Convergence of $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$

Question

Does $\int_{x=0}^\infty x^8 e^{-\sqrt x} dx$ converge?

I've tried splitting it into $\int_{x=0}^1 x^8 e^{-\sqrt x} dx$ + $\int_{x=1}^\infty x^8 e^{-\sqrt x} dx$. The first one is a continuous function over a bounded interval so that converges however I'm struggling with the second one. I have tried using limit comparison with $e^{- \frac{\sqrt x}{2}}$ as a test function but ended up going in circles. I have also tried direct comparison and integration by parts.

Thanks

Morgzy

Answer 1

Let $g(x)=\sqrt{x}-10\ln(x)$. Then $g$ is increasing for $x>400$.

$g(e^9)=e^{4.5}-90>0$. So $g$ is positive and increasing for $x>e^9>400$.

So, for $x>e^9$ we have that \begin{eqnarray} \sqrt{x}-10\ln(x)>0\\ \ln(x^{10})<\sqrt{x}\\ x^{10}<e^{\sqrt{x}}\\ x^8e^{-\sqrt{x}}<\frac{1}{x^2} \end{eqnarray}

You should be able to finish from there.

Answer 2

For the second one you can use direct comparison as follows:

Note that for $x>0$ you have

$$e^{-\sqrt x} = \frac 1{e^{\sqrt x}} = \frac 1{\sum_{n=0}^{\infty}\frac{x^{\frac n2}}{n!}} \stackrel{n=20}{<} \frac 1{\frac{x^{10}}{20!}}$$

Hence,

$$\int_{x=1}^\infty x^8 e^{-\sqrt x} dx< 20! \int_{x=1}^\infty \frac 1{x^2} dx < \infty$$

Answer 3

Put $t=\sqrt{x}$ or $t^2=x$, which yields $2t\,dt=dx$ and $x^8=t^{16}$. Then we have $$ \int_0^{\infty}x^8 e^{-\sqrt{x}}\,dx = 2\int_0^{\infty} t^{17}e^{-t}\,dt $$This is the well-known Gamma function. It converges as long as the power of $t$ is non-negative. You can figure out the exact value yourself by doing integration by parts once or twice until you see the pattern; you should get $711374856192000$.