自适应辛普森(复数)-1/3法则

  1/3法则-自适应辛普森数值积分 复数版，没有进行过多的限制和检查，该方法仅针对我自己遇到的一些问题进行简单的快速迭代，适用于简单函数的快速积分。

def adaptive_simpson(func, lower, upper, tolerance=1e-6,):
    if not callable(func):
        raise TypeError(f"Function type error: {func}")
        
    result = 0.0j
    length = upper - lower

    if abs(length) < tolerance:
        return result

    left = func(lower)
    right = func(upper)
    midpoint = (upper + lower) * 0.5
    middle = func(midpoint)

    stack = [(lower, upper, left, middle, right, (length * (left + 4.0 * middle + right) * 0.16666666666666666), tolerance)]
    
    while stack:
        (stack_lower, stack_upper, stack_left, stack_middle, stack_right, stack_integral, stack_tolerance) = stack.pop()
        
        stack_midpoint = (stack_upper + stack_lower) * 0.5
        stack_left_middle = func((stack_lower + stack_midpoint) * 0.5)
        stack_right_middle = func((stack_midpoint + stack_upper) * 0.5)
        stack_left_integral = ((stack_midpoint - stack_lower) * (stack_left + 4.0 * stack_left_middle + stack_middle) * 0.16666666666666666 )
        stack_right_integral = ((stack_upper - stack_midpoint) * (stack_middle + 4.0 * stack_right_middle + stack_right) * 0.16666666666666666)
        stack_sum = stack_left_integral + stack_right_integral

        error = (stack_sum - stack_integral) * 0.06666666666666666

        if abs(error) < stack_tolerance:
            result += stack_sum + error
        else:
            epsilon = stack_tolerance * 0.5
            stack.append((stack_midpoint, stack_upper, stack_middle, stack_right_middle, stack_right, stack_right_integral, epsilon))
            stack.append((stack_lower, stack_midpoint, stack_left, stack_left_middle, stack_middle, stack_left_integral, epsilon))

    return result