I am trying to do the following integral $$I = \int_{x}^{\infty}{ \frac{1}{\sqrt{u}} k^{\frac{2}{c}\ln u} \left(c \ln \left(\frac{k}{u}\right) - 2 c + 2 \ln^2 \left(\frac{k}{u}\right)\right) \exp \left\{-\frac{ \left[c^2 + 4 \ln^2 \left(ku\right)\right]}{8 c}\right\} {\kern 1pt} \,du}$$ where $k > 0$ and $c > 0$ are some constants.
Upon completing the square in the power of $e$. $$I = \sqrt{k}\int_{x}^{\infty}{k^{\frac{2}{c}\ln u} \left(c \ln \left(\frac{k}{u}\right) - 2 c + 2 \ln^2 \left(\frac{k}{u}\right)\right) \exp \left\{-\frac{ \left[c + 2 \ln \left(ku\right)\right]^2}{8 c}\right\} {\kern 1pt} \,du}$$
Let $\ln u = v$. Then the integral becomes $$I = \sqrt{k}\int_{x}^{\infty}{k^{\large \frac{2v}{c}} \left(c \ln k - cv - 2 c + 2 \left(\ln k - v\right) ^2 \right) \exp \left\{v -\frac{ \left[c + 2 \ln \left(k\right) + 2v\right]^2}{8 c}\right\} {\kern 1pt} \,dv}$$ I am not sure how to proceed from here. Your help will be greatly appreciated.
Thank you.