Improper integral - show convergence/divergence

Question

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.

Answer 1

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.

Answer 2

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.)

Answer 3

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 $n$, $\Gamma(n+1) = n \Gamma(n) = n!$.