Thread: Integration with Respect to a Derivative

1. Integration with Respect to a Derivative

In general, what is the integral $\int_{0}^{\infty} f^n(z) dn$ where n is the nth derivative of f(z)?

2. your answer is probably 0 because n is a constant. integration with respect to a constant(ie n) is 0,isn't it???

3. Originally Posted by PaulDirac2
In general, what is the integral $\int_{0}^{\infty} f^n(z) dn$ where n is the nth derivative of f(z)?
I mean? What do you expect? If this function is integrable, which it has to be, I would say that it equals $\lim_{z\to\infty}\left[f^{n-1}(z)-f^{n-1}(0)\right]$

Originally Posted by Pulock2009
your answer is probably 0 because n is a constant. integration with respect to a constant(ie n) is 0,isn't it???
No. Read more carefully, it is not a constant function.

4. Interation with Respect to a Derivative

In the integral I posted z is not the variable that I wanted to integrate with respect to.

5. Originally Posted by PaulDirac2
In general, what is the integral $\int_{0}^{\infty} f^n(z) dn$ where n is the nth derivative of f(z)?
What, exactly, does $f^n(z)$ mean? Is it the nth power of f or the nth derivative (more commonly written $f^{(n)}(z)$)? If it means the nth derivative, and you are integrating with respect to its nth derivative that is exactly the same as $\int x dx= (1/2)x^2$ evaluated at $f^{(n)}(z)$, $\frac{1}{2}\left[\lim_{x\to\infty} (f^{(n)}(x))^2- f^{(n)}(0)\right]$.

If you mean f to the power n, then it will depend more strongly on exactly what f is.

6. Z is a Constant

In the integral, the only variable is n, the nth derivative. z is a constant.

7. Integration HELP

I am really having trouble with the concept of this problem. If you integrated with respect to the variable n by taking the antiderivative of n, and the limits of the integral are set to 1 and 0, you come up with a 1/2 derivative which makes no sense at all. I cannot take the antiderivative of the function as a whole because I am not integrating with respect to z, since z is a constant. If anyone could point me in the right direction I would appreciate it very much.