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General formula for the nth integral

  1. Apr 17, 2014 #1
    For my own use and consumption, I created a generalization of the nth integral of a function f and I'm posting it here for you look: $$\int f(x) dx = f^{(-1)}(x) + C_1$$ $$\iint f(x) dxdx = f^{(-2)}(x) + xC_1 + C_2$$ $$\iiint f(x) dxdxdx = f^{(-3)}(x) + \frac{1}{2}x^2C_1 + xC_2 + C_3$$
    $$\int^n f(x) dx^n = f^{(-n)}(x) + \sigma_n(x)$$
    being ##\sigma_n(x)##:

    $$\sigma_0(x) = 0$$
    $$\sigma_1(x) = \frac{1}{0!}x^0C_1$$
    $$\sigma_2(x) = \frac{1}{1!}x^1C_1 + \frac{1}{0!}x^0C_2$$
    $$\sigma_3(x) = \frac{1}{2!}x^2C_1 + \frac{1}{1!}x^1C_2 + \frac{1}{0!}x^0C_3$$
    $$\sigma_4(x) = \frac{1}{3!}x^3C_1 + \frac{1}{2!}x^2C_2 + \frac{1}{1!}x^1C_3 + \frac{1}{0!}x^0C_4$$
    Properties
    $$\int^n \sigma_k(x) dx^n = \sigma_{k+n}(x)$$ $$\frac{d^n}{dx^n} \sigma_k(x) = \sigma_{k-n}(x)$$
    Identities
    $$f^{(-1)}(x) = \int_{x_0}^{x} f(x) dx + \frac{1}{0!}f^{(-1)}(x_0)(x-x_0)^0$$
    $$f^{(-2)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx + \frac{1}{1!}f^{(-1)}(x_0)(x - x_0)^1 + \frac{1}{0!}f^{(-2)}(x_0)(x-x_0)^0$$
    $$f^{(-3)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx dx + \frac{1}{2!}f^{(-1)}(x_0)(x - x_0)^2 + \frac{1}{1!}f^{(-2)}(x_0)(x - x_0)^1$$
    $$ + \frac{1}{0!}f^{(-3)}(x_0)(x-x_0)^0$$
    $$\int\limits_{x_0}^{x} \! {\;}^{n} f(x) dx^n = \frac{1}{(n-1)!} \int_{x_0}^{x} (x-u)^{(n-1)} f(u)du$$

    What you think about, rox?
     
    Last edited by a moderator: Apr 17, 2014
  2. jcsd
  3. Apr 17, 2014 #2

    pwsnafu

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  4. Apr 17, 2014 #3
    Yeah! I appreciated your comment. But note that Cauchy formula for repeated integration is a formula for definite integrals, I'm proposing a general formula for indefinite integrals that involve the cauchy's formula, the series taylor and plus a function that I call of sigma.
     
  5. Apr 18, 2014 #4

    pwsnafu

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    Consider ##f(x) = \frac{-1}{x^2}##. Then we define ##F(x)## as ##F(x) = \frac{1}{x}## for ##x<0## and ##F(x) = \frac{1}{x} + 1## for ##x > 0##. Clearly ##F'(x) = f(x)## but ##F## is not of the form ##\frac{1}{x} + C_1## for some real number. Instead ##F(x) = \frac{1}{x} + H(x)## where H is the Heaviside step function.
     
  6. Apr 18, 2014 #5
    You are using different functions with the same name (f, or F). If F(x) = 1/x + H(x), so F'(x) = -1/x² + δ(x).
     
  7. Apr 18, 2014 #6

    pwsnafu

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    One is a capital F, the other is lower case f. They are different.

    No, F is not defined at x=0. Under the standard derivative ##H'(x) = 0## for ##x \in (-\infty,0) \cup (0, \infty)##. With the generalized derivative we have ##H' = \delta##, but you never said you were using the generalized derivative.
     
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