# Choosing branch cuts for complex integration

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.

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}

• 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.
– user54031
Aug 18, 2017 at 16:07
• 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. 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

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.

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

• this makes me curious, can a rectangular contour with a cut-out semicircle on the top right quadrant also work or no 14 hours ago
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.962782084941864, "perplexity": 309.2104238222284}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662561747.42/warc/CC-MAIN-20220523194013-20220523224013-00657.warc.gz"}