Switch it with a summation operator and see if it makes sense. The problem isn’t the operation by itself, but the fact that the operator implies an argument application, like a function.
In the case of dx as an infinitesimal it makes sense. You are making a sum of all the values of the function in the integral range and multiplying with a constant dx.
Hum… I don’t think the integral “operator” applies by multiplication.
You can put the dx at the beginning of the integral, but not before it.
Physicists be like: whitness me
Nobody on your link is treating the integral “operator” as multiplicative.
dx \int f(x)is blatantly different from\int f(x) dxIf you were using nonstandard analysis with dx an infinitesimal you could put it outside I guess. Maybe with differential forms too?
Switch it with a summation operator and see if it makes sense. The problem isn’t the operation by itself, but the fact that the operator implies an argument application, like a function.
In the case of dx as an infinitesimal it makes sense. You are making a sum of all the values of the function in the integral range and multiplying with a constant dx.
In the context of differential forms, an integral expression isn’t complete without an integral symbol and a differential form to be integrated.