Problem calculating the integral $\int_{0}^{\infty}x^ne^{-xt}dx$

Question

I want to calculate the Laplace transform of the function $ x ^ n $ but I have a problem with the limit that I get from applying integration by parts:

\begin{align*} \int_{0}^{\infty}x^ne^{-xt}dx&=\int_{0}^{\infty}x^n\left(-\cfrac{1}{t}e^{-xt}\right)'dx\\ &=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}-\int_{0}^{\infty}-\cfrac{1}{t}e^{-xt}(nx^{n-1})dx\\ &=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}\int_{0}^{\infty}x^{n-1}e^{-xt}dx\\ &=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\ &=\left[-\cfrac{x^n}{t}e^{-xt}\right]_{0}^{\infty}+\cfrac{n}{t}y_{n-1}(t)\\ &=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}-\left[-\cfrac{0^n}{t}e^{-(0)t}\right]+\cfrac{n}{t}y_{n-1}(t)\\ &=\lim_{x\to\infty}-\cfrac{x^n}{t}e^{-xt}+\cfrac{n}{t}y_{n-1}(t)\\ \end{align*}

But the problem is that I don't know how to take this limit, my intuition tells me that it is $ 0 $ because lowering the term $ e ^ {- xt} $ presents an indeterminacy of the type $ \cfrac{\infty}{\infty} $ so if we apply l'Hôpital's rule of thumb enough times to eliminate the term "$ x $" from above the fraction and since the term below is never affected by the derivatives other than by the term $ t ^ m $ that multiplies it, then the limit would be 0 but I don't know how to prove it properly.

Although I also tried that it should be possible to demonstrate this limit by induction, but in that case I do not know how to make such a demonstration, especially I do not know how to use the fact that it is true for $ k $ and with it demonstrate that it is true for $ k + 1 $. I appreciate any help.

Answer 1

You may avoid dealing with the limits by integrating as follows

$$\int_{0}^{\infty}x^ne^{-xt}dx =(-1)^n \frac{d^n}{dt^n} \int_{0}^{\infty}e^{-xt}dx = (-1)^n \frac{d^n}{dt^n} \left(\frac1t\right)=\frac{n!}{t^{n+1}} $$

Answer 2

To prove $\lim_{x\to\infty}x^ne^{-xt}=0$ for all $n>0,\,t>0$, it suffices viz. $y:=x/n\implies x^ne^{-xt}=(nye^{-yt})^n$ to check the case $n=1$, and then it suffices viz. $z=yt\implies ye^{-yt}=t^{-1}ze^{-z}$ to check the case $n=t=1$, i.e. $\lim_{z\to\infty}\tfrac{z}{e^z}=0$. You can do this with e.g. L'Hôpital's rule.