Complex Integral final steps

Question

How do we evaluate the Cauchy Principal value for:

$$ \int_{-\infty}^\infty\frac{\cos kx}{x-a}dx $$

Given, a is real, k >${\ 0}$?

I thought of integrating from ${-\infty}$ to ${\ a}$ and then from ${\ a}$ to ${+\infty}$

Any help will be appreciated.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\mbox{P.V.}\int_{-\infty}^{\infty}{\cos\pars{3x} \over x^{2} - 25}\,\dd x = \Re\int_{-\infty}^{\infty} {\cos\pars{3x} \over x^{2} - \pars{25 + \ic 0^{+}}}\,\dd x \\[5mm] = &\ \Re\int_{-\infty}^{\infty}{\cos\pars{3x} \over \pars{x + \root{25 + \ic 0^{+}}}\pars{x - \root{25 + \ic 0^{+}}}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Re\int_{-\infty}^{\infty}{\expo{3\ic x} \over \pars{x + 5 + \ic 0^{+}}\pars{x - 5 - \ic 0^{+}}}\,\dd x + {1 \over 2}\,\Re\int_{-\infty}^{\infty}{\expo{-3\ic x} \over \pars{x + 5 + \ic 0^{+}}\pars{x - 5 - \ic 0^{+}}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Re\pars{2\pi\ic\,{\expo{3\ic\bracks{5 + \ic 0^{+}}} \over 10 + \ic 0^{+}}} + {1 \over 2}\,\Re\pars{-2\pi\ic\,{\expo{-3\ic\bracks{-5 - \ic 0^{+}}} \over -10 - \ic 0^{+}}} = {1 \over 10}\,\pi\,\bracks{\Re\pars{\ic\expo{15\,\ic}} + \Re\pars{\ic\expo{15\,\ic}}} \\[5mm] = &\ \bbx{\ds{-\,{1 \over 5}\,\pi\sin\pars{15}}} \end{align}

Answer 2

First note that the integral of interest $\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx$ diverges. Its Cauchy principal value ("CVP") does exist. We will now present an approach to evaluate the CVP.

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To apply Cauchy's Integral Theorem correctly we have

$$\begin{align} \oint_C\frac{e^{i3z}}{z^2-25}\,dz&=\int_{-R}^{-5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{-5+\epsilon}^{5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{5+\epsilon}^{R} \frac{e^{i3x}}{x^2-25}\,dx\\\\ &+\int_{\pi}^{0} \frac{e^{i3(-5+\epsilon e^{i\phi})}}{(-5+\epsilon e^{i\phi})^2-25}\,i\epsilon e^{i\phi}\,d\phi+\int_{\pi}^{0} \frac{e^{i3(5+\epsilon e^{i\phi})}}{(5+\epsilon e^{i\phi})^2-25}\,i\epsilon e^{i\phi}\,d\phi\\\\ &+\int_0^\pi \frac{e^{i3Re^{i\phi}}}{(Re^{i\phi})^2-25}\,iRe^{i\phi}\,d\phi\tag 1\\\\ &=0 \end{align}$$

As $R\to \infty$, it is easy to see that the last integral on the right-hand side of $(1)$ approaches $0$. And as $\epsilon \to 0$, the fourth and fifth integrals approach $i(\pi/10) e^{-i15}$ and $-i(\pi/10)e^{i15}$, respectively.

Hence, letting $R\to \infty$ an d$\epsilon\to 0$, and taking the real part of $(1)$, we find that

$$\text{PV}\left(\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx\right)=-\frac{\pi \sin(15)}{5}$$

where

$$\text{PV}\left(\int_{-\infty}^\infty \frac{\cos(3x)}{x^2-25}\,dx\right)=\lim_{R\to \infty}\left(\int_{-R}^{-5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{-5+\epsilon}^{5-\epsilon} \frac{e^{i3x}}{x^2-25}\,dx+\int_{5+\epsilon}^{R} \frac{e^{i3x}}{x^2-25}\,dx\right)$$