$$\int_{0}^\infty \frac{x^{\frac{1}{2}}}{x^2+1} \ dx$$
by taking the branch cut of $x^{\frac{1}{2}}$ along the positive real axis.

I wasn't sure what contour to choose, so I chose this keyhole contour:
enter image description here

Where the upper line segment is $C_1$ and the lower line segment is $C_2$, with the circular arc of radius $R$ being $C_R$ and the smaller circular arc of radius $r$ be $C_r$.

Can someone please show me how to use my contour to evaluate this?
This is my attempt

Note that $$\int_{C} \frac{z^{\frac{1}{2}}}{z^2+1} \ dz = 2\pi i \mathrm{Res}(f,i) + 2\pi i \mathrm{Res}(f,-i)$$ where $C = C_1 + C_2 + C_R + C_r$.
First consider $$\left|\int_{C_R} \frac{z^{\frac{1}{2}}}{z^2+1}\right| \ dz \leq ML$$
where $M = \mathrm{max}\left\{\left|\frac{z^{\frac{1}{2}}}{z^2+1}\right|:z\in C_R\right\} = \frac{R^{\frac{1}{2}}}{R^2 - 1}$
and $L = \mathrm{length}{C_R} = 2\pi R$.
So that expression goes to zero as $R\rightarrow \infty$.
$$\left|\int_{C_r} \frac{z^{\frac{1}{2}}}{z^2+1}\right| \rightarrow 0$$ as $r\rightarrow 0$.

So it seems that all that's left is to parametrize the remaining segments... but I can't seem to have an expression which has upper limits going to infinity.

  • $\begingroup$ You might find \operatorname{Res} looks slightly better \max is a bit easier $\endgroup$ – gen-ℤ ready to perish Nov 1 '17 at 12:48
  • $\begingroup$ Using the substitution $x = t^2$ transforms the integral into $$\int_{-\infty}^{\infty} \frac{t^2}{t^4+1} \, dt$$ Is that easier? $\endgroup$ – md2perpe Nov 1 '17 at 15:52

You have by the definition of your brach cut $$(x+ir)^{\frac12}\approx\sqrt{x}+\frac{ir}{2\sqrt{x}}$$ and $$(x-ir)^{\frac12}\approx-\sqrt{x}+\frac{ir}{2\sqrt{x}}$$ for $x>r$. Thus the integrals over the segments from $r+ir$ to $R+ir$ and from $R-ir$ to $r-ir$ approximate twice the searched for integral value.

For a more boring region substitute $x=e^u$ to get the transformed integral $$ \int_{-\infty}^\infty \frac{e^{\frac32 u}}{1+e^{2u}}du $$ Now integrate along the contour of the box $[-R,R]+i[0,\pi]$. The integration along the upper side has the integrand for $w=u+i\pi$ so that $$ f(w)=\frac{-i\,e^{\frac32 u}}{1+e^{2u}} $$ plus orientation reversal, the left and right sides are of size $e^{-R/2}$, so that the limit $R\to\infty$ of the contour integral is $(1+i)$ times the required value, and the only pole is at $w=i\frac\pi2$.


As an alternative approach, you may just remove the branch point by enforcing the substitution $x=z^2$, leaving you with $$ \int_{-\infty}^{+\infty}\frac{z^2}{z^4+1}\,dz = 2\pi i\sum_{k\in\{1,3\}}\text{Res}\left(\frac{z^2}{z^4+1},z=e^{\pi i k/4}\right)=\color{blue}{\frac{\pi}{\sqrt{2}}}. $$ For a real-analytic approach, the substitution $\frac{1}{x^2+1}=u$ directly leaves you with the value of a Beta function.

  • $\begingroup$ Thank you, is it always valid to 'remove the branch point' using substitutions like this? I'm not sure if it is mathematically valid or not in every case $\endgroup$ – Twenty-six colours Nov 3 '17 at 5:06
  • $\begingroup$ @Twenty-sixcolours: it is just a change of variable in a improper integral over $\mathbb{R}^+$. $\endgroup$ – Jack D'Aurizio Nov 3 '17 at 12:42

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