# Confused by a Laplace transform of $f(t)=t^ne^{at}$

Having looked at my lecture notes I was confused by the following part of a derivation of a Laplace transform for the function $\;f(t)=t^ne^{at} ,\quad n\ge0,\; a \in \mathbb{C}, \; f(t)=0 \;\forall \;t<0$: $$F(s)=\mathcal{L}[f(t)](s)=\int_0^\infty t^ne^{at}e^{-st}dt=\int_0^\infty t^ne^{(-(s-a))t}dt=:I_n$$

Using partial integration we get: $$\left. t^n{1 \over a-s}e^{-(s-a)t}\right|_0^\infty+{n \over s-a}\int_0^\infty t^{n-1}e^{-(s-a)t}dt={n \over s-a}I_{n-1}\quad\text{for } n \ge1$$

What I don't understand is how $\left.t^n{1 \over a-s}e^{-(s-a)t}\right|_0^\infty$ gets evaluated. I have tried the following: $$\left.t^n{1 \over a-s}e^{-(s-a)t}\right|_0^\infty={1 \over a-s}\lim_{R\to\infty}R^ne^{(a-s)R}={1 \over a-s}\lim_{R\to\infty}\sum_{j=0}^\infty{(a-s)^j\over j!}R^{j+n}\\={1 \over a-s}\lim_{R\to\infty}\left(R^n+{a-s \over1!}R^{1+n}+ {(a-s)^2 \over2!}R^{2+n}+\dots\right)$$ And according to the above integration process this limit should equal $0$. I am stuck on trying to show that. Of course assuming that $a>s$, the sum won't converge to $0$, so the essential condition is that $a<s$, thus we have a chance for sum convergence. How can I show that for any $R>0$ the sum turns $0$? Or maybe there's another way without power using series representation.

• I would apply L'Hospital's rule. It's a much cleaner approach. Commented Jul 18, 2014 at 17:30
• Forget the power series. For every $\varepsilon > 0$ and every $\alpha\in\mathbb{R}$, you have $$\lim_{x\to+\infty} x^\alpha e^{-\varepsilon x} = 0.$$ Commented Jul 18, 2014 at 17:33
• @CameronWilliams I see, applying L'Hospital $n$ times yields $${n! \over (s-a)^n} \lim_{R \to \infty}{1 \over e^{(s-a)R}}=0$$ Commented Jul 18, 2014 at 17:34
• An alternative route is to take the $n=0$ case and repeatedly differentiate with respect to $a$. Commented Jul 18, 2014 at 17:35
• @DanielFischer I will note that, thank you Commented Jul 18, 2014 at 17:38