We have the follwing PDE:
$$ \frac{\partial W}{\partial \tau} = {1 \over 2}\sigma^{2}\, \frac{\partial^{2} W}{\partial^{2}x}\quad \mbox{where}\quad x = \epsilon + \left(r - {1 \over 2}\sigma^{2}\right)\tau $$ We look for a solution of the following form:
$W(x,\tau)=\tau^af((x-c)/\tau^b)$
We have
$\mu=(x-c)/\tau^b$
We want to ensure that the special solution has the property that its integral over all $\epsilon$ is independent of $\tau$. To ensure that we require that
$\int_{-\infty}^\infty \tau^af((x-c)/\tau^b)dx$
is a constant. The part that I don't understand is how you get to the next step which is
$\int_{-\infty}^\infty \tau^{a+b}f(\mu)d\mu$
It is just a change of variables. Let $\mu = \frac{x-c}{\tau^b}$. As $x \to \infty$, $\mu \to \infty$, and as $x \to -\infty$, $\mu \to -\infty$. Moreover, $$\frac{d\mu}{d x} = \tau^{-b}. $$ Hence, \begin{align*} \int_{-\infty}^\infty \tau^a f \big ( \tau^{-b}(x-c)\big ) dx = \int_{-\infty}^\infty \tau^a f (\mu ) \frac{d x}{d \mu}d\mu =\int_{-\infty}^\infty \tau^{a+b} f (\mu ) d\mu . \end{align*}