 2], [-mu / 2], t ** 2 / 4), (t, 0, oo)
).is_Piecewise
s = symbols("s", positive=True)
assert integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo)) == gamma(s + 1)
assert integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo), meijerg=True) == gamma(s + 1)
assert isinstance(integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo), meijerg=False), Integral)
assert meijerint_indefinite(exp(x), x) == exp(x)
# TODO what simplifications should be done automatically?
# This tests "extra case" for antecedents_1.
a, b = symbols("a b", positive=True)
assert simplify(meijerint_definite(x ** a, x, 0, b)[0]) == b ** (a + 1) / (a + 1)
# This tests various conditions and expansions:
meijerint_definite((x + 1) ** 3 * exp(-x), x, 0, oo) == (16, True)
# Again, how about simplifications?
sigma, mu = symbols("sigma mu", positive=True)
i, c = meijerint_definite(exp(-((x - mu) / (2 * sigma)) ** 2), x, 0, oo)
assert simplify(i) == sqrt(pi) * sigma * (2 - erfc(mu / (2 * sigma)))
assert c == True
i, _ = meijerint_definite(exp(-mu * x) * exp(sigma * x), x, 0, oo)
# TODO it would be nice to test the condition
assert simplify(i) == 1 / (mu - sigma)
# Test substitutions to change limits
assert meijerint_definite(exp(x), x, -oo, 2) == (exp(2), True)
# Note: causes a NaN in _check_antecedents
assert expand(meijerint_definite(exp(x), x, 0, I)[0]) == exp(I) - 1
assert expand(meijerint_definite(exp(-x), x, 0, x)[0]) == 1 - exp(-exp(I * arg(x)) * abs(x))
# Test -oo to oo
assert meijerint_definite(exp(-x ** 2), x, -oo, oo) == (sqrt(pi), True)
assert meijerint_definite(exp(-abs(x)), x, -oo, oo) == (2, True)
assert meijerint_definite(exp(-(2 * x - 3) ** 2), x, -oo, oo) == (sqrt(pi) / 2, True)
assert meijerint_definite(exp(-abs(2 * x - 3)), x, -oo, oo) == (1, True)
assert meijerint_definite(exp(-((x - mu) / sigma) ** 2 / 2) / sqrt(2 * pi * sigma ** 2), x, -oo, oo) == (1, True)
assert meijerint_definite(sinc(x) ** 2, x, -oo, oo) == (pi, True)
# Test one of the extra conditions for 2 g-functinos
assert meijerint_definite(exp(-x) * sin(x), x, 0, oo) == (S(1) / 2, True)
# Test a bug
def res(n):
return (1 / (1 + x ** 2)).diff(x, n).subs(x, 1) * (-1) ** n
for n in range(6):
assert integrate(exp(-x) * sin(x) * x ** n, (x, 0, oo), meijerg=True) == res(n)
# This used to test trigexpand... now it is done by linear substitution
assert simplify(integrate(exp(-x) * sin(x + a), (x, 0, oo), meijerg=True)) == sqrt(2) * sin(a + pi / 4) / 2
# Test the condition 14 from prudnikov.
# (This is besselj*besselj in disguise, to stop the product from being
# recognised in the tables.)
a, b, s = symbols("a b s")
from sympy import And, re
assert meijerint_definite(
meijerg([], [], [a / 2], [-a / 2], x / 4) * meijerg([], [], [b / 2], [-b / 2], x / 4) * x ** (s - 1), x, 0, oo
) == (
4
* 2 ** (2 * s - 2)
* gamma(-2 * s + 1)
* gamma(a / 2 + b / 2 + s)
/ (gamma(-a / 2 + b / 2 - s + 1) * gamma(a / 2 - b / 2 - s + 1) * gamma(a / 2 + b / 2 - s + 1)),
And(0 < -2 * re(4 * s) + 8, 0 < re(a / 2 + b / 2 + s), re(2 * s) < 1),
)
# test a bug
assert integrate(sin(x ** a) * sin(x ** b), (x, 0, oo), meijerg=True) == Integral(
sin(x ** a) * sin(x ** b), (x, 0, oo)
)
# test better hyperexpand
assert (
integrate(exp(-x ** 2) * log(x), (x, 0, oo), meijerg=True) == (sqrt(pi) * polygamma(0, S(1) / 2) / 4).expand()
)
# Test hyperexpand bug.
from sympy import lowergamma
n = symbols("n", integer=True)
assert simplify(integrate(exp(-x) * x ** n, x, meijerg=True)) == lowergamma(n + 1, x)
# Test a bug with argument 1/x
alpha = symbols("alpha", positive=True)
assert meijerint_definite((2 - x) ** alpha * sin(alpha / x), x, 0, 2) == (
sqrt(pi)
* alpha
* gamma(alpha + 1)
* meijerg(((), (alpha / 2 + S(1) / 2, alpha / 2 + 1)), ((0, 0, S(1) / 2), (-S(1) / 2,)), alpha ** S(2) / 16)
/ 4,
#.........这里部分代码省略.........
