Evaluate the integral $\int_0^{\infty} x^3e^{-x^2}$

Question

I used integration by parts since we have two functions multiplied by each other.

$u=x^3$ $du=3x^2$ $dv=e^{-x^2}$ $v=-e^{-x^2}$

setting up got me:

$-x^3e^{-x^2}-\int -3x^2e^{-x^2}$ I then repeated the same steps over again and integrated by parts again

setting $u=-3x^2$ $du=-6x$ $v=e^{-x^2}$ $dv=-e^{-x^2}$

and obtained:

$-x^3e^{-x^2}--3x^2e^{-x^2}-\int -6xe^{-x^2}$

Doing so again gave me:

$-x^3e^{-x^2}+3x^2e^{-x^2}-6xe^{-x^2}-\int -6e^{-x^2}$

$-x^3e^{-x^2}+3x^2e^{-x^2}-6xe^{-x^2}+6e^{x^2}$

evaluating this final answer gives something that doesn't make sense, so I'm guessing that the approach to this problem was wrong but why? I was always under the impression that if you have two functions multiplied by each other then you integrate by parts, why on earth does that not work in this situation?

Answer 1

Write $u(x)=x^2$ and $v(x)=-{1\over 2}e^{-x^2}$, $\int u(x)v'(x)dx=\int x^3e^{-x^2}dx= uv]-\int -xe^{-x^2}dx$ $=uv]_0^{\infty}-{1\over 2}e^{-x^2}]_0^{\infty}={1\over 2}.$

Answer 2

First do a substitution, $$u=x^2.$$

Answer 3

You did a mistake here

$u=x^3$ $du=3x^2$ $dv=e^{-x^2}$ $v=-e^{-x^2}$ $$v=-e^{-x^2} \implies dv=2xe^{-x^2}dx$$ It's not the best choice for an integration by part..

It's far better to choose $u=x^2$

Hint

Substitute $u=x^2$ $$I=\int x^3e^{-x^2}=\frac 1 2\int ue^{-u}du=\frac 1 2(-ue^{-u} -e^{-u})=-\frac 1 2e^{-u}(u+1)$$ $$I=-\frac 1 2e^{-x^2}(x^2+1)$$

Answer 4

Your choice of $dv$ does not work in your very first step, because $$\int e^{-x^2} \, dx \ne -e^{-x^2}.$$ You will find that taking the derivative of $e^{-x^2}$ yields, by the chain rule, $-2x e^{-x^2}$, and this in turn suggests that a suitable choice of $dv$ should be $$dv = x e^{-x^2} \, dx, \quad v = -\frac{1}{2} e^{-x^2}.$$ This then motivates the choice $$u = x^2, \quad du = 2x \, dx,$$ yielding $$\int x^3 e^{-x^2} \, dx = -\frac{1}{2} x^2 e^{-x^2} + \int x e^{-x^2} \, dx.$$ Then we observe that this remaining integral is one we have already done, namely it was our choice of $dv$ above; thus $$\int x^3 e^{-x^2} \, dx = -\frac{1}{2} x^2 e^{-x^2} - \frac{1}{2} e^{-x^2} + C.$$

An alternative approach that still involves integration by parts but is perhaps computationally easier, is to first write the integrand as $$x^3 e^{-x^2} = \frac{1}{2} x^2 \cdot 2x \cdot e^{-x^2},$$ and now we choose the substitution $y = x^2$, $dy = 2x \, dx$, to obtain $$\int x^3 e^{-x^2} \, dx = \frac{1}{2} \int y e^{-y} \, dy.$$ This is now amenable to integration by parts with the simpler choice $u = y$, $du = dy$, $dv = e^{-y} \, dy$, $v = -e^{-y}$.

Answer 5

It's actually more simple than that. To evaluate an improper integral, you must use limits to take care of the upper limit as such:

Set $u=x^2$ and $du=2xdx$.

$$\lim\limits_{b\rightarrow\infty}\frac{1}{2}\int_{0}^b e^{-u}u\text{ }du$$

Now we use integration by parts

$$\lim\limits_{b\rightarrow\infty}\Bigl(-\frac{1}{2}e^{-u}u\Bigr)\Big|_{0}^{b}- \lim\limits_{b\rightarrow\infty}\frac{1}{2}\int_{0}^b e^{-u}\text{ }du$$

Where $\lim\limits_{b\rightarrow\infty}\Bigl(-\frac{1}{2}e^{-u}u\Bigr)\Big|_{0}^{b}=\lim\limits_{b\rightarrow\infty}\bigl(0+0\bigr)=0$ then we are left with

$$\lim\limits_{b\rightarrow\infty}\frac{1}{2}\int_{0}^b e^{-u}\text{ }du =\frac{1}{2}\lim\limits_{b\rightarrow\infty}\bigl(-e^{-b}+e^{0}\bigr)=\frac{1}{2}\bigl(0+1\bigr)=\frac{1}{2}$$

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TIP: If you are integrating a function $\int f(x)g(x)dx$ using integration by parts, then you should always make sure to check which one of $$u=f(x);\text{ } dv=g(x)dx$$ and $$u=g(x)dx;\text{ } dv=f(x)$$ is the better choice for your substitution. Usually, you will know which one will lead to an easier integration just by checking.

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