Python meijerint.meijerint_definite函数代码示例

            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

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.........

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

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