Show the convergence of an integral

Question

The question is asking to show the convergence and calculate the value of the integral $$ \int_0^{\infty} x^3 e^{-x} dx$$

I calculated the integral using the integral by parts 3 times and the result is 6. In the other hand, I don't have an idea how to show the convergence! any help?

Answer 1

Hint: Show that $\frac {x^{3}e^{-x}} {e^{-x/2}}$ is bounded and use the fact that $e^{-x/2}$ is integrable.

Answer 2

You need one of the standard estimates to justify the convergence. For the present task do the following (also standard, hence Community Wiki). Let $\sqrt{e}=1+a$. By the binomial formula we have for all integers $n\ge4$ $$ \begin{aligned} e^{n/2}&=(1+a)^n\\ &=\sum_{k=0}\binom nk a^k\\ &>\binom n4a^4\\ &=\frac{n(n-1)(n-2)(n-3)}{24}a^4. \end{aligned} $$ Dividing this by $n^3$ gives the bound $$ \frac{e^{n/2}}{n^3}>n(1-\frac1n)(1-\frac2n)(1-\frac3n)a^4\to \infty $$ when $n\to\infty$. It follows that $x^3/e^{x/2}$ is bounded.