How to find the value of $e^{-x} x^n$ at x = $\infty$ ? Actually while proving that $\Gamma(n+1) = n\Gamma(n)$ there is a step where I need to evaluate ($e^{-x} x^n$)$\big|_0^\infty$ . Now, since this is not an improper integral but the expression after calculating an improper integral(by parts) for which I need to find the value by substituting upper and lower limit, I donot know how to calculate the value at infinity since infinity is not a finite value. Thanks for any help.
Find the value of $e^{-x} x^n$ at x = $\infty$
gamma functionimproper-integralsinfinityreal-analysis
Best Answer
There is a result that $\lim_{x \rightarrow \infty}\frac{p(x)}{e^x} = 0$ for any polynomial $p(x)$. You can see this intuitively looking at Taylor's Series expansion of $\exp(x)$.
Then, $$\frac{x^n}{e^x} |_0^\infty = \lim_{\alpha \rightarrow \infty} \frac{\alpha^n}{e^\alpha} - \frac{0}{1} = 0. $$
Otherwise, you could use L'Hopital since numerator and denominator are going to $\infty$ and are continuous and differentiable in all $\mathbb R$.