# Why do keyhole contours work?

When evaluating integrals on $$(0,\infty)$$, we often extend the integral into the complex plane by using a "keyhole contour" with inner radius $$\epsilon$$ and outer radius $$R$$:

Credit for the image goes to Linda J. Cummings.

My question is - why do we do this? When we take $$R\to\infty$$ and $$\epsilon\to 0$$, the contribution from the circular arcs typically vanish, but what we have remaining is two integrals, i.e $$\int\limits_0^\infty$$ and $$\int\limits_\infty^0$$. But these will just cancel out right? When we integrate over the same line in the opposite direction, the result is negated, is it not?

I have not seen anyone rigorously justify this. More worrying is a section on Wikipedia where they conclude $$\int\limits_R^\epsilon \frac{\sqrt{z}}{z^2+6z+8}\mathrm{d}z=\int\limits_R^\epsilon \frac{-\sqrt{z}}{z^2+6z+8}\mathrm{d}z=\int\limits_\epsilon^R \frac{\sqrt{z}}{z^2+6z+8}\mathrm{d}z$$ But from this surely the integral is trivially zero?!

Why is this not a completely pointless endeavor? Does it have to do with the fact we are taking a limit as the upper bound of the integral goes $$\to\infty$$? Or does domain reversal work differently in the complex plane? I'd appreciate if someone could explain this.

• No, the integrals do NOT cancel out, precisely because the integrand gets modified by a certain factor $e^{2\pi i \alpha}$ (if you have a $z^{\alpha}$ like term). May 9, 2021 at 2:07
• It's been a long while since I've done complex analysis, so I won't write an answer, but I think this contour is rarely used for integrating things on $(0,\infty)$ precisely for the reason you note - it cancels! The Wikipedia example you cite is tricky - it's considering $\sqrt{z}$ to be defined everywhere except the positive real line - with one contour above this discontinuity and the other below - so they have entirely different values, so no arguments about cancellation apply. May 9, 2021 at 2:27
• @MiloBrandt are there any nice contours that we can use to evaluate integrals on $(0,\infty)$ ? May 9, 2021 at 2:52
• The answer is that domain reversal works differently in the complex plane and they are abusing notation in the wikipedia page. A good contour for finding a real integral over $(0,\infty)$ is to take a semicircle centered at $r+0i$ with radius $r$. You can decompose this into the integral along $(0,2r)$ and the integral along the arc of the semicircle and then try argue the arc part vanishes as $r\to\infty$. May 9, 2021 at 2:59
• I think it would help if you could see the plot of the complex function in 3D; the complex function is multi-valued - which is why there's talk about branches; the two parts of the contour that run parallel to the x axis are not on the same branch, so you're not integrating along the same set of numbers: see here. It's a plot of the Wikipedia example; tap the 3 bars in the lower right corner, click on the "3D view" button, then rotate. May 9, 2021 at 14:18

Let $$R(x)=\frac{P(x)}{Q(x)}$$ be a rational function with no poles on the interval $$[0,\infty)$$, and let $$-1<\alpha<1$$ be any real number. We wish to calculate the real improper integral \begin{align} \int_0^{\infty}x^{\alpha}R(x)\,dx. \end{align} For this improper integral to converge, note that since $$R(x)$$ has no poles on the non-negative real axis, we only have to worry about the appropriate decay conditions at $$\infty$$. A necessary and sufficient condition for this to occur is that \begin{align} \deg (Q)-\deg (P)-\alpha >1. \end{align} Now, to use the method of residues, we should declare the branch of $$z^{\alpha}$$ we are using. So, consider the simply connected open set $$U=\Bbb{C}\setminus [0,\infty)$$, and here we choose the branch of the logarithm such that $$0<\arg(z)<2\pi$$, and we define $$z^{\alpha}:=e^{\alpha \log(z)}$$, and we consider the meromorphic function $$f(z):= z^{\alpha}R(z)$$ defined in $$U$$. To apply the residue theorem, we have to take a path $$\gamma$$ lying completely inside the open set $$U$$. An explicit description of a keyhole contour is as the union of 4 pieces: for large $$r>0$$ and small $$\epsilon>0$$, we define

\begin{align} \begin{cases} \sigma_1(x):= x +i\epsilon& \text{for x\in [0,r]}\\\\ \gamma_1(t):= \sqrt{r^2+\epsilon^2}\,\,e^{it} & \text{for t\in [\arctan(\epsilon/r), 2\pi - \arctan(\epsilon/r)]}\\\\ \sigma_{2}(x):= x-i\epsilon&\text{for 0\leq x \leq r}\\\\ \gamma_{2}(t):= \epsilon e^{it}& \text{for \frac{\pi}{2}\leq t \leq \frac{3\pi}{2}} \end{cases} \end{align} And we define $$\gamma_{r,\epsilon}:=\sigma_1 + \gamma_1 - \sigma_2 - \gamma_2$$, where the signs indicate the orientations of the curves. Here, $$\sigma_1$$ and $$\sigma_2$$ are straight line segments just above and below the non-negative real axis, and $$\gamma_2$$ is a small semicircle of radius $$\epsilon$$. $$\gamma_1$$ is then a large circular arc, which "closes up" the curve; this is why it has a funny radius $$\sqrt{r^2+\epsilon^2}$$ and the parametrization is from $$\arctan(\epsilon/r)$$ to $$2\pi - \arctan(\epsilon/r)$$ (be sure to draw a picture).

