Show that the function $$ f(t)=\int\limits_0^{\infty}e^{-tx}\, dx=\frac{1}{t}, t>0 $$ can be differentiated indefinit times under the integral and use this to show the formula $$ \int\limits_0^{\infty}x^ne^{-x}\, dx=n! $$
Hello, my problem is to show
1) that $f$ can be differentiated under the integral. Is this an improper parameter integral? What do I have to show?
2) that $f$ can be differentiated under the integral indefinit times.
The idea is to use dominated convergence theorem on $$ \frac{\partial g}{\partial t} = \lim_{n\to \infty} \frac{g(x,t_n) - g(x,t)}{t_n - t} $$ where $g(x,t) = e^{-tx}$. Find an $L^1$ function $h(x)$ such that $$ |\frac{\partial g}{\partial t}(x,t)| \leq h(x) \quad \forall t $$ (Hint: $e^{-tx} \leq e^{-x}$ for all $t \geq 1$) From this you will see that $$ \frac{df}{dt} = \int_0^{\infty} -xe^{-tx}dx = \frac{-1}{t^2} $$ Now repeat the process inductively and plug in $t=1$.
Bartle's book on Measure Theory has a very nice section on "integration w.r.t. a parameter" which might be worth a read.