# 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

 $\int_{0}^{1}x^{m}\,dx=\frac{1}{m+1}$

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

 $\int_{0}^{1}\frac{\partial}{\partial m}e^{m\ln{x}}\,dx=\int_{0}^{1}e^{m\ln{x}}% \ln{x}\,dx=\int_{0}^{1}x^{m}\ln{x}\,dx=\frac{-1}{(m+1)^{2}}$
 $\int_{0}^{1}\frac{\partial}{\partial m}x^{m}\ln{x}\,dx=\int_{0}^{1}x^{m}(\ln{x% })^{2}\,dx=\frac{+1\cdot 2}{(m+1)^{3}}$
 $\int_{0}^{1}\frac{\partial}{\partial m}x^{m}(\ln{x})^{2}\,dx=\int_{0}^{1}x^{m}% (\ln{x})^{3}\,dx=\frac{-1\cdot 2\cdot 3}{(m+1)^{4}}$
 $\cdots$

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

 $\int_{0}^{1}x^{m}(\ln{x})^{n}\,dx=\frac{(-1)^{n}n!}{(m+1)^{n+1}}\qquad(m>-1).$
Title example of differentiation under integral sign ExampleOfDifferentiationUnderIntegralSign 2013-03-22 17:01:56 2013-03-22 17:01:56 pahio (2872) pahio (2872) 4 pahio (2872) Example msc 26A24 msc 26B15