2
$\begingroup$

$$\int_{0}^\infty \frac{x^{\frac{1}{2}}}{x^2+1} \ dx$$
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 $R\rightarrow \infty$.
Similarly,
$$\left|\int_{C_r} \frac{z^{\frac{1}{2}}}{z^2+1}\right| \rightarrow 0$$ 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.

$\endgroup$
2
  • $\begingroup$ 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
2
$\begingroup$

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 $w=i\frac\pi2$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.