# Real integrals with two poles in the complex plane

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.

• Ah, I'm so used to thinking that $| e^{ix} | = 1$ that I forgot it's not true for complex $x$. Thanks! – 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 $$R\to\infty$$.

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