I'm studying the Cauchy Integral Theorem / Formula, but realised I have a misunderstanding.

Consider an integral over the function $f: \mathbb{R} \to \mathbb{C}$ $$ I = \int^\infty_{-\infty} f(x) \, dx = \int^\infty_{-\infty} \frac{e^{ix}}{x^2 + 1} \, dx \quad. $$

This can be considered in the complex plane, where $f(z)$ is holomorphic except at its two poles at $z = \pm i$. Choosing to consider the positive case, we therefore factorise as

$$ I = \int_C \frac{\frac{e^{iz}}{z+i}}{z-i} \, dz = 2\pi i \frac{e^{-1}}{2i} =\pi e^{-1} \quad , $$

where the contour $C = C_R + C_+$ is taken anticlockwise, $C_R$ is the real axis between $\pm R$ and $C_+$ is the positive semicircle in the complex plane with $\left|z\right| = R$. It's clear from inspection that $C_+$ does not contribute to the path integral if we let $R \to \infty$, so we can equate the integral in the complex plane to the real integral $I$.

Now my question: I could equally well have chosen the negative half of the complex plane ($C'=C_R + C_-$) to evaluate my integral, negating the answer since the anticlockwise integral would otherwise calculate $\int_{\infty}^{-\infty}$. This would give

$$ I = - \int_{C'} \frac{\frac{e^{iz}}{z-i}}{z+i} \, dz = - 2\pi i \frac{e^{+1}}{-2i} =\pi e^{+1} \quad . $$

But clearly, that's different to my previous calculation. I thought these two methods should give the same answer, so what went wrong?


No, you cannot choose the the low half-plane of $\mathbb C$. If $z=x+yi$ with $y\leqslant0$, then$$\lvert e^{iz}\rvert=e^{\operatorname{Re}(-y+xi)}=e^{-\operatorname{Re}(y)}\geqslant1$$and, in fact, as $y$ goes to $-\infty$, $\lvert e^{iz}\rvert$ goes to $+\infty$. So, the integral along the negative semicircle most definitely will contribute to the path integral. You will not have this problem if you work with the highest half-plane.

  • $\begingroup$ Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks! $\endgroup$ – CharlieB Dec 28 '18 at 12:03

If you choose the contour in the upper half-plane, the integral over the semicircle tends to zero as $R\to\infty$. This is because $|e^{iz}|\le1$ on the upper half-plane. The bound $\pi R/(R^2-1)$ easily follows for the absolute value of the integral, when $R>1$.

But on the lower half-plane, the integral over the semicircle does not tend to zero as $R\to\infty$.

  • 1
    $\begingroup$ In hindsight, I should have suspected the part of my argument that was "clear from inspection"... $\endgroup$ – CharlieB Dec 28 '18 at 12:04

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