I am trying to use integration by parts and evaluate

$$\int x^3 e^{-ax^2}dx$$

So far I have gotten it to

\begin{align}

u&= x^3\\

du &= 3x^2 dx \\

dv & = e^{-ax^2}dx\\

v &= \int e^{-ax^2}dx

\end{align}

$$\int x^3 e^{-ax^2}dx = -\frac{x^2}{2a} e^{-ax^2} – \int \bigg \lbrace \int e^{-ax^2} dx \bigg \rbrace 3x^2 dx $$

I keep screwing up this integral

$$\int e^{-ax^2}dx $$

and Wolfram Alpha gives me something unintelligible for it.

Can someone help me understand what I am missing? I think if someone could just evaluate

$$\int e^{-ax^2}dx $$ for me, I could complete the problem.

## Best Answer

The problem with your current choice of $u$ and $dv$ is that $\displaystyle\int e^{-ax^2}\,dx$ is not expressible in terms of elementary functions.

Fortunately, there is more than one way to break up $x^3e^{-ax^2}\,dx$ into $u$ and $dv$.

Try setting $u = x^2$ and $dv = xe^{-ax^2}\,dx$. Then, $du = 2x\,dx$ and $v = -\dfrac{1}{2a}e^{-ax^2}$.