Can we apply dimensional analysis for variables inside integrals? Ex: if we have integral $$\int \frac{\text{d}x}{\sqrt{a^2 – x^2}} = \frac{1}{a} \sin^{-1} \left(\frac{a}{x}\right),$$ the LHS has no dimensions, while the RHS has dimensions of $\frac{1}{length}$. So let me know , whether I am correct?
[Physics] Is dimensional analysis valid for integrals
dimensional analysisintegration
Related Solutions
Your last suggestion is correct : the use which you are making of dimensional analysis is not justified.
Dimensional Analysis has two purposes : (1) to check that equations or terms in equations are commensurate; and (2) to find combinations of quantities with particular dimensions or no dimensions at all. It is not able to derive physically meaningful formulas, and certainly not a unique formula, which is what you seem to be expecting it to do.
The 2nd purpose can be achieved using Rayleigh's Method, or Buckingham's Pi Theorem which is a more formal version of it. For this purpose the method works, and it is easy to see why it works. That to me is sufficient justification.
In your 1st example, Rayleigh's Method tells you that the formula $\Delta P=k\rho gh$ (where $k$ is a dimensionless constant) is dimensionally consistent, but it cannot tell you that this formula is physically meaningful or correct for the application you have in mind. It cannot tell you what value of $k$ you should use. It can only tell you what dimensions $k$ must have for the formula $\Delta P=k\rho gh$ to be dimensionally consistent.
In the 2nd example, the method cannot tell you that the force of gravitational attraction between two masses should be $F=Gm_1m_2/r^2$. The method does not generate any dimensionless combinations of the quantities $F, m_1, m_2, r$. This is not a failure of Rayleigh's Method because no dimensionless combinations of these variables can be derived by any method. As with your 1st example, dimensional analysis can only tell you what dimensions the constant $G$ must have for the formula $F=Gm_1m_2/r^2$ to be dimensionally correct.
The strengths and limitations of the method are stated in the wikipedia article :
The Buckingham π theorem provides a method for computing sets of dimensionless parameters from given variables, even if the form of the equation remains unknown. However, the choice of dimensionless parameters is not unique; Buckingham's theorem only provides a way of generating sets of dimensionless parameters and does not indicate the most "physically meaningful".
Two systems for which the [dimensionless] parameter [are equal] are called similar. As with similar triangles, they differ only in scale. They are equivalent for the purposes of the [unknown] equation. The experimentalist who wants to determine the form of the equation can choose the most convenient [system to investigate].
Most importantly, Buckingham's theorem describes the relation between the number of variables and [the number of] fundamental dimensions.
The problematic equation is $f_{salt}=Kw^\alpha$. It is problematic because $w$ has dimensions. The simplest way to solve this problem is to split your constant $K$ into $k/(w_0)^\alpha$ for some arbitrary chosen reference value $w_0$. Then your equation becomes
$$f_{salt}=k\left(\frac{w}{w_0}\right)^\alpha$$
This has two advantages: first, in contrast to the original equation (where the dimensions of $K$ depended on $\alpha$), you now know exactly what dimensions your proportionality constant is supposed to have; namely, $k$ has the same dimensions as $f_{salt}$. Second, now the quantity being exponentiated is dimensionless, which eliminates any effect that $\alpha$ would have on the dimensional analysis.
Best Answer
Yes, you can apply dimensional analysis to integrals. You count differentials like $dx$ as having the units of the associated variable, because $dx$ can be interpreted as an infinitesimal change in $x$.
In your example, if $a$ and $x$ have units of distance, then checking the units in your result shows that it's incorrect. The correct integral does not have the factor of $1/a$.