$\int \frac{\cos x}{x} dx$ using contour integration

I saw a solution of how to use complex analysis to integrate $$\int_{-\infty}^{\infty} \frac{\cos x}{x} dx$$ which uses the contour that traverses an upper semi circle of radius $$R$$ in the positive direction followed by the line segment from $$-R$$ to $$-\epsilon$$ then the upper semi circle of radius $$\epsilon$$ in the negative direction followed by the line segment from $$\epsilon$$ to $$R$$. Everything makes sense to me except computing $$\lim_{\epsilon \rightarrow 0} \int_{C_\epsilon} \frac{e^{iz}}{z}dz$$. Everywhere I have looked, the limit is computed by taking the limit inside the integral. Why can this be done? I think it has to do with dominated convergence but what is the dominating integrable function?

• Can you provide the solution you saw? Or post a link to it? – Nicholas Roberts Dec 27 '20 at 17:21
• Here's one place I saw it: web.williams.edu/Mathematics/sjmiller/public_html/372Fa15/…. The example on page 4. – BR_math Dec 27 '20 at 17:25
• I think perhaps the dominating integrable function is the constant function $1$? This works because we are on a finite measure space: the interval $[-\pi,0]$. – Nicholas Roberts Dec 27 '20 at 17:32
• I just want to make sure I understand why 1 is the dominating function. If we let $f_r(\theta) = e^{ire^{i\theta}}$ then $f_r(\theta) = e^{ir(\text{cos}\theta +i\text{sin}\theta)}$ so $|f_r(\theta)| = |e^{-r\text{sin}\theta}| < e^r <1$ for small $r$. Does that sound right? I guess from this logic I could also use the constant function $e$ as a dominating function. – BR_math Dec 27 '20 at 17:52
• @NicholasRoberts, yes, please post as a complete response so I can accept it as an answer. – BR_math Dec 27 '20 at 18:33

To be completely rigorous and apply the dominated convergence theorem, we need to convert the function family $$\{e^{ire^{i\theta}}\}_{r>0}$$ to a sequence $$\{e^{\frac{i}{n}e^{i\theta}}\}_{n=1}^{\infty}$$ via the change of variable $$r=\frac{1}{n}$$. Keeping in mind that we are on the measure space $$[-\pi,0]$$, it is not too hard to see that $$|e^{-\frac{1}{n}\sin\theta}| \leq e^{-\sin\left(\frac{-\pi}{2}\right)} = e$$ on our measure space. So now we are ready to show the crucial inequality which will allow us to apply the dominated convergence theorem:

$$|e^{\frac{i}{n}e^{i\theta}}| = |e^{-\frac{1}{n}\sin\theta}| < e \text{ for all } n \in \mathbb{N} \text{ and } \theta \in [-\pi,0]$$

Thus, we can use the constant function $$g \equiv e$$ as our dominating function. This is integrable because we are on a finite measure space.

This is a shameless adaptation from my own set of notes, but it does help (me) tie a few things together mentioned in the relevant PDF linked below the question.

Consider the principal value integral $$$$\label{principalval}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} \:,$$$$ where by shifting $$x \rightarrow z$$, we assume that $$f\left(z\right)$$ is analytic except for a finite number of poles, and that $$\left|f\right| \rightarrow 0$$ on the upper (or lower) infinite semicircle in the complex plane.

Since the pole $$x_0$$ lies on the real axis, the integration contour cuts directly through $$x_0$$. This is handled by regularization of the denominator, which entails introducing a small factor $$\delta>0$$ as $$I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \lim_{\delta\rightarrow 0} \int_{-\infty}^\infty \frac{\left(x-x_0\right)f\left(x\right) \: dx}{\left(x-x_0\right)^2 + \delta^2} = \lim_{\delta\rightarrow 0} \oint_C \frac{\left(z-x_0\right)f\left(z\right) \: dz}{\left(z-x_0\right)^2 + \delta^2} \:.$$ After a little complex algebra, find $$I = \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} + \lim_{\delta\rightarrow 0} \oint_C i\delta \frac{f\left(z\right) \: dz}{\left(z-x_0-i\delta\right)\left(z-x_0+i\delta\right)} \:,$$ which indicates one simple pole $$z_0 = x_0 + i\delta$$ inside the upper-half plane. The first integral in fact excludes the pole, so $$x_0$$ is skipped in subsequent residue calculations. (Use the $$\delta$$-term as a reminder to skip $$x_0$$.) The second integral is solved by standard residue calculus, i.e., let $$g\left(z\right)=f\left(z\right)/\left(z-x_0+i\delta\right)$$, resulting in $$\pi i f\left(x_0\right)$$.

Pulling the results together, we write $$$$\label{principalvaltwoterms}I^+ = \pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} \:,$$$$ where if we started with $$\delta<0$$ instead, the integration contour would flip to the lower-half plane, resulting in $$I^- = -\pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0-i\delta} \:.$$ In tighter notation (regardless of path or the sign of $$\delta$$), one may write $$$$\label{principalvalcomplex}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \text{P} \oint_C \frac{f\left(z\right) \: dz}{z-x_0} \:,$$$$ reminding us to include $$x_0$$ inside integration contour, but take the residue with a factor of $$1/2$$.

So for instance, to evaluate $$I = \int_{-\infty}^\infty \frac{\sin x}{x} \: dx \:,$$ the answer is almost too trivial to show off the method, but here it is:

$$I = \text{Im} \left( \pi i \: e^{i \cdot 0} + \lim_{\delta\rightarrow 0} \oint_C \frac{e^{iz} \: dz}{z+i\delta} \right) = \text{Im} \left( \pi i \: e^{i \cdot 0} + 0 \right) = \pi$$

By the same apparatus, we can show the cosine-version to be zero.