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While evaluating the integral $\int_{0}^{\infty}\frac{\sqrt{z}}{1+z^2}$ we use branch cut.Such a shame that I couldn't draw the complex plane with diagram, but I hope you would understand my explanation to the problem. The two horizontal parallel lines above branch cut would result the same integral which we can add, the bigger circle (contour) would vanish because of jordan's lemma if we take the limit as radius tends to infinity, but why the smaller circle with centre at branch point 0 would vanish as its radius tends to 0.

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  • $\begingroup$ By the estimation lemma. $\endgroup$ – dustin Nov 17 '14 at 2:11
  • $\begingroup$ But, how would I prove it for this integral. $\endgroup$ – Roshan Shrestha Nov 17 '14 at 2:12
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    $\begingroup$ $\left| \int_\Gamma f(z)\; dz\right| \le M L$ where $|f(z)| \le M$ on $\Gamma$ and $\Gamma$ has length $L$. $\endgroup$ – Robert Israel Nov 17 '14 at 2:19
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For the smaller circle about the origin, set $z=\epsilon \, e^{i \phi}$. Then the integral about the smaller circle is

$$i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\sqrt{\epsilon} \, e^{i \phi/2}}{1+\epsilon^2 e^{i 2 \phi}}$$

As $\epsilon \to 0$, this integral goes to zero as $\frac{4}{3} \epsilon^{3/2}$.

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  • $\begingroup$ Now, that's the explanation I needed, Thanks....... $\endgroup$ – Roshan Shrestha Nov 17 '14 at 2:31
  • $\begingroup$ Shouldn't the integral be through an incomplete circle? $\endgroup$ – Pedro Tamaroff Nov 17 '14 at 2:37
  • $\begingroup$ @PedroTamaroff: Technically, yes, but the correction gained by introducing such a complication would result in a smaller order error. $\endgroup$ – Ron Gordon Nov 17 '14 at 2:38
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Estimation Lemma:

Let $M=\sup_{z\in\gamma}|f(z)|$ and let $L = \int|\gamma'(t)|dt$ be the length of the curve $\gamma$. Then $$ \biggl\lvert\int_{\gamma} f(z)dz\biggr\rvert \leq ML $$

By definition $$ \biggl\lvert\int_{\gamma} f(z)dz\biggr\rvert = \biggl\lvert\int_{\gamma} f(\gamma(t))\gamma'(t)dt\biggr\rvert\leq\int\lvert f(\gamma(t))\rvert\lvert\gamma'(t)\rvert dt\leq\int M\lvert\gamma'(t)\rvert dt = ML $$

As $\epsilon\to 0$, what does the length go to?

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  • $\begingroup$ Sorry, but I am not a maths student, so I don't think I will be able to understand the topology statements, so it would be great if there are any other simple explanation for it with regard to this problem. $\endgroup$ – Roshan Shrestha Nov 17 '14 at 2:25
  • $\begingroup$ @RoshanShrestha the sup is the supremum which is the least upper bound. Let's call it M. Next let's let the length of the small circle be L. We have that the integral is less than or equal to ML where L is the length. When $\epsilon\to 0$, the length goes to zero. So the integral goes to what? $\endgroup$ – dustin Nov 17 '14 at 2:27
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    $\begingroup$ @RoshanShrestha if you aren't a math student, then professor should be okay with you saying integral goes to zero when epsilon goes to zero by the estimation lemma since he is a professor of physics, engineering etc. $\endgroup$ – dustin Nov 17 '14 at 2:30
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    $\begingroup$ Thanks @dustin and everyone others, since I am a student of physics $\endgroup$ – Roshan Shrestha Nov 17 '14 at 2:35

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