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.

  • $\begingroup$ this would mean cutting through the singularities which are at $ia,ib,ic$ which is not allowed $\endgroup$ Commented Feb 7 at 11:18

1 Answer 1


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 $$I(\alpha)\sin\pi\alpha=-\pi i\alpha\frac{(b-c)\ln a+(c-a)\ln b+(a-b)\ln c}{(a-b)(b-c)(c-a)}+O(\alpha^2)$$ $$\Rightarrow I_0=I(0)=-i\frac{(b-c)\ln a+(c-a)\ln b+(a-b)\ln c}{(a-b)(b-c)(c-a)}$$

  • $\begingroup$ Thank you for your suggestion. Would you mind explaining why you expand (for the lack of a better word) with $i$ and then introduce the parametrization in $\alpha$? $\endgroup$ Commented Feb 11 at 13:41
  • $\begingroup$ $a+ix=i(x-ia)$ - for the convenience (to make clear what poles are). Parametrization $x^\alpha$ allows to use complex integration (the keyhole contour) and provides, in my opinion, a shortcut to the answer $\endgroup$
    – Svyatoslav
    Commented Feb 11 at 13:57
  • $\begingroup$ you seem to imply that the function $f(x)$, as is, can not be extended to the complex plane. Is that correct? If so, why? $\endgroup$ Commented Feb 11 at 14:16
  • $\begingroup$ It can - this is only the question of convenience. Of course you can leave $a+iz$ as it is in the denominator, but you have to define the pole and evaluate the residue correctly. $\endgroup$
    – Svyatoslav
    Commented Feb 11 at 14:24
  • $\begingroup$ Sorry, what I really meant was, that it sounds like you are saying that for analytic continuation the introduction of $z^{\alpha}$ is required, is that right? $\endgroup$ Commented Feb 11 at 14:37

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