Now, I went through all this effort to give a very explicit description of the contour $$\gamma_{r,\epsilon}$$ so that first of all, we can legitimately apply the residue theorem, and second so that it is clear what exactly happens when we take the limits $$\epsilon\to 0^+$$ and $$r\to \infty$$. If $$\epsilon>0$$ is small enough and $$r>0$$ is large enough then all the poles of $$f(z)=z^{\alpha}R(z)$$ will lie "inside" the contour, so \begin{align} 2\pi i \sum \text{Res}(f(z), \text{poles}) &=\int_{\gamma_{r,\epsilon}}f(z)\,dz\\ &=\int_{\sigma_1}f(z)\,dz - \int_{\sigma_2}f(z)\,dz + \int_{\gamma_1}f(z)\,dz- \int_{\gamma_2}f(z)\,dz\\ &=\int_0^r(x+i\epsilon)^{\alpha}R(x+i\epsilon)\,dx- \int_0^r(x-i\epsilon)^{\alpha}R(x-i\epsilon)\,dx\\ &+\int_{\gamma_1}f(z)\,dz-\int_{\gamma_2}f(z)\,dz \end{align} I hope you're convinced that as $$\epsilon\to 0^+$$ and $$r\to \infty$$, the intergals over the circular arcs $$\gamma_1,\gamma_2$$ vanish. So, let us focus on the other two terms.

First, lets focus our attention on $$\int_0^r(x+i\epsilon)^{\alpha}R(x+i\epsilon)\,dx$$. What happens to the integrand as $$\epsilon\to 0^+$$? Well it approaches $$x^{\alpha}R(x)$$, but let's not be too quick and really justify why: \begin{align} (x+i\epsilon)^{\alpha}R(x+i\epsilon)&=e^{\alpha \cdot \log(x+i\epsilon)}R(x+i\epsilon)\\ &=e^{\alpha \cdot [\ln |x+i\epsilon|+ i \arg(x+i\epsilon)]}R(x+i\epsilon)\\ &=e^{\alpha\cdot [\ln|x+i\epsilon| + i\arctan(\epsilon/x)]}R(x+i\epsilon), \end{align} where in the last line, the argument is as such because by definition it has to lie between $$0$$ and $$2\pi$$ (and since $$\epsilon>0$$, $$x>0$$). Now, we can use continuity of $$\exp$$, $$\ln$$, of $$|\cdot|$$ and of $$\arctan$$ to see that as $$\epsilon\to 0^+$$, the RHS approaches $$e^{\alpha\cdot [\ln|x|+0]}R(x)=e^{\alpha \ln x}R(x)=:x^{\alpha}R(x)$$, since $$x>0$$ (we also used the fact that $$R$$ is a rational function with no poles on the non-negative real axis hence it is continuous there).

For the second integral $$\int_0^r(x-i\epsilon)^{\alpha}R(x-i\epsilon)$$, we need to be slightly careful: \begin{align} (x-i\epsilon)^{\alpha}R(x-i\epsilon)&=e^{\alpha\log(x-i\epsilon)}R(x-i\epsilon)\\ &=e^{\alpha\cdot [\ln|x-i\epsilon|+i(2\pi - \arctan(x/\epsilon))]}R(x-i\epsilon) \end{align} As $$\epsilon\to 0^+$$, we can use continuity of $$\exp,\ln,|\cdot|,\arctan$$ and of $$R$$ on the non-negative real axis to deduce that the RHS approaches $$e^{\alpha \cdot[\ln x + 2\pi i \alpha]}R(x)=e^{2\pi i \alpha}\cdot x^{\alpha}R(x)$$.

So, you see it is this extra factor of $$e^{2\pi i \alpha}$$ which is what prevents the cancellation (unless $$\alpha=0$$, in which case our original integral is just $$\int_0^{\infty}R(x)\,dx$$, and this intergal cannot be evaluated by this method). Now, we can for instance appeal to dominated convergence to justify the interchange of limits and integrals to finally deduce that \begin{align} 2\pi i \sum\text{Res}(z^{\alpha}R(z),\text{poles}) &=\int_0^{\infty}x^{\alpha}R(x)\,dx - \int_0^{\infty}e^{2\pi i \alpha}x^{\alpha}R(x)\,dx\\ &=(1-e^{2\pi i \alpha})\int_0^{\infty}x^{\alpha}R(x)\,dx \end{align} Or equivalently, if $$\alpha\neq 0$$, we can divide both sides to get \begin{align} \int_0^{\infty}x^{\alpha}R(x)\,dx &=\frac{2\pi i}{1-e^{2\pi i \alpha}}\sum\text{Res}(z^{\alpha}R(z)\,\text{poles}) \end{align} Also, at this stage one should be mindful when calculating residues that $$z^{\alpha}$$ has been defined such that $$0<\arg(z)<2\pi$$.

