Difficulty finding the principal value of $\int_{-\infty}^{\infty}\frac{\cos3x}{x-1}\,\mathrm{d}x$

Question

I am having problems in evaluating the following Cauchy principal value

$$\int_{-\infty}^{+\infty} \frac{\cos(3x)}{x-1}d x$$

I know it is supposed to show my work, but I got barely nothing. I tried with using $\cos(x) = (z + z^{-1})/2$ (clearly adapting it to $\cos(3x)$, but I think something went wrong because I got $0$ (I have the solution, it shall be $-\pi \sin(3)$.

Also, I would like a "true proof", like with contour integration explained. Sometimes theory comes easier with an example.

Thank you!

Answer 1

I am unable to draw the contour lines in the books-fancy way, so I will avoid it.

Let's proceed by considering the following integral instead:

$$\int \frac{e^{3iz}}{z-1}\ \text{d}z$$

steppint into the complex plane, remembering that $e^{ix} = i\sin(x) + \cos(x)$, integrated along a closed contour (the "typical one", the upper semi-circle, to be clear).

We have to modify it, though, because the above function has a pole at $z = 1$ and this point must be avoided by means of a semicircular indentation of a small radius, say $\epsilon$, in the contour.

Notice that $\frac{e^{3iz}}{z-1}$ is analytic at all points lying on, and interior to, this contour.

Integrating this function around the chosen path and putting $z = x$ where appropriate we have

$$\int_{-R}^{1 - \epsilon} \frac{e^{3ix}}{x-1} \text{d}x + \int_{|z-1| = \epsilon} \frac{e^{3iz}}{z-1} \text{d}z + \int_{1 + \epsilon}^R \frac{e^{3ix}}{x-1}\text{d}x + \int_{|z| = R} \frac{e^{3iz}}{z-1}\text{d}z = 0$$

Allowint $R\to +\infty$ and invoking Jordan's lemma, we can easily argue that the integral around the semicircle of radius $R$ goes to zero.

Taking the limit $\epsilon \to 0$ we can evaluate the integral over the semicircular indentation at $z = 1$.

Now we invoke the following:

THEOREM

Let $f(z)$ have a simple pole at $z_0$. An arc $C_0$ of radius $r$ is constructed using $z_0$ as its center. The arc subtends an angle $\alpha$ at $z_0$. Then

$$\lim_{r\to 0} \int_{C_0} f(z) \text{d}z = 2\pi i\left[\frac{\alpha}{2\pi}\text{Res}[f(z), z_0]\right]$$

where the integration is done in the counterclockwise direction (for a clockwise direction, a factor $-1$ is placed on the right).

---

We use that Theorem with $\epsilon = r$, $f(z) = \frac{e^{3iz}}{z-1}$ and $z_0 = 1$. The angle $\alpha = \pi$ (for a semicircle) and we find the second integral on the left becomes in the limit: $-i\pi e^{3i}$.

The minus sign appears because of the clockwise direction of integration.

With $R\to +\infty$ and $\epsilon \to 0$ we rewrite:

$$\lim_{R\to +\infty} \lim_{\epsilon \to 0} \left[\int_{-R}^{1 - \epsilon} \frac{e^{3ix}}{x-1} \text{d}x + \int_{1 + \epsilon}^R \frac{e^{3ix}}{x-1}\text{d}x \right] + i\pi e^{3i} = 0$$

The sum of the two integral in the brackets becomes, with the limits indicates, the Cauchy principal value of

$$\int_{-\infty}^{+\infty} \frac{e^{3ix}}{x-1}\text{d}x = \int_{-\infty}^{+\infty} \frac{\cos(3x) + i\sin(3x)}{x-1}\ \text{d}x$$

Thus

$$\int_{-\infty}^{+\infty} \frac{\cos(3x)}{x-1}\text{d}x + i\int_{-\infty}^{+\infty}\frac{\sin(3x)}{x-1}\ \text{d}x = i\pi e^{3i}$$

Remembering indeed that

$$i\pi e^{3i} = i\pi (\cos(3) + i\sin(3)]$$

We obtain eventually the real and the imaginary part:

$$\int_{-\infty}^{+\infty} \frac{\cos(3x)}{x-1}\text{d}x = -\pi \sin(3)$$

and

$$\int_{-\infty}^{+\infty}\frac{\sin(3x)}{x-1}\ \text{d}x = \pi \cos(3)$$

as wanted.

Answer 2

An even shorter solution: for any $n\in\mathbb{N}^+$ we have $\int_{-\infty}^{+\infty}\frac{\sin(nx)}{x}\,dx=\pi$ and

$$\begin{eqnarray*} \text{PV}\int_{\mathbb{R}}\frac{\cos(3x)}{x-1}\,dx &\stackrel{x\mapsto x+1}{=}& \text{PV}\int_{\mathbb{R}}\frac{\cos(3x+3)}{x}\,dx\\&=&\cos(3)\cdot\text{PV}\int_{\mathbb{R}}\frac{\cos(3x)}{x}\,dx-\sin(3)\cdot\text{PV}\int_{\mathbb{R}}\frac{\sin(3x)}{x}\,dx\\&=&\color{red}{0 }\,-\color{blue}{\pi\sin 3} \end{eqnarray*}$$ since $\text{PV}\int_{\mathbb{R}}\frac{\cos(3x)}{x}\,dx$ vanishes by the oddness of the integrand function.