I came across a problem where I need to show that $$ \Gamma'(1)-\frac{\Gamma'(y)}{\Gamma(y)} = \int_0^1\frac{(1-t)^{y-1}-1}{t}dt $$ for all $y>0$.
I proved the case for $y>1$, where the integrated function is bounded, by finding a power series of the Beta function. Now I'm considering to show that both sides are analytic on $(0, \infty)$ so that the proof can be completed.
My thoughts are that,
by replacing $(1-t)^{y-1}$ with its binomial series I can show the right hand side is analytic on $(0,\infty)$
for the left hand side, I proved a result in previous problems that $\log \Gamma(y)$ is a smooth function and $$ \frac{d^k}{dy^k}\log\Gamma(y) = (-1)^k (k-1)!\sum_{n=0}^\infty(y+n)^{-k}, k>1 $$ Note that $\frac{\Gamma'(y)}{\Gamma(y)} = \frac{d}{dy}\log\Gamma(y)$.
Did I miss some simple ideas to show that $\log\Gamma$ is indeed real analytic? I haven't studied complex analysis yet but any method is welcome.