TITLE: Evaluate $\int_0^\infty e^{-x}\sin e^{-x}\cos e^{-x} \ln e^{-x}\;dx$
QUESTION [7 upvotes]: How to evaluate the following integral?
$$\int_0^\infty e^{-x}\sin e^{-x}\cos e^{-x} \ln e^{-x}\;dx$$

REPLY [5 votes]: Considering $$I=\int e^{-x}\sin (e^{-x})\cos( e^{-x}) \ln (e^{-x})\;dx$$the most natural change of variable seems to be $e^{-x}=t$. So $$I=-\int\log (t) \sin (t) \cos (t)\;dt=-\frac 12\int\log (t) \sin (2t) \;dt$$ Integrating by parts once to remove the logarithm then leads to $$I=\frac{1}{4} \log (t) \cos (2 t)-\frac{\text{Ci}(2 t)}{4}$$ and to the integral the result is $$\frac{1}{4} (\text{Ci}(2)-\gamma -\log (2))$$