How to handle the complex integration of this function around a branch point

Question

I have this complex integral to which I don't know if it's possible to assign a value:

The integral is on a small circle around the origin. The function is $\frac{1}{(z-1)\sqrt{z}}$.

The fact is that $z=0$ is a branch point of $\frac{1}{\sqrt{z}}$ and I don't know how to manage it.

Somebody can help me?

Thanks

Answer 1

I assume that the integral over a "small circle" is a circle of radius $r$. To evaluate the integral properly, you need to avoid the branch point at the origin. In this case, it is common to create a branch cut along the negative real axis. Along these lines, consider the contour integral

$$\oint_C \frac{dz}{\sqrt{z} (z-1)} $$

where $C$ is a keyhole contour of radius $0 \lt r \lt 1$ about the negative real axis, with an inner radius of $\epsilon$. Thus, the contour integral is

$$\int_{\text{circle}} \frac{dz}{\sqrt{z} (z-1)} + e^{i \pi} \int_r^{\epsilon} \frac{dx}{i \sqrt{x} (-x-1)}\\+ i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac1{\sqrt{\epsilon}\, e^{i \phi/2} (\epsilon \, e^{i \phi}-1)}+ e^{-i \pi} \int_{\epsilon}^r \frac{dx}{-i \sqrt{x} (-x-1)}$$

As $\epsilon \to 0$, the third integral vanishes. Also, by Cauchy's theorem, the contour integral is zero as there are no poles in the integrand. Thus, we have for the integral about the circle,

$$\int_{\text{circle}} \frac{dz}{\sqrt{z} (z-1)} = -2 i \int_0^r \frac{dx}{\sqrt{x} (1+x)} = -4 i \int_0^{\sqrt{r}} \frac{dy}{1+y^2} = -4 i \arctan{\sqrt{r}}$$

Answer 2

Taking the branch cut along the positive real axis and computing the integral along the counterclockwise circular path, we get $$ \begin{align} \oint\frac{\mathrm{d}z}{\sqrt{z}(z-1)} &=\int_0^{2\pi}\frac{ire^{i\theta}\,\mathrm{d}\theta}{\sqrt{r}e^{i\theta/2}(re^{i\theta}-1)}\tag{1}\\ &=-i\sqrt{r}\int_0^{2\pi}\frac{e^{i\theta/2}\,\mathrm{d}\theta}{1-re^{i\theta}}\tag{2}\\ &=-2i\sqrt{r}\int_0^\pi\frac{e^{i\phi}\,\mathrm{d}\phi}{1-re^{i2\phi}}\tag{3}\\ &=-2i\sqrt{r}\sum_{k=0}^\infty\int_0^\pi r^ke^{i(2k+1)\phi}\,\mathrm{d}\phi\tag{4}\\ &=-2i\sqrt{r}\sum_{k=0}^\infty r^k\frac{2i}{2k+1}\tag{5}\\ &=4\sum_{k=0}^\infty\frac{\sqrt{r}^{2k+1}}{2k+1}\tag{6}\\ &=2\log\left(\frac{1+\sqrt{r}}{1-\sqrt{r}}\right)\tag{7} \end{align} $$ Explanation:
$(1)$: $z=re^{i\theta}$
$(2)$: cancelling and factoring
$(3)$: $\theta=2\phi$
$(4)$: $\frac1{1-x}=\sum\limits_{k=0}^\infty x^k$
$(5)$: $\int_0^\pi e^{i(2k+1)\phi}\,\mathrm{d}\phi=\frac{2i}{2k+1}$
$(6)$: collecting and combining
$(7)$: $\sum\limits_{k=0}^\infty\frac{x^{2k+1}}{2k+1}=\frac12\log\left(\frac{1+x}{1-x}\right)$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ I believe you want to compute $${\mathscr I} \equiv \lim_{\epsilon\ \to\ 0^{+}} \oint_{\verts{z}\ =\ R \atop {\vphantom{\Huge A} \epsilon\ <\ {\rm Arg}\pars{z}\ <\ 2\pi\ -\ \epsilon}} {\dd z \over \root{z}\pars{z - 1}}\,,\qquad\qquad R < 1 $$

The integration is 'closed' along a contour $\ds{\gamma}$ which takes care of the $\ds{\root{z}}$-branch cut and a 'small semi-circle' around the origin ( this one yields a zero contribution when its radius goes to zero ). Obviously, the integral over $\ds{\gamma}$ vanishes out. Then, \begin{align} \color{#66f}{\Large{\mathscr I}}&= -\int_{0}^{R}{x^{-1/2}\pars{\expo{\ic 0}}^{-1/2}\,\dd x \over x - 1} -\int_{R}^{0}{x^{-1/2}\pars{\expo{2\pi\ic}}^{-1/2}\,\dd x \over x - 1} =-2\ \overbrace{\int_{0}^{R}{x^{-1/2}\,\dd x \over x - 1}}^{\ds{x \equiv t^{2}}} \\[3mm]&=4\int_{0}^{\root{R}}{\dd t \over 1 - t^{2}} =2\int_{0}^{\root{R}}\pars{{1 \over 1 - t} + {1 \over 1 + t}} =\left.2\ln\pars{1 + t \over 1 - t}\right\vert_{0}^{\root{R}} \\[3mm]&=\color{#66f}{\large 2\ln\pars{1 + \root{R} \over 1 - \root{R}}} \end{align}