Contour Integral of $1/(x^2+1)^{3/2}$

Question

If I want to find the contour integral of $$ \oint_{|x-i|=\rho} \frac{1}{(x^2+1) \sqrt{ x^2+1}} dx $$ where $$ \rho \rightarrow 0 $$ How can I do that? Is there any useful approximation I can used for this integral so that the residue theorem can be applied? I have tried the generalized binomial theorem, but it seems that it dose not work.

Answer 1

Let $x=z+i$, then $$ \begin{align} &\lim_{\rho\to0}\int_{|x-i|=\rho}\frac1{(x^2+1)\sqrt{x^2+1}}\,\mathrm{d}x\\ &=\lim_{\rho\to0}\int_{|z|=\rho}\frac1{(z^2+2iz)^{3/2}}\,\mathrm{d}z\\ &=\lim_{\rho\to0}\int_{|z|=\rho}\frac1{z^{3/2}}\frac1{(z+2i)^{3/2}}\,\mathrm{d}z\\ &=\frac{-1-i}4\lim_{\rho\to0}\int_{|z|=\rho}\frac1{z^{3/2}}\left(1+O(z)\right)\,\mathrm{d}z\\ &=\frac{-1-i}4\int_{|z|=1}\frac1{z^{3/2}}\,\mathrm{d}z\lim_{\rho\to0}\rho^{-1/2}+\lim_{\rho\to0}O\!\left(\rho^{1/2}\right)\int_{|z|=1}\frac1{z^{1/2}}\,\mathrm{d}z\\ \end{align} $$ If we put the branch cut along $\operatorname{Im}(z)=0$ and $\operatorname{Re}(z)\gt0$ $$ \begin{align} \int_{|z|=1}\frac1{z^{3/2}}\,\mathrm{d}z &=\int_0^1e^{-3\pi it}\,\mathrm{d}e^{2\pi it}\\ &=2\pi i\int_0^1e^{-\pi it}\,\mathrm{d}t\\ &=2\left(1-e^{-\pi i}\right)\\[6pt] &=4 \end{align} $$ Thus, $$ \begin{align} \lim_{\rho\to0}\int_{|x-i|=\rho}\frac1{(x^2+1)\sqrt{x^2+1}}\,\mathrm{d}x &=(-1-i)\lim_{\rho\to0}\rho^{-1/2} \end{align} $$ which diverges. This diverges no matter where the branch cut is.

Answer 2

That's what I did, but I am not sure whether it's correct or not. First, I draw a contour with a branch cut as shown below.

Based on the residue theorem, $$ \oint_C\frac{1}{(1+x^2)^{3/2}}dx = \int_{C_R} + \int_{C_\rho} +\int_{L_1} +\int_{L_2} +\int_{L_3} = 0 $$

$C_R$: $ Z=Re^{i\theta} $ $$ \lim_{R \rightarrow \infty} \int_{C_R} \frac{iRe^{i\theta}}{(R^2e^{i2\theta}+1)^{3/2}} d\theta <\lim_{R \rightarrow \infty} \int_{C_R} \frac{R}{\vert (R^2e^{i2\theta}+1)^{3/2} \vert} d\theta =0$$

$L_1$: $$ \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^{3/2}}dx = 2$$

$L_2$: $ Z = Re^{-i3\pi/2}, R:1\sim \infty$ $$ \int_1^\infty \frac{e^{-i3\pi/2}}{(R^2e^{-i3\pi}+1)^{3/2}} dR = \int_1^\infty \frac{i}{(1-R^2)^{3/2}}dR$$

$L_3$: $ Z = Re^{i\pi/2}, R:\infty\sim 1$ $$ \int^1_\infty \frac{e^{i\pi/2}}{(R^2e^{i\pi}+1)^{3/2}} dR = \int^1_\infty \frac{i}{(1-R^2)^{3/2}}dR$$

Then, $$ \int_{L_2}+\int_{L_3} = 0$$ Therefore, $$ \lim_{\rho \rightarrow0}\oint_{|x-i|=\rho}\frac{1}{(1+x^2)^{3/2}}dx = -\int_{-\infty}^{\infty} \frac{1}{(1+x^2)^{3/2}}dx =-2 $$

Is my idea correct? If not, please tell me why please.