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 Another question! tommy1729 Ultimate Fellow Posts: 1,380 Threads: 338 Joined: Feb 2009 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 « Next Oldest | Next Newest »

 Messages In This Thread Another question! - by JmsNxn - 08/22/2013, 05:54 PM RE: Another question! - by tommy1729 - 08/22/2013, 10:46 PM RE: Another question! - by JmsNxn - 08/25/2013, 03:00 PM RE: Another question! - by mike3 - 08/25/2013, 10:21 PM RE: Another question! - by JmsNxn - 08/27/2013, 06:57 PM

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