Improper convergent integral: differentiation of parameter I'm interested in the following integral:
$$I = -\int \limits_0^\infty \frac{\sin (\pi x)}{\log (x)} \mathrm{d}x$$
It can be shown by the courtesy of Dirichlet, that this integral converges; there are no singularities, the function $\sin \pi x/\log x$ is in fact continuous, so the only limit that might be the problem is $x \to \infty$, but, from the Dirichlet test, we know that $\sin \pi x$ has a bounded primitive function and $1/\log(x)$ is monotonically decreasing for $x > 1$.
The only thing that've come to my mind as to transform this is to exploit the following property:
$$\frac{\partial}{\partial a} x^a = x^a \log x$$
So I defined the following function:
$$G (a, b) \equiv - \int \limits_0^\infty e^{-b x} x^a \frac{\sin \pi x}{\log x}$$
And now we can see that
$$F (a, b) \equiv \frac{\partial}{\partial a} G (a, b) \equiv - \int \limits_0^\infty e^{-b x} x^a \sin \pi x$$
which is computable and the result is:
$$F (a, b) = -\frac{\Gamma (1+a)}{\left( b^2 + \pi^2 \right)^{(1+a)/2}} \sin \left( (1+a) \arctan \frac{\pi}{b} \right)$$
Now I hope, that the following is true:
$$I = \lim_{b \to 0^+} G(0,b) = \lim_{b \to 0^+} \left[ \int \limits_C^0 F(a, b) \mathrm{d} a \right] = \int \limits_C^0 \lim_{b \to 0^+} F(a, b) \mathrm{d} a $$
$$\lim_{b \to 0^+} F(a, b) = - \frac{\Gamma (1+a)}{\pi^{1+a}} \cos \frac{\pi a}{2}$$
The final step is to find $C$ such that the calculation is easy. The actual value is $C = -1$, as this is Dirichlet integral, yielding $\pi/2$:
$$I = - \frac{\pi}{2} - \int \limits_{-1}^0 \frac{\Gamma (1+a)}{\pi^{1+a}} \cos \frac{\pi a}{2} \mathrm{d} a $$
The result is, however, wrong, because $I > 0$ and the right hand side is negative.
My goal is to express $I$ in terms of a simpler integral (not necessarily analytically) with a better asymptotic behaviour and/or over a finite interval.
Is it possible to compute $I$ efficiently?
Just a side note: it seems to me that it's impossible to practically compute $I$ by the definition, as the $\log(x)$ is increasing crazily slowly - that means the "$\infty$" would have to be very, very big to get a good approximation.
 A: Using the contour

contour integration gives
$$
\begin{align}
\overbrace{\color{#C00}{\text{PV}\int_0^\infty\frac{e^{i\pi x}}{\log(x)}\,\mathrm{d}x}}^{\substack{\text{integral along the line}\\\text{minus an infinitesimal}\\\text{interval centered at $1$}}}+\overbrace{\vphantom{\int_0^\infty}\ \ \ \ \ \color{#00F}{\pi i}\ \ \ \ \ }^{\substack{\text{integral along}\\\text{an infinitesimal}\\\text{semicircular arc}\\\text{centered at $1$}}}
&=\overbrace{\color{#090}{\int_0^\infty\frac{e^{-\pi x}}{\frac\pi2-i\log(x)}\,\mathrm{d}x}}^{\substack{\text{integral along the}\\\text{positive imaginary axis}}}\\
&=\int_0^\infty\frac{e^{-\pi x}\left(\frac\pi2+i\log(x)\right)}{\frac{\pi^2}4+\log(x)^2}\,\mathrm{d}x\tag{1}
\end{align}
$$
since there are no singularities inside the contour and the integral along the black arc vanishes as the radius tends to $\infty$.
Taking the imaginary part of $(1)$, we get an integral that is far easier to evaluate numerically:
$$
\begin{align}
\int_0^\infty\frac{\sin(\pi x)}{\log(x)}\,\mathrm{d}x
&=-\pi+\int_0^\infty\frac{e^{-\pi x}\log(x)}{\frac{\pi^2}4+\log(x)^2}\,\mathrm{d}x\\[6pt]
&\doteq-3.2191900386476588051\tag{2}
\end{align}
$$
