Gamma like integral with power in exponent

Question

I came across an integral in my stat mech book of the form
$$\int_0^\infty x^{d-1}e^{x^s}dx$$
The book claims without proof that this is
$$\frac 1s\Gamma(d/s)$$
I tried doing a change of variables $y=x^s, dy=sx^{s-1}$ but I couldn't get my integrand to come out right. Is this the correct approach?

Answer 1

The integral you have provided is divergent $$ \int_0^\infty x^{d-1}e^{x^s}dx. $$ since your exponential term is positive. For the integral to converge to the value you wrote, we must have a decaying exponential $$ I\equiv\int_0^\infty x^{d-1}e^{-x^s}dx=\frac{1}{s}\Gamma\big(\frac{d}{s}\big),\ \Re(s)>0, \ \Re(d)>0. $$ For this we use your idea, $y=x^s$, $dy=s x^{s-1} dx$, $dx={s^{-1}dy x^{1-s}}$. We also have $x=y^{1/s}$, $x^{d-1}=y^{(d-1)/s}$. Thus we have $$ I=\int_0^\infty y^{(d-1)/s}e^{-y}\frac{y^{(1-s)/s}}{s}dy=\frac{1}{s}\int_0^\infty y^{d/s-1} e^{-y}dy. $$ We can choose to define $\alpha\equiv d/s-1$ to obtain $$ I=\frac{1}{s}\int_0^\infty y^{\alpha} e^{-y} dy=\frac{1}{s}\Gamma(1+\alpha)=\frac{1}{s}\Gamma\big(\frac{d}{s}\big). $$ This concludes the proof.