# General formula for the nth integral

1. Apr 17, 2014

### Jhenrique

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
$$f^{(-1)}(x) = \int_{x_0}^{x} f(x) dx + \frac{1}{0!}f^{(-1)}(x_0)(x-x_0)^0$$
$$f^{(-2)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx + \frac{1}{1!}f^{(-1)}(x_0)(x - x_0)^1 + \frac{1}{0!}f^{(-2)}(x_0)(x-x_0)^0$$
$$f^{(-3)}(x) = \int_{x_0}^{x} \int_{x_0}^{x} \int_{x_0}^{x} f(x) dx dx dx + \frac{1}{2!}f^{(-1)}(x_0)(x - x_0)^2 + \frac{1}{1!}f^{(-2)}(x_0)(x - x_0)^1$$
$$+ \frac{1}{0!}f^{(-3)}(x_0)(x-x_0)^0$$
$$\int\limits_{x_0}^{x} \! {\;}^{n} f(x) dx^n = \frac{1}{(n-1)!} \int_{x_0}^{x} (x-u)^{(n-1)} f(u)du$$

Last edited by a moderator: Apr 17, 2014
2. Apr 17, 2014

### pwsnafu

3. Apr 17, 2014

### Jhenrique

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.

4. Apr 18, 2014

### pwsnafu

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.

5. Apr 18, 2014

### Jhenrique

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).

6. Apr 18, 2014

### pwsnafu

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.