(08/22/2013, 05:54 PM)JmsNxn Wrote:  < e^{-t})
 = \int_0^\infty t^{s-1} \frac{d^s f}{dt^s}(-t) dt)
Integrate by parts, and for
we get the spectacular identity that:
 = \phi(s+1))
I do not even need to use integrate by parts to see a problem.
Its funny you say
 < 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
)
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 "
 (-1)^{-M})
if you take the derivative

times.
This is similar to

.
Hence by the very definition of the gamma function you also get
 = \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