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

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

• integrate by parts a few times (maybe let $x=u^2$ first) – yoyo Nov 25 '20 at 0:04
• You could use the fact that $0\leq x^8 \leq e^{\sqrt{x}/2}$ whenever $x\geq 2$ – Matthew Pilling Nov 25 '20 at 0:11
• So if we know $x^8 \leq e^\frac{\sqrt x}{2}$ can we say $x^8 e^{-\sqrt x} \leq e^{-\sqrt x}e^\frac{\sqrt x}{2} = e^{-{\frac{\sqrt x}{2}}}$ and this is a standard test function we know converges so original integral must converge by direct comparison? I have looked at the graph and it looked as if it doesn't converge. – Morgzy Nov 25 '20 at 0:31

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}

You should be able to finish from there.

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,

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

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 $$711374856192000$$.