(ctx, f, x, n=1, x0=0)
return ctx.diff(g, x, m) / ctx.gamma(m-n)
r"""
    Calculates the Riemann-Liouville differintegral, or fractional
    derivative, defined by

    .. math ::

        \,_{x_0}{\mathbb{D}}^n_xf(x) \frac{1}{\Gamma(m-n)} \frac{d^m}{dx^m}
        \int_{x_0}^{x}(x-t)^{m-n-1}f(t)dt

    where `f` is a given (presumably well-behaved) function,
    `x` is the evaluation point, `n` is the order, and `x_0` is
    the reference point of integration (`m` is an arbitrary
    parameter selected automatically).

    With `n = 1`, this is just the standard derivative `f'(x)`; with `n = 2`,
    the second derivative `f''(x)`, etc. With `n = -1`, it gives
    `\int_{x_0}^x f(t) dt`, with `n = -2`
    it gives `\int_{x_0}^x \left( \int_{x_0}^t f(u) du \right) dt`, etc.

    As `n` is permitted to be any number, this operator generalizes
    iterated differentiation and iterated integration to a single
    operator with a continuous order parameter.

    **Examples**

    There is an exact formula for the fractional derivative of a
    monomial `x^p`, which may be used as a reference. For example,
    the following gives a half-derivative (order 0.5)::

        >>> from sympy.mpmath import *
        >>> mp.dps = 15; mp.pretty = True
        >>> x = mpf(3); p = 2; n = 0.5
        >>> differint(lambda t: t**p, x, n)
        7.81764019044672
        >>> gamma(p+1)/gamma(p-n+1) * x**(p-n)
        7.81764019044672

    Another useful test function is the exponential function, whose
    integration / differentiation formula easy generalizes
    to arbitrary order. Here we first compute a third derivative,
    and then a triply nested integral. (The reference point `x_0`
    is set to `-\infty` to avoid nonzero endpoint terms.)::

        >>> differint(lambda x: exp(pi*x), -1.5, 3)
        0.278538406900792
        >>> exp(pi*-1.5) * pi**3
        0.278538406900792
        >>> differint(lambda x: exp(pi*x), 3.5, -3, -inf)
        1922.50563031149
        >>> exp(pi*3.5) / pi**3
        1922.50563031149

    However, for noninteger `n`, the differentiation formula for the
    exponential function must be modified to give the same result as the
    Riemann-Liouville differintegral::

        >>> x = mpf(3.5)
        >>> c = pi
        >>> n = 1+2*j
        >>> differint(lambda x: exp(c*x), x, n)
        (-123295.005390743 + 140955.117867654j)
        >>> x**(-n) * exp(c)**x * (x*c)**n * gammainc(-n, 0, x*c) / gamma(-n)
        (-123295.005390743 + 140955.117867654j)
r"""
    Calculates the Riemann-Liouville differintegral, or fractional
    derivative, defined by
r Calculates the Riemann - Liouville differintegral or fractional derivative defined by
def differint(ctx, f, x, n=1, x0=0):
    r"""
    Calculates the Riemann-Liouville differintegral, or fractional
    derivative, defined by

    .. math ::

        \,_{x_0}{\mathbb{D}}^n_xf(x) \frac{1}{\Gamma(m-n)} \frac{d^m}{dx^m}
        \int_{x_0}^{x}(x-t)^{m-n-1}f(t)dt

    where `f` is a given (presumably well-behaved) function,
    `x` is the evaluation point, `n` is the order, and `x_0` is
    the reference point of integration (`m` is an arbitrary
    parameter selected automatically).

    With `n = 1`, this is just the standard derivative `f'(x)`; with `n = 2`,
    the second derivative `f''(x)`, etc. With `n = -1`, it gives
    `\int_{x_0}^x f(t) dt`, with `n = -2`
    it gives `\int_{x_0}^x \left( \int_{x_0}^t f(u) du \right) dt`, etc.

    As `n` is permitted to be any number, this operator generalizes
    iterated differentiation and iterated integration to a single
    operator with a continuous order parameter.

    **Examples**

    There is an exact formula for the fractional derivative of a
    monomial `x^p`, which may be used as a reference. For example,
    the following gives a half-derivative (order 0.5)::

        >>> from sympy.mpmath import *
        >>> mp.dps = 15; mp.pretty = True
        >>> x = mpf(3); p = 2; n = 0.5
        >>> differint(lambda t: t**p, x, n)
        7.81764019044672
        >>> gamma(p+1)/gamma(p-n+1) * x**(p-n)
        7.81764019044672

    Another useful test function is the exponential function, whose
    integration / differentiation formula easy generalizes
    to arbitrary order. Here we first compute a third derivative,
    and then a triply nested integral. (The reference point `x_0`
    is set to `-\infty` to avoid nonzero endpoint terms.)::

        >>> differint(lambda x: exp(pi*x), -1.5, 3)
        0.278538406900792
        >>> exp(pi*-1.5) * pi**3
        0.278538406900792
        >>> differint(lambda x: exp(pi*x), 3.5, -3, -inf)
        1922.50563031149
        >>> exp(pi*3.5) / pi**3
        1922.50563031149

    However, for noninteger `n`, the differentiation formula for the
    exponential function must be modified to give the same result as the
    Riemann-Liouville differintegral::

        >>> x = mpf(3.5)
        >>> c = pi
        >>> n = 1+2*j
        >>> differint(lambda x: exp(c*x), x, n)
        (-123295.005390743 + 140955.117867654j)
        >>> x**(-n) * exp(c)**x * (x*c)**n * gammainc(-n, 0, x*c) / gamma(-n)
        (-123295.005390743 + 140955.117867654j)


    """
    m = max(int(ctx.ceil(ctx.re(n)))+1, 1)
    r = m-n-1
    g = lambda x: ctx.quad(lambda t: (x-t)**r * f(t), [x0, x])
    return ctx.diff(g, x, m) / ctx.gamma(m-n)
def differint ( ctx , f , x , n = 1 , x0 = 0 ) : m = max ( int ( ctx . ceil ( ctx . re ( n ) ) ) + 1 , 1 ) r = m - n - 1 g = lambda x : ctx . quad ( lambda t : ( x - t ) ** r * f ( t ) , [ x0 , x ] ) return ctx . diff ( g , x , m ) / ctx . gamma ( m - n )