When calculating integrals like $$\int_0^\infty \frac{x^\alpha}{1 + x^2}dx$$ for $\alpha \in (-1,1 )$, it is convenient to take the branch cut of the integrand along the positive real axis and then use the keyhole contour.

I was wondering is there a way to use the principal branch cut, which runs along the negative real axis, to calculate this kind of integrals? I'm assuming some trivial manipulation of the integrand for $x>0$ would do the trick, but I fail to see it.


2 Answers 2


Using the principal branch, we can write the integral $\oint_C \frac{z^a}{z^2+1}\,dz$, where $C$ is comprised of (i) the real line segment from $-R$ to $R$ and (ii) the semi-circle in the upper-half plane, centered at the origin and with radius $R$, as

$$\begin{align} \oint_C \frac{z^a}{z^2+1}\,dz&=\int_{-R}^0 \frac{x^a}{x^2+1}\,dx+\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\\\\ &=(1+e^{i\pi a})\int_0^R \frac{x^a}{x^2+1}\,dx+\int_0^\pi \frac{(Re^{i\phi})^a}{(Re^{i\phi}))^2+1}\,(iRe^{i\phi}))\,d\phi\tag1 \end{align}$$

As $R\to \infty$ the second integral on the right-hand side of $(1)$ approaches $0$. Hence, taking this limit and invoking the reside theorem we find that

$$\begin{align} \int_0^R \frac{x^a}{x^2+1}\,dx&=\frac1{1+e^{i\pi a}}\,(2\pi i) \text{Res}\left(\frac{z^a}{z^2+1}, z=i\right)\\\\ &=\frac{\pi e^{i\pi a/2}}{1+e^{i\pi a}}\\\\ &=\frac{\pi}{2\cos(\pi a/2)} \end{align}$$

  • $\begingroup$ Thanks. Your method works because there's an $x^2$ in the denominator which simplifies the problem. What if there was an $x^3$ instead? I will edit the question accordingly. $\endgroup$
    – user54031
    Aug 18, 2017 at 16:07
  • $\begingroup$ You're welcome. My.pleasure. If the denominator were $x^3$, there is a pole at $-1$. So, we need to deform the contour around that pole. $\endgroup$
    – Mark Viola
    Aug 18, 2017 at 19:03

Usually I recommend two different approaches to solve the following problem

$$\int^\infty_0 \frac{x^{a}}{p(x)}dx$$

where $p(x)$ is polynomial of order 2 or more and $p(x) \neq 0$ for $x\geq 0$. Also we assume that $p(x)$ is an even function $p(-x) = p(x)$. Then we integrate the following function

$$f(z) = \frac{z^a}{p(z)}$$

around one of the following

No negative zero

Suppose that $p(z)$ has no zeros in the negative real axis then I would follow @Mark approach because it is the easiest. Why ? you would just use a sermi-circle contour and avoid evaluating unnecessary residues. Look at the figure below

enter image description here

The only residues that are continbuting to the integral are those above the x-axis.

Negative zeros

Suppose that $p(z)$ has zeros in the negative real axis then I would suggest a key-hole contour to avoide making a detour around the pole at the negative zero. Of course this comes at a price where we have to evaluate more poles. enter image description here

But this approach is more versatile and we don't need the assumption that $p(x)$ must be an even function.

  • $\begingroup$ this makes me curious, can a rectangular contour with a cut-out semicircle on the top right quadrant also work or no $\endgroup$
    – Max0815
    14 hours ago

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