The task is to evaluate the following integral:
$$ C = \int_{0}^{\infty} \frac{\ln{t}~\sin{t}}{t}~dt.$$
It can be shown that:
$$ C = f'(0) ,$$
where
$$ f(a) = \int^{\infty}_{0} t^{a - 1}\sin{t}~dt = \sin{\bigg(\frac{\pi a}{2} \bigg)}\Gamma(a), $$ $$ f'(a) = \frac{\pi}{2}\cos{\bigg(\frac{\pi a}{2} \bigg)}\Gamma(a) + \sin{\bigg(\frac{\pi a}{2} \bigg)}\Gamma'(a). $$
What I'm struggling with is calculating the limit
$$ \lim_{a\to 0} f'(a) \; = \;? $$
Note: I know the answer is $\frac{-\gamma \pi}{2}$, but I just don't see why.
Hint: $f(a)=g(a)h(a)$ with $g(a)=\Gamma(a+1)$ and $h(a)=\sin(\pi a/2)/a$. Now use $$g(0)=1,\ h(0)=\pi/2,\ g'(0)=-\gamma,\ h'(0)=0.$$