example of differentiation under integral sign

Differentiation with respect to a parameter under the integral sign may sometimes yield useful formulae.  One example is given here.

We know that the equation

$${\int }_0^1{x}^m𝑑x=\frac{1}{m+1}$$

is valid for all  $m>-1$.  If one differentiates with respect to $m$ under the integral sign (http://planetmath.org/DifferentiationUnderTheIntegralSign) in succession, one gets

$${\int }_0^1\frac{\partial }{\partial m}{e}^{m\ln x}𝑑x={\int }_0^1{e}^{m\ln x}\ln xdx={\int }_0^1{x}^m\ln xdx=\frac{-1}{{(m+1)}^2}$$

$${\int }_0^1\frac{\partial }{\partial m}{x}^m\ln xdx={\int }_0^1{x}^m{(\ln x)}^2𝑑x=\frac{+1\cdot 2}{{(m+1)}^3}$$

$${\int }_0^1\frac{\partial }{\partial m}{x}^m{(\ln x)}^2𝑑x={\int }_0^1{x}^m{(\ln x)}^3𝑑x=\frac{-1\cdot 2\cdot 3}{{(m+1)}^4}$$

$$\mathrm{\cdots }$$

It’s evident that repeating the differentiation $n$ times the final result is the

$${\int }_0^1{x}^m{(\ln x)}^{n}dx=\frac{{(-1)}^{n}n!}{{(m+1)}^{n+1}}\mathit{\hspace{1em}\hspace{1em}}(m>-1).$$