How to evaluate the integral $e^{-(c\ln(\frac{1}{x}))^s} dx$? 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.
 A: 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:
$$
\begin{align*}
\int_{\alpha}^{1}{x^{cs}}dx &= \frac{1^{c} - \alpha^{cs + 1}}{cs + 1} \\
\end{align*}
$$
