How do we indicate the integrating variable?

Question

Integrating the following simple equation $te^x$ w.r.t. $t$ using stokes' theorem, $\int_{\partial \Omega} \omega = \int_{\Omega} \mathrm d\omega$, yields the following equation

$$ \int_0^5 te^x\,\mathrm dt = \color{red}{\int_0^5\mathrm d(\frac12t^2 e^x)} = \frac{25}2e^x $$

Integrating w.r.t to $x$ yields

$$ \int_0^5 te^x\,\mathrm dx = \color{purple}{\int_0^5\mathrm d(t e^x)} = t (e^5 - 1) $$

At the first step, it is obvious that we are integrating with respect to $t$ and $x$, because it is the only variable present in the differential. But as soon as we deal with differentials of multiple variables, this becomes ambiguous. There is nothing in the second step indicating what variable we integrate over. They would also appear if we integrated the following expressions:

$$ \int_0^5 \frac12 t^2e^x\,\mathrm dx = \color{red}{\int_0^5\mathrm d(\frac12t^2 e^x)} = \frac12 t^2 (e^5 - 1) $$

or

$$ \int_0^5 e^x\,\mathrm dt = \color{purple}{\int_0^5\mathrm d(t e^x)} = 5e^x $$

Is there an unambiguous way of writing these integrals, such that the integrating variable is clear without context? Preferably one that appears in literature. My immediate thought would be to write $\int_{t=0}^5$ (as in sums), but I haven't seen this use anywhere.

Answer 1

Yes, you can very well enhance the integration range.

$$\int_{t=0}^5 d\left(\frac{t^2}2e^x\right)$$

or

$$\int_{x=0}^5 d(te^x). $$

Answer 2

Alright, so there a couple of things. First you are using two variables $x$ and $t$ so it seems that you have a 2 dimensional space. Then specifying two numbers is not a meaningful way to give a region of integration.

There are several things that make sense. Just to try to keep you example. consider the function (0-form) $$ \omega =t^2 e^x $$ Then $$ d\omega =2te^x dt + t^2e^x dx $$ Note that is the full differential, somehow above you decided that x will be constant in one side and t will be constant in the other side, so it doesn't make much sense.

Now suppose we want to integrate along the line in the $x,t$ plane given by $\gamma(s)=(s,s^2)$, $0\leq s\leq $1, the 1-form $d\omega$. Then $$ \int_\gamma d\omega=\int_0^1 2s^2e^s2sds+s^4e^s ds=\int_0^1 e^ss^3(4+s) ds=e $$ Then what stokes theorem says is that this is the same as ($\partial \gamma = (1,1)- (0,0)$) $$ \int_{\partial \gamma}\omega=t^2e^x|_{(0,0)}^{(1,1)}=e $$

There are several other paths you can try, but for instance to get the result of the first integral you put, if you integrate along the straight line from $(x,0)$ to $(x,5)$, i.e. $\gamma(s)=(x,s)$ (note that $x$ is some fixed constant) we get

$$ \int_\gamma te^x dt =\int_\gamma d\left (\frac{1}{2}t^2e^x\right)=\frac{25}{2}e^x $$

I will let you figure out what path makes the second integral sensible as well. But I just point out that the confusion in the second set of equations is precisely because you haven't given any path, so the integral doesn't make sense.