Is this a valid argument for the dimensions of this integral? Given the integral:
$I = \int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^2} dx$
If I say that if:
$$[I] = [L]^{2n+1}$$
And the dimensions of alpha must be $\frac{1}{[L]^2}$ since the exponent must be dimensionless
Then the integral must scale with $[\alpha]^{-n+1}$ since
$$[\alpha] = [x]^{-2}=[L]^{-2}$$
$$[\alpha]^{-n} = [L]^{2n}$$
$$[\alpha]^{-n+1} = [L]^{2n+1}$$
$$[\alpha]^{-n+1} = [I]$$
but If I do u substitution, then I get something like:
$$\int_{-\infty}^{\infty} x^{2n}e^{-\alpha x^2} dx $$
$$= \int_{-\infty}^{\infty} x x^{2n-1}e^{-\alpha x^2} dx$$
using:
$$u = -\alpha x^2$$
$$du = -2\alpha x dx$$
i get:
$$\int_{-\infty}^{\infty} \frac{1}{-2 \alpha} \left(\sqrt{\frac{-u}{\alpha}}\right)^{2n-1}e^{u} du$$
edit: fixed algebra error below:
so the integral is proportional to 
$$\alpha^{-1}(\alpha^{-1/2})^{2n-1} = \alpha^{-n-\frac{1}{2}}$$
Is there something wrong with the first method? How do you infer how the integral varies with alpha from dimensional analysis without u substitution here?
 A: The problem is that you are treating the units $[\alpha]$ and $[L]$ like equivalent "variables", when they are not. For example, let's say the units of $[L]$ are meters and the units of $[\alpha]$ are something, say $a$. Then:
$$1\space a=1\space m^{-2}$$
$$1\space a^2=1\space m^{-4}$$
$$1\space a^3=1\space m^{-6}$$
So you can see, adding $1$ to the units of $\alpha$ subtracts $2$ from the units of $L$. But in arriving at the line $[\alpha]^{-n+1}=[L]^{2n+1}$ you did something similar by adding one to both sides. 
To relate this to normal algebra without thinking about units, this is similar to saying if $x^2 = y$, then $x^3=y^2$, which is not the case. The correct thing to do is $x^3=xy=y^{3/2}$
Therefore, we see that for every $1$ we add to the exponent of $[\alpha]$, we must subtract $2$ to the exponent of $[L]$. This is the fix we need. Except we want to subtract $1/2$ from $[\alpha]$ and then subsequently add $1$ to $[L]$. This is essentially the same thing as saying either "multiply both sides by $[\alpha]^{-1/2}$" or "multiply both sides by $[L]$", and then using the relation $[\alpha]=[L]^{-2}$ to only deal with $[\alpha]$ or $[L]$ on each corresponding side.
$$[\alpha]^{-n}=[L]^{2n}$$
$$[\alpha]^{-n-1/2}=[L]^{2n+1}=[I]$$
Your work at the end of your question gets to this exact conclusion.
