How to show that this integral is independent of a constant?

In a statistics problem, I got the following integral of a joint distribution function of two random variables that arose from a bivariate transformation:

$$ \int_{-\infty}^{\infty} \alpha^2 \exp \left( \alpha uv+\alpha u-e^{\alpha uv}-e^{\alpha u} \right) |u| \, \mathrm du, $$

where $\alpha,v \in \mathbb{R}$ and $\alpha > 0$.

Now, it has to be shown that the value of this integral does not depend on the value of $\alpha$. After some numerical evaluations it seems that indeed it does not depend on $\alpha$. As far as I can see, the integrand goes to 0 much quicker in one half of the domain (relative to $u=0$) than the other half, but it does indeed always seems to decay to zero.

I'm looking for some vague direction to head towards to prove that that's the case. Maybe not an outright hint - I want to have a crack at working it out, but I don't know enough about such "weird" functions to know what to look for.

After playing with it a bit, it looks that the general form of such integrals is

$$ \int_{-\infty}^{\infty} \alpha^2 \exp \left( \sum_i \alpha uv_i-\sum_i e^{\alpha uv_i} \right) |u| \, \mathrm du, $$ where $\alpha,v_i \in \mathbb{R}, \alpha>0$,

and it seems to be independent of $\alpha$ for both $\int \dots |u| \mathrm{d}u$ as well as $\int \dots u \mathrm du$, i.e. the absolute value of $u$ is optional.

Wolfram Alpha runs out of computation time even for the simplest form of this integral (with just one $v$: $v_1 = 1$).

It looks like I'm missing something very obvious...

Let $$f(u,v;\alpha) = \alpha^2 \exp\left( \alpha uv + \alpha u - e^{\alpha uv} - e^{\alpha u} \right) |u|.$$

Since $\alpha > 0$, the substitution $$t = \alpha u, \quad dt = \alpha \, du,$$ gives $$\int_{u = -\infty}^\infty f(u,v;\alpha) \, du = \int_{t=-\infty}^\infty f(t/\alpha, v; \alpha)/\alpha \, dt = \int_{t=-\infty}^\infty f(t, v; 1) \, dt.$$