Examine convergence of $\int_0^{+\infty} x^{17} e^{-\sqrt {x}} dx$ 
Examine convergence of $\int_0^{+\infty} x^{17} e^{-\sqrt {x}} dx$

My try:
Let: $$u=\sqrt {x}, du=\frac{1}{2\sqrt {x}} dx$$ Then:$$\int_0^{\infty} x^{17} e^{-\sqrt {x}} dx=2\int_0^{\infty} u^{35} e^{-u} du$$However I don't know what I can do the next. Can you help me?
 A: Hint: Use
$$\int_{0}^{\infty}u^{35}e^{-u}{\rm d}u=\int_{0}^{L}u^{35}e^{-u}{\rm d}u+\int_{L}^{\infty}u^{35}e^{-u}{\rm d}u$$
For $0\leq x\leq L$ you have $u^{35}e^{-u}\leq u^{35}$. For $L<x$ (with $L$ high enough) you have $u^{35}<e^{\frac{1}{2}u}$ since the exponent diverges faster than a polynomial. Can you take it from here?
A couple of remarks: Note that the proof remains valid for other positive powers. The resulting integral for a general power is related to the Gamma function.
A: We have for some $M > 0$ that:  $\displaystyle \int_{M}^{\infty} 2u^{35} e^{-u}du< \displaystyle \int_{M}^{\infty} \dfrac{2}{u^2}du= \dfrac{2}{M}$. Also $0 < \displaystyle \int_{0}^{M} 2u^{35}e^{-u}du < \infty $ since the integrand is a continuous function over the closed interval $[0, M]$. Thus Using the fact that $\displaystyle \int_{0}^\infty = \displaystyle \int_{0}^{M} + \displaystyle \int_{M}^{\infty}$ we conclude the convergence of the given improper integral.
A: Note that for large enough $x$, we have $e^{\sqrt x} > x^{19}$ since exponentials grow faster than powers. After this $x$, we have $x^{17} e^{-\sqrt x} < \frac{1}{x^2}$, the integral of which converges.
A: Write $$ 2\int_0^\infty u^{35}e^{-u}du=2\int_0^1u^{35}e^{-u}du+2\int_1^\infty u^{35}e^{-u}du $$
Note that $0\le u^{35}e^{-u}\le 1$ when $u\in [0,1]$, and $u^{35}e^{-u}\le u^{-2}$ when $u\to\infty$ (To see this, note that $\lim_{n\to\infty}u^{37}e^{-u}=0<1$). And we know that $\int_0^\infty u^{-2}<\infty.$ Hence the integral converges.
In fact, using Gamma function, we have $2\int_0^\infty u^{36-1}e^{-u}du=2\Gamma (36)=2\cdot 35!$.
