$$\int_0^\infty x^{\nu-1}\ e^{-\mu x}\ \ln(x)\ dx = \frac1{\mu^\nu}\Gamma(\nu)\left[\psi(\nu)-\ln(\mu)\right] \quad\qquad [\Re\,\mu \gt 0, \quad \Re\,\nu\gt 0]$$
Hello, I found above equation on Table of Integrals, Series, and Products by I.S. GradshteynI.M. Ryzhik. I want extend the problem from above integral, by changing $\ln(x)$ to $(\ln(x+a))$, which $a$ is a variable. Anybody can help me to solve it ?
$$\int_0^{\infty } x^{v-1} \exp (-\mu x) \log (x+a) \, dx=\\\mathcal{L}_x\left[x^{v-1} \log (x+a)\right](\mu )=\\\mathcal{M}_x[\exp (-\mu x) \log (x+a)](v)$$
Integral is equivalent to Laplace Transform or Mellin Transform.
With Maple help:
$\int_0^{\infty } x^{v-1} \exp (-\mu x) \log (x+a) \, dx={\frac {\mu\,{\mbox{$_2$F$_2$}(1,1;\,2,2-v;\,\mu\,a)}\Gamma \left( v \right) a}{{\mu}^{v} \left( v-1 \right) }}-{\frac {{a}^{v}\pi\,\Gamma \left( v,-\mu\,a \right) }{\sin \left( \pi\,v \right) \left( -\mu\,a \right) ^{v}}}+{\frac {{a}^{v}\pi\,\Gamma \left( v \right) v}{\sin \left( \pi\,v \right) \left( -\mu\,a \right) ^{v} \left( v-1 \right) }}-{\frac {{a}^{v}\pi\,\Gamma \left( v \right) }{\sin \left( \pi\,v \right) \left( -\mu\,a \right) ^{v} \left( v-1 \right) }}+{ \frac {\Gamma \left( v \right) \Psi \left( v \right) v}{{\mu}^{v} \left( v-1 \right) }}-{\frac {\Gamma \left( v \right) \Psi \left( v \right) }{{\mu}^{v} \left( v-1 \right) }}-{\frac {\Gamma \left( v \right) v\ln \left( \mu \right) }{{\mu}^{v} \left( v-1 \right) }}+{ \frac {\Gamma \left( v \right) \ln \left( \mu \right) }{{\mu}^{v} \left( v-1 \right) }} $
where:$\, _2F_2(1,1;2,2-v;a \mu )$ is the generalized hypergeometric function
where: $\Psi \left( v \right)$ is PolyGamma function
Edited 11.04.2018:
On Maple 2018: