The following result is obtained through Mathematica: $$\int_0^\infty \exp\left(-C\left(\frac{s}{\theta}\right)^\lambda\right)\,\mathrm{d}s = C^{-\lambda^{-1}}\theta\Gamma(1 + \lambda^{-1})$$
where $\Gamma(.)$ is the Gamma function.
Can you show the intermediary steps to achieve this result?
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Hint : You might want to pose :
$$u = \left(\frac{s}{\theta C^{1/\lambda}}\right)$$
It might help you to make $\Gamma$ function appear...
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I think the only issue here is the cluster of arbitrary variables $C$, $\theta$, and $\lambda$. The correct substitution is to let$$u=C\left(\frac s{\theta}\right)^{\lambda}$$and differentiate with respect to $s$ on the right-hand side. The work gets rather messy, so I'll walk you through the work. First thing's first: solve for $s$. This can be easily done by dividing both sides by $C$ and multiplying by $\theta^{\lambda}$. Hence$$s=\frac {u^{1/\lambda}\theta}{C^{1/\lambda}}$$ Now differentiate. Everything on the right-hand side can be viewed as a constant except for $u^{1/\lambda}$ where the power rule will come in handy.$$\mathrm ds=\frac {\theta}{C^{1/\lambda}}\frac {u^{1/\lambda-1}}{\lambda}\mathrm du$$Replacing $\mathrm ds$ with the new change of variables gives$$\mathfrak{I}=\frac {\theta}{C^{1/\lambda}\lambda}\int\limits_0^{\infty}\mathrm du\, u^{1/\lambda-1}e^{-u}=\frac {\theta}{C^{1/\lambda}\lambda}\Gamma\left(\frac 1{\lambda}\right)$$ The result given in the question is just a step away by using the identity$$\Gamma(1+s)=s\Gamma(s)$$
Hint. One may recall the following integral representation of the gamma function $$ \int_0^\infty x^se^{-x}\:dx=\Gamma(s+1),\quad s>-1, $$ then, in the given integral, $$\int_0^\infty \exp\left(-C\left(\frac{s}{\theta}\right)^\lambda\right)\,\mathrm{d}s$$ one may just perform the change of variable $$ x=C\left(\frac{s}{\theta}\right)^\lambda,\qquad s=\theta \left(\frac{x}{C}\right)^{1/\lambda},\qquad ds=\frac{\theta}{\lambda C^{1/\lambda}} x^{1/\lambda-1}dx. $$Hope you can take it from here.