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

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?

• Your final integral calculation was incorrect: $\int 6 e^{-x^2} \, dx \neq 6 e^{-x^2}$. In fact, your calculation of $v$ and $dv$ for $e^{-x^2}$ earlier was also incorrect. – B. Mehta Apr 15 '18 at 0:59

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 $u = y$, $du = dy$, $dv = e^{-y} \, dy$, $v = -e^{-y}$.

• +1 for addressing the OP's question, which was about what made their approach invalid – pjs36 Apr 15 '18 at 1:09

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

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

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

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

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.

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