Can anyone help me evaluate
$$\int_{\alpha}^1 \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} dx$$,
Where $0 \leq \alpha \leq 1$ and $s \in \mathbb{R}$.
I tried changing variables, integration by parts etc. and got nowhere. Any clues on how to handle this would be appreciated.
I tried integration by parts with $dv=1$ and $u = \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}}$. This gives us:
$$x \exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} {\Huge|}_{\alpha}^1 - s c\int_{\alpha}^1 \left(c\ln\left(\frac{1}{x}\right)\right)^{s-1}\exp{\left\{-\left(c\ln\left(\frac{1}{x}\right)\right)^s\right\}} dx .$$
Note that the second term reminds the incomplete Gamma function (with $ln(1/x)$ instead of $t$), however, i couldn't reach further than this.
Consider: $$ \begin{align*} e^{-(c\ln\frac{1}{x})^s} &= e^{-(\ln(\frac{1}{x})^c)^s} \\ &= e^{-(\ln(\frac{1}{x^c}))^s} \\ &= \frac{1}{e^{(\ln(\frac{1}{x^c}))^s}} \\ &= \frac{1}{e^{\ln(\frac{1}{x^c})} \cdot e^{\ln(\frac{1}{x^c})} \ldots \text{s times} } \\ &= \frac{1}{\frac{1}{x^c} \cdot \frac{1}{x^c} \ldots \text{s times}} \,\,\,\,\, \text{[exponent and log cancel out] } \\ &= \frac{1}{(\frac{1}{x^c})^s} \\ &= \frac{1}{(\frac{1}{x^{cs}})} \\ &= x^{cs} \end{align*} $$
Now, your integral gets reduced to:
$$ \begin{align*} \int_{\alpha}^{1}{x^{cs}}dx &= \left[\int{x^{cs}}dx\right]_{\alpha}^{1} \\ &= \left[\frac{x^{cs + 1}}{cs + 1} + C\right]_{\alpha}^{1} \\ &= \left[\frac{1^{cs + 1}}{cs + 1} + C - \frac{\alpha^{cs + 1}}{cs + 1} - C \right] \\ &= \left[\frac{1^{cs + 1}}{cs + 1} - \frac{\alpha^{cs + 1}}{cs + 1}\right] \\ &= \left[\frac{1^{cs + 1} - \alpha^{cs + 1}}{cs + 1}\right] \\ \end{align*} $$
Since, $ s \in \mathbb{R}$ but you haven't given any information about $c$, we can simplify it up to: $$ \begin{align*} \int_{\alpha}^{1}{x^{cs}}dx &= \frac{1^{c} - \alpha^{cs + 1}}{cs + 1} \\ \end{align*} $$