What Is the Integral of the nth Derivative from Zero to Infinity?

seanhbailey
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Homework Statement



In general, what is \int_{0}^{\infty} f^{(n)}(z) dn?

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The Attempt at a Solution


Is the answer as simple as taking the antiderivative of n?
 
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n is the nth derivative of f(z), and z is a constant. I want to integrate with respect to the variable n, the nth derivative. If I simply took the aniderivative of n, and if the values I wanted to evaluate the integral over were 1 and 0, I would obtain f^{(1/2)}(z) which does not make any sense. What am I doing wrong?
 
The whole thing seems screwy to me. Your limits of integration are 0 to infinity, but derivatives make sense only for integer values of n. E.g., the "one-halfth" derivative doesn't make any sense.
 
Does the integral even have an antiderivative, forgetting about the limits?
 
This is the problem I am encountering in the Euler- MacLaurin expansion in my proof. The proof is given in another one of my posts called Zeta Function Proof. If anyone would like to point me in the right direction, I would appreciate it very much.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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