08/22/2013, 10:46 PM
(08/22/2013, 05:54 PM)JmsNxn Wrote: \( \frac{d^s f}{dt^s}(-t) < e^{-t} \)
\( \phi(s) = \int_0^\infty t^{s-1} \frac{d^s f}{dt^s}(-t) dt \)
Integrate by parts, and for \( \Re(s) > 0 \) we get the spectacular identity that:
\( s \phi(s) = \phi(s+1) \)
I do not even need to use integrate by parts to see a problem.
Its funny you say \( \frac{d^s f}{dt^s}(-t) < e^{-t} \)
because its more like an equality when we differentiate a given amount of times with respect to t.
You see : s is considered a constant with respect to t since s is not a function OF t NOR f.
There is big difference between a function , an operator , a variable and a constant.
ALthough that may sound belittling or trivial , your example shows this is an important concept !!
If you consider \( \frac{d^s f}{dt^s}(-t) \) as a function F(s,f) then it is no surprise that taking the derivative with respect to f leaves s unchanged.
By the chain rule you then get the " wrong " / " correct "
\( \frac{d^s f}{dt^s}(-t) (-1)^{-M} \) if you take the derivative \( M \) times.
This is similar to \( D^m [ s e^{-t}] dt \).
Hence by the very definition of the gamma function you also get
\( s \phi(s) = \phi(s+1) \) here which you already showed yourself with the - overkill - method integrate by parts.
This might not answer all your questions yet but I assume it helps.
It not completely formal either sorry.
It might affect your other posts about integral representations for fractional calculus , tetration and continuum sum.
Im still optimistic though and hope I did not discourage you to much.
regards
tommy1729