Let $f(z)$ be analytic in the upper half plane $\operatorname{Im} \geq z$, except for a finite number of poles on the real axis.

For certain integrals $I :=\int_{-\infty}^{\infty} f(x) dx$, we can use contour integration, along the real interval $(-R,R)$ as $R\to \infty$ and a semicircle $\Gamma$ joining the ends of this interval. Our functions $f(x)$ are chosen such that the contribution along $\Gamma$ can be shown to vanish as $R\to \infty$; leaving us to use the residue theorem to compute $I$.

If we introduce a simple pole $z_0$ along the real axis, we can treat it as a separate case by excluding it from our initial contour by defining the principal value $P$

$$ P \equiv \lim_{\rho \to 0}\ \int_{-\infty}^{z_0-\rho} f(x)dx\ + \int_{z_0+\rho}^{\infty} f(x)dx \ . $$

So far, so good.

Our contour $\mathcal C$ now consists of $\Gamma$ and $P$, with an open end at the pole. We close $\mathcal C$ by introducing an indentation in the form of a semicircle $\gamma$ of radius $\rho$ around $z_0$ in the upper half plane, where $\rho$ is the same as in $P$.

I'm wondering why the text requires $\gamma$ to be in the upper half-plane.

My guess is that it's because we want to exclude $z_0$ from the interior of $\mathcal C$; so we can use the residue theorem to compute the contribution of $P$ and $\gamma$ to $\oint_{\mathcal C} f(z)dz$.

But, say, instead, we require $\gamma$ to be in the lower half-plane, thereby including $z_0$ in the interior of $\mathcal C$. Accordingly, the residue theorem now includes the contribution of $z_0$ to $I$, without the need to consider the separate case of $\gamma$.

But, since we defined our $f$ to be analytic only in the upper half-plane, the residue theorem doesn't need to apply for my previous line.


The contribution of $\gamma$, $\int_{\gamma}f(z)dz$, can be shown to equal $-ia_{-1}\pi$, as $\rho \to 0$.

TL;DR: I'm wondering why the text requires $\gamma$ to be in the upper half-plane.


1 Answer 1


Your read on the situation seems good. In general you don't have to put the circle in the upper half plane, but it might be more convenient. Your remark about how you might need to do so if the function is not analytic in the lower half plane is correct as well.

To see how things work when there's a choice, let's to compute $$I= P\int_{-\infty}^\infty\frac{e^{ix}}{x}dx.$$

As in your set-up, you can say $$ \int_\mathcal{C}\frac{e^{iz}}{z}dz = I +B + L $$ where $I$ is the contribution from the principal part, $B$ is the contribution from the big circle and $L$ is the little one. In this situation we choose the big circle to be in the upper half plane because $e^{iz}$ decays in that direction so that $B\rightarrow0.$ Note that the integrand is analytic in both half-planes, we just chose the upper because it only goes to zero in that direction.

We can choose the little circle in one of two ways. First, do it in the upper half-plane. Then $L$ is an integral over the semicircle going clockwise around the origin from angle $\pi$ to $0$. Since the integrand looks like $1/z$ near the origin, we have $$L\rightarrow-\pi i$$ as the circle gets small. There are no poles inside the contour so we have $$ 0 =\int_\mathcal{C}\frac{e^{iz}}{z}dz = I-\pi i$$ so that $$I = \pi i.$$

If we instead make the little circle go into the lower half plane (which, again, is legal since the integrand is analytic there), then we now have a counter-clockwise circle from $\pi$ to $2\pi,$ instead of a clockwise one. So we have $$L \rightarrow \pi i.$$

However, now there is a pole inside the contour, with residue $1$, so the integral over the contour is $2\pi i.$ Then we have $$ 2\pi i =\int_\mathcal{C}\frac{e^{iz}}{z}dz = I+\pi i$$ so that, again, $$I = \pi i.$$

It's no accident that the effect of the pole and the circle cancel each other out.


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