To summarize: the integrals do NOT cancel (unless $$\alpha=0$$) because although we have an integral $$\int_0^{\infty}+\int_{\infty}^0=\int_0^{\infty}-\int_0^{\infty}$$ in the end, the integrands are DIFFERENT (unless $$\alpha=0$$).

Also, contour integrals work the same in the complex plane as they do in $$\Bbb{R}^3$$ or in $$\Bbb{R}^n$$. In every case, integrating over a path $$\gamma$$ and then adding the integral over the oppositely oriented path $$-\gamma$$ will of course yield $$0$$, provided you're integrating the SAME object (differential form technically). Otherwise, your argument is equivalent to saying something along the lines of $$\int_0^1(1)\,dx + \int_1^0(-1)\,dx = 0$$ simply because we're integrating in opposite directions. This is blatantly false; yes we're integrating in opposite directions, but we're integrating different things in different directions.

What made the above computation work out for us is that we took a one-sided limit $$\epsilon\to 0^+$$, so it MATTERS whether we have $$(x+i\epsilon)^{\alpha}$$ or $$(x-i\epsilon)^{\alpha}$$; afterall, that's the whole point of branches of the argument and logarithm: they cannot be continuously extended to $$\Bbb{C}\setminus\{0\}$$, so they will of course in general yield different limits (in our case we picked up an extra factor of $$e^{2\pi i \alpha}$$) along different paths.

The wikipedia page on the other hand is sloppily written: in their first line $$\int_R^{\epsilon}\frac{\sqrt{z}}{z^2+6z+8}\,dz$$ is very poor notation. As written this means an integral along the real axis from $$R$$ to $$\epsilon$$; it doesn't matter that they use the letter $$z$$, no matter what letter they use this means a real integral and this will have to equal $$-\int_{\epsilon}^R\frac{\sqrt{z}}{z^2+6z+8}\,dz$$. But this is not what they meant. They're trying to justify why integrating over the paths $$\sigma_1$$ and $$\sigma_2$$ (which is different from integrating on the real axis) yields a different factor (in their case $$\alpha=\frac{1}{2}$$, so $$e^{2\pi i \alpha}=e^{i\pi}=-1$$, and this is the $$-1$$ you're asking about).

• you've done it again. Yet another beautifully written and very enlightening answer. Thank you for the crisp and detailed explanation. (I hate no nitpick, about halfway through your answer you wrote $R(x+i\alpha)$ when I think you meant to write $R(x+i\epsilon)$. ) May 9, 2021 at 13:05
• I'm glad you found this useful, and thanks for pointing out the typo :) May 9, 2021 at 13:32

Imagine that in the wild we found a holomorphic function $$f$$ defined in an open neighborhood of the black contour in picture $$A$$, but not defined at either of the red points, where there are poles of $$f$$, and we would like to compute the integral of $$f$$ around the black contour in picture $$A$$.

To evaluate the integral of $$f$$ around the black contour, draw neat centered circles around the red points, following the keyhole contour. (Why centered circles? Because when we began studying complex analysis, line integrals around neat centered circles were amongst the first of the contours we learned how to compute directly.)

In the limit as the "corridors" connecting the circles to the surrounding contour narrow, the integrals over the segments of the keyhole contour that connect the black contour to the blue circles in picture $$A$$ will cancel each other because of their opposite orientations, as you noted, leaving us with the statement that $$\int_{\text{black contour}}f(z)\,dz + \int_{\color{blue}{\text{blue circles}}} f(z)\,dz = 0,$$ summarized in picture $$B$$. Moving the integral over the blue circles to the other side of the equation and using the fact that the orientation change switches a minus sign to plus sign, we find that $$\int_{\text{black contour}}f(z)\,dz = \int_{\text{black circles}} f(z)\,dz,$$ which is depicted in picture $$C$$. Because we know how to compute the right-hand side, we have proven the

Residue Theorem. Suppose that $$f$$ is holomorphic in an open set containing a nice enough contour $$\Gamma$$ and its interior, except for a pole at $$z_0$$ in the interior of $$\Gamma$$. Then $$\int_\Gamma f(z)\,dz = 2\pi i\,\mathrm{res}_{z_0}f.$$

To summarize, we can turn hopeless integrals over rather arbitrary and hard-to-understand contours $$\Gamma$$ into hopefully more tractable integrals over neat circles with the keyhole family of contours.