Does an integral of the form $\int f(x) \, \sqrt{dx}$ have any meaning? If $f(x)$ is a Riemann-integrable function, what meaning is there to an integral of the form $$\displaystyle\int f(x) \, \sqrt{dx}~?$$ I have read that stochastic processes like Brownian motion may be described by integrals of somewhat unusual forms as $\displaystyle\int dW^2$. 
I suppose that $\displaystyle\int f(x) \, \sqrt{dx}$ doesn't represent a random process since $f(x)$ here is defined and deterministic. 
Any insight would be helpful.
 A: Good question! Think of differentials like $dx^2$ existing because integrals undo derivatives:
$$\begin{align}
\frac{d}{dx}\frac{d}{dx}\,f(x) &= f’’(x) \\[2ex]
\frac{d^2f}{dx^2} &= f’’(x) \\[2ex]
d^2f &= f’’(x)\, dx^2 \\[2ex]
\end{align}$$
Let’s integrate once (ignoring the $+c$). Remember how integrating always eliminates the differential $d$ and undoes the derivative, like how $\int2x\,dx=x^2$, except we’re using a two-letter symbol that isn’t $x$:
$$\begin{align}
\int d(df) &= \int f’’(x)\,dx\,dx \\
\int 1\, d(df) &= \left(\int f’’(x)\,dx\right)\,dx \\
df &= f’(x) \, dx
\end{align}$$
Now we can do a normal integration (once again ignoring $+c$), but this time on the left we’ll integrate with respect to $f$—but don’t worry; the same rules apply, it’s just a different letter:
$$\begin{align}
\int 1\,df &= \int f’(x) \, dx \\
f &= f(x)
\end{align}$$
So, what does this mean? It means that the power tells us how many times to integrate, just like the power in the derivative tells us how many times to take the derivative. We can’t do either of these $1/2$ times or $\sqrt{\text{once}}$ (unless you make up some crazy version of calculus).
A: It seems to me that you can look at this as the square-root of a differential $~1-$form ${(dx)}^{1/2}$ which has been asked and answered here:
What are half forms?
