I would like to solve the following integral for $f(x)$
$$\int_{0}^{\infty} \frac{1}{(a+ix)(b+ix)(c+ix)} dx $$
by expanding it to the complex and then using a contourlike the half circle, i.e. if $C$ is the half circle I can decompose it into the real axis $C_1$ and the half circle around 0, $C_R$:
$$ \oint_C f(z) dz = \int_{C_R} f(z) dz + \int_{C_1} f(z) dz $$
where I can show that the integral over $C_R$ is zero in the limit $R\rightarrow \infty$.
Now I can deform $C$ integral to encircle the 3 singularities which then reads:
$$ \oint_C = \oint_{C_{a}} + \oint_{C_b} + \oint_{C_c} $$
which gives me almost the result I am looking for using the residual theorem.
However $C_1$ runs from $-\infty$ to $+\infty$, and since $f(z)$ does not have even symmetry I can not just multiply by $1/2$ and be done with it.
I was wondering if there was another routinely used contour that could be applied here, or if I had another oversight.
Too long for a comment $$I_0=\int_0^{\infty}\frac{dx}{(a+ix)(b+ix)(c+ix)} =i\int_0^{\infty}\frac{dx}{(x-ia)(x-ib)(x-ic)}$$ We can consider instead $$I(\alpha)=i\int_0^{\infty}\frac{x^\alpha}{(x-ia)(x-ib)(x-ic)}dx$$ then $I_0=I(\alpha=0)$
To evaluate $I(\alpha)$ we go in the complex plane and consider the keyhole contour (with the cut along the positive part of the axis $X$). The integrals along the big circle and small circle (around $z=0$) tend to zero.
Then $$\oint_C\frac{z^\alpha}{(z-ia)(z-ib)(z-ic)}dz=I(\alpha)-I(\alpha)e^{2\pi i\alpha}=i\cdot2\pi i\underset{z=ia, ib, ic}{\operatorname{Res}}\frac{z^\alpha}{(z-ia)(z-ib)(z-ic)}$$ where we take $ia=ae^{\frac{\pi i}2}, ib=be^{\frac{\pi i}2}, ic=ce^{\frac{\pi i}2}$, turning in the positive (counter-clockwise direction) from the upper bank of the cut.
The residues evaluation is straightforward and gives $$I(\alpha)\big(1-e^{2\pi i\alpha}\big)=2\pi e^{\frac{\pi i\alpha}2}\left(\frac{a^\alpha}{(a-b)(a-c)}+\frac{b^\alpha}{(b-a)(b-c)}+\frac{c^\alpha}{(c-b)(c-a)}\right)$$ Leading $\alpha\to 0$ and decomposing $a^\alpha=1+\alpha\ln a$ (and the same with $b, c$), we finally get $$I(\alpha)\sin\pi\alpha=-\pi i\alpha\frac{(b-c)\ln a+(c-a)\ln b+(a-b)\ln c}{(a-b)(b-c)(c-a)}+O(\alpha^2)$$ $$\Rightarrow I_0=I(0)=-i\frac{(b-c)\ln a+(c-a)\ln b+(a-b)\ln c}{(a-b)(b-c)(c-a)}$$