Determine divergence of $\frac{x^3}{ e^x}$ over [0 $\infty$)

Question

$$\int_{0}^{\infty} \frac{x^3}{e^x}\mathrm{d}x $$

The function is checked to converge in the interval $ ( x= 0,1), $ but how can we check whether it diverges for the full interval?

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Answer 1

Integrate by parts three times and you can evaluate it exactly.

Answer 2

Actually, this is the gamma function.

$\Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt$

You can easily check that $\Gamma(1)=1$.

After this, prove by induction that $\Gamma(n+1)=n\Gamma(n)$.

This integral is thus, $\Gamma(4)=\int_0^\infty x^{4-1}e^{-x}dx$.

Thus, it is convergent to the value $3!=6$.

Edit:

As for the pretty simple proof that $\Gamma(n+1)=n\Gamma(n)$, you integrate by parts with $t^{n}=u, e^{-n}=v'$

Answer 3

Let us split at $1$ like you did, and look at the interval $[1,\infty)$. By the Maclaurin expansion for $e^x$ we have for positive $x$ that $$e^x\gt 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}\gt \frac{x^5}{5!}.$$

It follows that on $[1,\infty)$ we have $\frac{x^3}{e^x}\lt \frac{120}{x^2}$.

But we know that $\int_1^\infty \frac{120}{x^2}\,dx$ converges. By Comparison, so does our integral.

Remark: Much more informally, $e^x$ in the long run grows far faster than any polynomial. So in the long run $\frac{x^3}{e^x}$ approaches $0$ very fast.