I have the following integral that I try to solve with integration by parts;
$$\int_{0}^{L}z^{\frac{1}{\phi}-1}p\left(z\right)dz$$
What I have tried is as follows ;
I know from my calculations (some other calculations that are not relevant for the my question) that I have ;
$$\frac{dp\left(z\right)}{dz}=-\frac{p\left(z\right)}{z\epsilon}$$
I make some simple manipulations, in order to replace $\frac{dp\left(z\right)}{dz}$ in the integral and I find ;
$$\epsilon\int_{0}^{L}z^{\frac{1}{\phi}-1}z\frac{p\left(z\right)}{z\epsilon}dz=\epsilon\int_{0}^{L}z^{\frac{1}{\phi}}\frac{p\left(z\right)}{z\epsilon}dz$$
I write $z$ as following ;
$$z=-\frac{p\left(z\right)}{\frac{dp\left(z\right)}{dz}\epsilon}$$
So, I write my first integral as follows ;
$$-\int_{0}^{L}\left(\frac{p\left(z\right)}{\frac{dp\left(z\right)}{dz}\epsilon}\right)^{\frac{1}{\phi}}\frac{dp\left(z\right)}{dz}dz=-\int_{0}^{L}\left(p\left(z\right)\right)^{\frac{1}{\phi}}\left(\frac{dp\left(z\right)}{dz}\right)^{1-\frac{1}{\phi}}dz$$
I nominate $u(z)=p\left(z\right)^{\frac{1}{\phi}}$ and $v'\left(z\right)=\left(\frac{dp\left(z\right)}{dz}\right)^{1-\frac{1}{\phi}}$ but I got stuck and I can not solve the integral.
Any hints or solutions ? Thanks in advance.
If $$\dfrac{d}{dz} p(z) = - \dfrac{p(z)}{\epsilon z} $$ then for some constant $c$ you must have $$ p(z) = c z^{-1/\epsilon}$$ and then you don't need to integrate by parts.