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I need help computing an integral. My motivation is to understand a standard way to build the holomorphic functional calculus for unbounded operators, though the actual question could probably be a homework problem in a complex analysis class. Anyway, here goes.

Let $\Omega \subseteq \mathbb{C}$ be the complement of the set $\{iy: y \geq 1, y \leq -1\}$. Let $\Gamma$ be the contour described by the equation $y^2 - x^2 = \frac{1}{4}$, whose "top half" is oriented from right to left and whose "bottom half" is oriented from left to right. (The exact equation of $\Gamma$ shouldn't be important, just the overall shape and orientation). Choose the branch of the holomorphic function $(1 + z^2)^{-1/2}$ on $\Omega$ which takes the value $+1$ at $z = 0$. I want to prove:

$$\frac{1}{(1+z^2)^{1/2}} = \frac{-1}{2 \pi i} \int_\Gamma \frac{1}{(z - w)(1 + w^2)^{1/2}} dw$$

Don't feel obligated to work out all of the details in your answer - I really should be able to do this on my own...

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I don't have the time to work it out, but perhaps you can use Dennis' solution by first closing the contour with vertical lines and showing that their contribution goes to zero when you send them off to infinity. Kind of like this. –  Antonio Vargas Mar 6 '12 at 5:15
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@AntonioVargas: to add the key estimate to make this work, observe that for fixed $z$, we can close the path by the two curves $x = \pm k$, where $k$ is large. Then, we integrate over $y$ between $\pm \sqrt{ x^2 + 1/4}$. Then, $| z - w |$ has approximately linear growth in $k$, $|1+w^2|$ has approx quadratic growth in $k$, so the integrand is on the order of $k^{-2}$. The length of the interval over which you are integrating is of order $k$, so the contribution over the two straight lines we added is about $1/k$, which goes to zero as necessary. –  Sam Lisi Mar 6 '12 at 10:01
    
If you are somewhat familiar with this kind of contour integration trick, I think this should get you started. If not, I am happy to provide a few more details. –  Sam Lisi Mar 6 '12 at 10:02
    
@AntonioVargas: sorry, I was confused when I wrote that comment. The offer for details was intended for Paul, not for you since you clearly don't need them. –  Sam Lisi Mar 6 '12 at 16:29
    
OK, I got this to work. Does somebody want to put this up as an answer? –  Paul Siegel Mar 8 '12 at 19:20

2 Answers 2

up vote 2 down vote accepted
+100

As in the original question, let $\Gamma$ be the contour described by the equation $y^2-x^2=1/4$ whose upper part is oriented right-to-left and whose lower part is oriented left-to-right. We will call the region between these two curves the region interior to $\Gamma$.

Let $k > 0$ and let $C_k = \Gamma_k + \Delta_k^+ + \Delta_k^-$ be the closed contour whose components are defined by

$$\begin{align} \Gamma_k &= \{x+iy \colon y^2-x^2=1/4 \text{ and } |x| \leq k\}, \\ \Delta_k^+ &= \left\{k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\ \Delta_k^- &= \left\{-k+iy \colon |y| \leq \sqrt{k^2+1/4}\right\}, \\ \end{align}$$

which is oriented as in the following picture.

enter image description here

For any $k>0$ the function $f(z) = (1+z^2)^{-1/2}$ is analytic interior to $C_k$, so Cauchy's integral formula tells us that, for $z$ interior to $C_k$,

$$\begin{align} \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \int_{C_k} \frac{dw}{(w-z)(1 + w^2)^{1/2}} \\ &= \frac{1}{2 \pi i} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right), \end{align} \tag{1}$$

where $g(w,z) = (w-z)^{-1} (1+w^2)^{-1/2}$. Note that for any $z$ interior to $\Gamma$, $z$ will be interior to $C_k$ when $k$ is large enough.

We will show that the integral over $\Delta_k^+$ goes to $0$ as $k \to \infty$. The calculation for the integral over $\Delta_k^-$ is identical.

Indeed, we have

$$\begin{align} \left|\int_{\Delta_k^+}g(w,z)\,dw\right| &\leq L(\Delta_k^+) \cdot \sup_{w \in \Delta_k^+} |g(w,z)| \\ &\leq 2 \sqrt{k^2+1/4} \cdot \frac{1}{(k-\text{Re}(z))\sqrt{1+k^2}} \\ &= O\left(\frac{1}{k}\right). \end{align}$$

Since $(1)$ holds for any $z$ interior to $\Gamma$ when $k$ is large enough, we can let $k \to \infty$. Because $\lim_{k \to \infty} C_k = \Gamma$, we get

$$\begin{align} \frac{1}{(1+z^2)^{1/2}} &= \frac{1}{2 \pi i} \cdot \lim_{k \to \infty} \left(\int_{\Gamma_k}g(w,z)\,dw + \int_{\Delta_k^+}g(w,z)\,dw + \int_{\Delta_k^-}g(w,z)\,dw\right) \\ &= \frac{1}{2 \pi i} \int_{\Gamma} \frac{dw}{(w-z)(1 + w^2)^{1/2}} \end{align}$$

for all $z$ interior to $\Gamma$.

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First, show that $f(z)=\frac{1}{(1+z^2)^{1/2}}$ is holomorphic (analytic) in the set $\Omega$. Then apply the Cauchy's integral formula on $f(z)$: $$f(z)=\frac{1}{2\pi i}\int_\Gamma \frac{f(w)}{w-z}dw$$

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The problem is that the usual Cauchy's integral formula only applies to closed curves, and my curve is not. It is natural to try to reparametrize the integral using some conformal map so that $\Gamma$ is a closed curve, but I haven't figured out a particularly convenient way to do this. –  Paul Siegel Mar 2 '12 at 5:04

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