Improper integral - show convergence/divergence I ran into this question:
show convergence/divergence of:
$$\int_{0}^{\infty}x^3e^{-x^2}.$$
I tried for a long time and I'm kind'a lost.
Thanks in advance,
yaron.
 A: Hint:
To find $\int x^3e^{-x^2}\, dx$, write
$$
\int x^3e^{-x^2}\,dx= \int x^2\cdot xe^{-x^2}dx
$$
and apply the integration by parts formula, $$\int u\,dv=uv-\int v\,du,$$ with $u=x^2$ and $dv=xe^{-x^2}\,dx$.  Note, then, using substitution if you like, we have $v={-1\over2} e^{-x^2}$. 
Once you've found the antiderivative, $F(x)$, you'll have to compute $\lim\limits_{b\rightarrow\infty} [F(b)-F(0)]$. 
(There are quicker ways, as in Américo's answer, using comparison.)
A: We have
\begin{equation*}
\int_{0}^{\infty
}x^{3}e^{-x^{2}}dx=\int_{0}^{1}x^{3}e^{-x^{2}}dx+\int_{1}^{\infty
}x^{3}e^{-x^{2}}dx.
\end{equation*}
The first integral has no singularities. The second one is convergent as can be seen by applying the limit test and using the fact that $\int_{1}^{\infty }\frac{dx}{x^{2}}$ is convergent:
$$\lim_{x\rightarrow \infty }\frac{x^{3}e^{-x^{2}}}{x^{-2}}=0.$$
Consequently the given integral is convergent.
A: As no one has mentioned it yet, 
$$
\int_0 ^\infty x^3 e^{-x^2} \mathrm{d} x = \frac{1}{2} \int_0 ^\infty \left( x^2 \right)^{2-1} e^{-\left(x^2 \right)} \mathrm{d} \left( x^2 \right) = \frac{1}{2} \Gamma (2) = \frac{1}{2} 1! = \frac{1}{2} .
$$
So, the integral is convergent. This is shown by using properties of the Gamma Function:
$$
\Gamma(z) = \int_0 ^\infty x^{z-1} e^{-x} \mathrm{d} x,
$$
and that $\Gamma(z+1) = z \Gamma(z)$ which gives for integers $n$, $\Gamma(n+1) = n \Gamma(n) = n!$.