开发者ID:Carreau，项目名称:sympy，代码行数:101，代码来源:test_meijerint.py
示例3: test_meijerint_definite
def test_meijerint_definite():
v, b = meijerint_definite(x, x, 0, 0)
assert v.is_zero and b is True
v, b = meijerint_definite(x, x, oo, oo)
assert v.is_zero and b is True
开发者ID:Carreau，项目名称:sympy，代码行数:5，代码来源:test_meijerint.py
示例4: test_meijerint
def test_meijerint():
from sympy import symbols, expand, arg
s, t, mu = symbols("s t mu", real=True)
assert integrate(
meijerg([], [], [0], [], s * t) * meijerg([], [], [mu / 2], [-mu / 2], t ** 2 / 4), (t, 0, oo)
).is_Piecewise
s = symbols("s", positive=True)
assert integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo)) == gamma(s + 1)
assert integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo), meijerg=True) == gamma(s + 1)
assert isinstance(integrate(x ** s * meijerg([[], []], [[0], []], x), (x, 0, oo), meijerg=False), Integral)
assert meijerint_indefinite(exp(x), x) == exp(x)
# TODO what simplifications should be done automatically?
# This tests "extra case" for antecedents_1.
a, b = symbols("a b", positive=True)
assert simplify(meijerint_definite(x ** a, x, 0, b)[0]) == b ** (a + 1) / (a + 1)
# This tests various conditions and expansions:
meijerint_definite((x + 1) ** 3 * exp(-x), x, 0, oo) == (16, True)
# Again, how about simplifications?
sigma, mu = symbols("sigma mu", positive=True)
i, c = meijerint_definite(exp(-((x - mu) / (2 * sigma)) ** 2), x, 0, oo)
assert simplify(i) == sqrt(pi) * sigma * (erf(mu / (2 * sigma)) + 1)
assert c is True
i, _ = meijerint_definite(exp(-mu * x) * exp(sigma * x), x, 0, oo)
# TODO it would be nice to test the condition
assert simplify(i) == 1 / (mu - sigma)
# Test substitutions to change limits
assert meijerint_definite(exp(x), x, -oo, 2) == (exp(2), True)
assert expand(meijerint_definite(exp(x), x, 0, I)[0]) == exp(I) - 1
assert expand(meijerint_definite(exp(-x), x, 0, x)[0]) == 1 - exp(-exp(I * arg(x)) * abs(x))
# Test -oo to oo
assert meijerint_definite(exp(-x ** 2), x, -oo, oo) == (sqrt(pi), True)
assert meijerint_definite(exp(-abs(x)), x, -oo, oo) == (2, True)
assert meijerint_definite(exp(-(2 * x - 3) ** 2), x, -oo, oo) == (sqrt(pi) / 2, True)
assert meijerint_definite(exp(-abs(2 * x - 3)), x, -oo, oo) == (1, True)
assert meijerint_definite(exp(-((x - mu) / sigma) ** 2 / 2) / sqrt(2 * pi * sigma ** 2), x, -oo, oo) == (1, True)
# Test one of the extra conditions for 2 g-functinos
assert meijerint_definite(exp(-x) * sin(x), x, 0, oo) == (S(1) / 2, True)
# Test a bug
def res(n):
return (1 / (1 + x ** 2)).diff(x, n).subs(x, 1) * (-1) ** n
for n in range(6):
assert integrate(exp(-x) * sin(x) * x ** n, (x, 0, oo), meijerg=True) == res(n)
# Test trigexpand:
assert integrate(exp(-x) * sin(x + a), (x, 0, oo), meijerg=True) == sin(a) / 2 + cos(a) / 2
开发者ID:kendhia，项目名称:sympy，代码行数:56，代码来源:test_meijerint.py
注：本文中的sympy.integrals.meijerint.meijerint_definite函数示例由纯净天空整理自Github/MSDocs等源码及文档管理平台，相关代码片段筛选自各路编程大神贡献的开源项目，源码版权归原作者所有，传播和使用请参考对应项目的License；未经允许，请勿转载。 |
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