Show that $\Gamma^{(n)}(z) = \int_0^\infty t^{z-1}(\log(t))^ne^{-t}dt$

Question

So, I need to show that the n-th derivative of Gamma function is equal to:

\begin{equation} \Gamma^{(n)}(z) = \int_0^\infty t^{z-1}(\log(t))^ne^{-t}dt \end{equation}

I already know that:

\begin{equation} \Gamma(z) = \lim_{n \to \infty}\int_0^n t^{z-1}(1-\frac{t}{n})^n dt = \lim_{n \to \infty} \gamma_n(z) \end{equation}

I also see that: \begin{equation} \gamma_n'(z_0) = \lim_{z \to z_0} \int_0^n (1-\frac{t}{n})^n \cdot \frac{t^{z-1} - t^{z_0-1}}{z - z_0} dt \end{equation}

Now, if I could show that \begin{equation} \frac{t^{z_n-1} - t^{z_0-1}}{z_n - z_0} \to t^{z_0 - 1}\log(t)\textrm{ uniformly, as } n \to \infty \; (z_n \to z_0) \; \; \; \; \; \;(1) \end{equation} then I would be able to pull the limit to the inside of the integral and get:

\begin{equation} \gamma_n'(z_0) = \int_0^n t^{z-1}\log(t)(1-\frac{t}{n})^n dt \end{equation}

From which I see how I can get $\Gamma'(z_0) = \int_0^\infty t^{z-1}\log(t)e^{-t}dt$. However, I don't know how to prove $(1)$. Could someone please help me?

Answer 1

Since$$\Gamma(z)=\int_0^\infty t^{z-1}e^{-t}\,\mathrm dz=\int_0^\infty e^{(z-1)\log(t)}e^{-t}\,\mathrm dt,$$it can be deduced from the Leibniz integral rule that$$\Gamma'(z)=\int_0^\infty\log(t)e^{(z-1)\log(t)}e^{-t}\,\mathrm dt=\int_0^\infty\log(t)t^{z-1}e^{-t}\,\mathrm dt.$$Now, by the same argument,$$\Gamma''(z)=\int_0^\infty\log^2(t)t^{z-1}e^{-t}\,\mathrm dt$$and so on.