Evaluate $\int_0^\infty e^{-x}\sin e^{-x}\cos e^{-x} \ln e^{-x}\;dx$ How to evaluate the following integral?
$$\int_0^\infty e^{-x}\sin e^{-x}\cos e^{-x} \ln e^{-x}\;dx$$
 A: 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 $$\frac{1}{4} (\text{Ci}(2)-\gamma -\log (2))$$
A: First substituting $z=e^{-x}$ and then integrating by parts, this integral may evaluated in terms of the cosine integral function, $\operatorname{Ci}{(x)}$:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\infty}e^{-x}\sin{\left(e^{-x}\right)}\cos{\left(e^{-x}\right)}\ln{\left(e^{-x}\right)}\,\mathrm{d}x\\
&=\int_{0}^{1}\sin{(z)}\cos{(z)}\ln{(z)}\,\mathrm{d}z\\
&=\left[\sin^2{(z)}\ln{(z)}\right]_{0}^{1}-\int_{0}^{1}\sin{(z)}\left(\frac{\sin{(z)}}{z}+\cos{(z)}\ln{(z)}\right)\,\mathrm{d}z\\
&=-\int_{0}^{1}\frac{\sin^2{(z)}}{z}\,\mathrm{d}z-\int_{0}^{1}\sin{(z)}\cos{(z)}\ln{(z)}\,\mathrm{d}z\\
&=-\int_{0}^{1}\frac{\sin^2{(z)}}{z}\,\mathrm{d}z-\mathcal{I}\\
\implies 2\mathcal{I}&=-\int_{0}^{1}\frac{\sin^2{(z)}}{z}\,\mathrm{d}z\\
\implies \mathcal{I}&=-\frac12\int_{0}^{1}\frac{\sin^2{(z)}}{z}\,\mathrm{d}z\\
&=-\frac14\int_{0}^{1}\frac{1-\cos{(2z)}}{z}\,\mathrm{d}z\\
&=-\frac14\int_{0}^{2}\frac{1-\cos{(t)}}{t}\,\mathrm{d}t\\
&=-\frac14\operatorname{Cin}{(2)}\\
&=-\frac14\left(\gamma+\ln{(2)}-\operatorname{Ci}{(2)}\right).\\
\end{align}$$

Note on Definitions:
A couple of different definitions of the cosine integral function, $\operatorname{Ci}{(x)}$, include either,
$$\operatorname{Ci}{(x)}:=-\int_{x}^{\infty}\frac{\cos{t}}{t}\,\mathrm{d}t,$$
or,
$$\operatorname{Ci}{(x)}:=\gamma+\ln{(x)}+\int_{0}^{x}\frac{\cos{t}-1}{t}\,\mathrm{d}t.$$
For convenience we may also define the related auxiliary function, $\operatorname{Cin}{(x)}$,
$$\operatorname{Cin}{(x)}:=\int_{0}^{x}\frac{1-\cos{t}}{t}\,\mathrm{d}t,$$
implying the relation,
$$\operatorname{Cin}{(x)}=\gamma+\ln{(x)}-\operatorname{Ci}{(x)}.$$
