How to solve indefinite integral of $e^{t\left(\pi \:-r\right)}\left(-\frac{ds}{dt}\right)^{1-\frac{1}{n}}$?

Question

I have the following integral: $$ \int e^{t\left(\pi -r\right)}\left(-\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{1-\frac{1}{n}}\mathrm{d}t. $$ I am not sure if this is solvable. I have tried integration by parts but do not know how to handle the term $$ \left(-\frac{\mathrm{d}s}{\mathrm{d}t}\right)^{1-\frac{1}{n}} $$ due to the presence of the exponent $1-\frac{1}{n}$. Also, I can't see if Feynman Integration (if this becomes a definite integral with bounds from $0$ to infinity) would apply here. I would be grateful for any help resolving this.

Answer 1

The integral actually has a a simple solution:

$$\int e^{t(\pi-r)}\left(-\frac{ds}{dt}\right)^{1-1/n}dt=(-1)^{1-1/n}\int e^{t(\pi-r)}(s'(t))^{1-1/n}dt$$

This is because $\frac{ds}{dt}$ is in a derivative form, and can be represented without raising $dt$ to a power.

For example: if $n=1$, $s(t)=t$, and $r=\pi-1$, then:

$$\int e^{t(\pi-r)}\left(-\frac{ds}{dt}\right)^{1-1/n}dt=\int e^tdt=e^t+C$$

Answer 2

Since $\;\dfrac{\text ds}{\text dt} = \left(\dfrac{\text dt}{\text ds}\right)^{-1},$ then the given integral can be presented via the inverse function $\;t(s):$ $$\int e^{t(\pi-r)}\left(-\dfrac{\text ds}{\text dt}\right)^{\large 1-^1/_n}\,\text dt =-\int e^{(\pi-r)\,t(s)}\;\sqrt[\Large n]{-\dfrac{\text dt}{\text ds}}\,\text ds.$$

However, detalization of $\;s(t)\;$ or $\;t(s)\;$ looks nesessary.