Hi this is the function:
G(a) = integral from 0 to infinity of [t^(a-1)*e^-t*dt]
What is G'(a)?
Thanks =D
Say a>0 then,
L{ t^(a-1) } = Gamma(a)/s^a
Note, your function is the Laplace transform evaluated at 1.
Thus,
G(a) = Gamma(a)/1^a = Gamma(a)
The derivative of the gamma function is.
The Psi(x)*Gamma(x)
Where "psi" is the Digamma function.