Assign set in probability generating functional say i have an integral $A = \lambda c_d \displaystyle\int_0^\sim (1-exp(-shr^{-\alpha}))dr^{d-1}\mathrm {d}r$ $A = \lambda c_d \displaystyle\int_0^\sim (1-exp(-shr^{-1/\delta}))\mathrm {d}r$ (subs. r $\leftarrow$ $r^d$ ) $A = \lambda c_d \displaystyle\int_0^\sim (1-exp(-sh/x))\delta x^{\delta -1}\mathrm {d}x$ (subs. x $\leftarrow$ $r^{1/\delta}$), where $\delta \triangleq d/\alpha $ the left arrow is assign symbol i pressume, but how can we get that result by assigning $r^d$ to r on the first line? I've try $exp(-shr^{-\alpha})dr^ddr^{-1}\mathrm{d}r$ (subs. r $\leftarrow$ $r^d$) $exp(-shr^{-\alpha})dr^ddr^{-d}\mathrm{d}r$ $exp(-shr^{-\alpha})\mathrm{d}r$ which is of course totally far-fetched Can anyone please guide me with the step-by-step solution to get the result from the first line to the third line

Since notations are confusing, I used different notations: $d\rightarrow b$, $\alpha\rightarrow a$. $$I=[1-e^{-shr^{-a}}]br^{b-1}dr$$ when $r^b=p\rightarrow br^{b-1}dr=dp$ and $r=p^{\frac{1}{b}}$, by substituting these, we get $$I=[1-e^{-shp^{-\frac{a}{b}}}]dp$$ Now use $\delta=\frac{b}{a}$ which gives $$I=[1-e^{-shp^{-\frac{1}{\delta}}}]dp$$ Now use $p^{\frac{1}{\delta}}=x\rightarrow p^{-\frac{1}{\delta}}=\frac{1}{x}$ and $p=x^\delta\rightarrow dp=\delta x^{\delta-1}dx$, sustituting these, you get $$I=[1-e^{-\frac{sh}{x}}]\delta x^{\delta-1}dx$$ which gives the third step.