Evaluate $$\int_0^\infty \frac{\operatorname{Si}(x)}{x(x^2+1)}\,dx.$$
I use the Feynman’s technique. Let $$I(a)=\int_0^\infty \frac{\operatorname{Si}(ax)}{x(x^2+1)}\,dx$$ then $I'(a)=\frac{\pi}{2a} (1-e^{-a})$. After some steps, I get: $$I(a)=\frac{\pi}{2} \ln(a) -\frac{\pi}{2} \operatorname{Ei}(-a)+C$$ Wolfram Alpha gives me the result $\frac{\pi \gamma}{2}-\frac{\pi}{2} \operatorname{Ei}(-1)$, but I have no idea how to find the constant $C$.
$\operatorname{Si}(x)$ is the Sine Integral, $\operatorname{Ei}(x)$ is the Exponential Integral, $\gamma$ is the Euler-Mascheroni Constant.
The hint given by J.G. is a good one!
From the definition of $I$ we have that $\lim_{a\to 0^+}I(a)=I(0)=0$. On the other hand $$\lim_{a\to 0^+}I(a)=\frac{\pi}{2}\lim_{a\to 0^+}(\ln(a) - \operatorname{Ei}(-a))+C=\frac{\pi}{2}\lim_{a\to 0^+}(-\gamma-\sum _{n=1}^{\infty } \frac{ (-a)^n}{n\cdot n!})+C=-\frac{\pi}{2}\gamma+C$$ where we used the the power series expansion for $\operatorname{Ei}$ at $x=0$: $$\operatorname{Ei}(x)= \gamma +\ln|x| + \sum _{n=1}^{\infty } \frac{ x^n}{n\cdot n!}.$$ Therefore $C=\frac{\pi}{2}\gamma$, and $$\int_0^\infty \frac{\operatorname{Si}(x)}{x(x^2+1)}\,dx=I(1)=-\frac{\pi}{2} \operatorname{Ei}(-1)+\frac{\pi}{2}\gamma$$ as expected. Note that the result can also be written as $$-\frac{\pi}{2} \operatorname{Ei}(-1)+\frac{\pi}{2}\gamma=\frac{\pi}{2} \sum _{n=1}^{\infty } \frac{(-1)^{n+1}}{n\cdot n!}\approx 1.2513.$$