Prove that

$\begin{array}{c}\frac{\mathbf{d}}{\mathbf{d}\mathbf{p}}\mathbf{\Gamma }\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{l}\mathbf{n}\mathbf{x}\mathbf{d}\mathbf{x}\\ \frac{{\mathbf{d}}^{\mathbf{n}}}{\mathbf{d}{\mathbf{p}}^{\mathbf{n}}}\mathbf{\Gamma }\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathbf{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}{\mathbf{\left(}\mathbf{l}\mathbf{n}\mathbf{x}\mathbf{\right)}}^{\mathbf{n}}\mathbf{d}\mathbf{x}\end{array}$

Here, we need to prove that,

$\begin{array}{l}\frac{d}{dp}\Gamma \left(p\right)=\underset{0}{\overset{\infty }{\int }}{x}^{p-1}{e}^{-x}\ln xdx\\ \frac{d^n}{d{p}^n}\Gamma \left(p\right)={\int }_0^{\infty }{x}^{p-1}{e}^{-x}{\left(\ln x\right)}^{n}dx\end{array}$

The Gamma Function is defined as $\mathbf{\Gamma }\mathbf{\left(}\mathit{p}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{p}\mathbf{>}\mathbf{0}$.

Start with the definition of the Gamma function.

$\Gamma \left(p\right)=\underset{0}{\overset{\infty }{\int }}{x}^{p-1}{e}^{-x}dx$

Differentiate this function with respect to.

$\frac{d}{dp}\Gamma \left(p\right)=\underset{0}{\overset{\infty }{\int }}\frac{d}{dp}\left(x^{p-1}{e}^{-x}\right)dx$

Take the equation.

$\frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)$

Define a new function $u=a^x$and simplify further.

$\begin{array}{c}\frac{d}{dx}x=1\\ \frac{du}{dx}=u\ln a\end{array}$

Use the base change rule of logarithms to simplify further.

$\begin{array}{c}{\log }_{f}g=\frac{{\log }_{h}g}{{\log }_{h}f}\\ {\log }_{a}u=\frac{\ln u}{\ln a}\end{array}$

Using above, simplify the equations.

$\begin{array}{c}\frac{d}{dx}\frac{\ln u}{\ln a}=\frac{1}{\ln a}\frac{d\ln u}{du}\frac{du}{dx}\\ =\frac{1}{\ln a}\frac{1}{u}\frac{du}{dx}\\ =1\end{array}$

Consider $\frac{du}{dx}=u\ln a$.

Replace $u=a^x$.

$\frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)$

That proves the result.

Differentiate the Gamma function with respect to p using results proved earlier.

$\begin{array}{c}\frac{d}{dp}\Gamma \left(p\right)=\underset{0}{\overset{\infty }{\int }}\frac{d}{dp}\left(x^{p-1}{e}^{-x}\right)dx\\ =\underset{0}{\overset{\infty }{\int }}{x}^{p-1}\ln x{e}^{-x}dx\end{array}$

$\frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)$Since differentiations are with respect to in the equation.

ln(a) is constant.

This implies that multiple differentiations result in $(\ln a{)}^n$.

Thus, the result is obtained.

$\frac{d^n}{d{p}^n}\Gamma \left(p\right)={\int }_0^{\infty }{x}^{p-1}{e}^{-x}{\left(\ln x\right)}^{n}dx$