# Evaluation of a cauchy Singular Integral

While evaluating the details of this paper by Taylor et al. I'm stuck at the following integral. In need to evaluate the following:

Regard $x$ as spatial x-coordinate & $t$ as time.

$I(x) = \int_{0}^{\infty} \{ \frac{cos(kx)-cos(k_0x)}{k^2-k_0^2}\}dk$ where $k_0\in (0,\infty)$ is some given fixed value and $x>0$. By tallying results of the paper & my calculations one can conclude that this must result in $Csin(k_0x)$ where $C$ is some constant.

• From trying a few test cases, it appears that $I(x) = -\dfrac{\pi}{2k_0}\sin(k_0x)$. Commented Jun 21, 2015 at 7:36

Instead of considering principal value, we can give a direct proof. Starting by applying partial fraction decomposition, we have

\begin{align*} I &= \frac{1}{2k_0} \int_{0}^{\infty} \left( \frac{1}{k - k_0} + \frac{1}{(-k) - k_0} \right) (\cos kx - \cos k_0 x) \, dk \\ (1) \cdots \quad &= \frac{1}{2k_0} \int_{-\infty}^{\infty} \frac{\cos kx - \cos k_0 x}{k - k_0} \, dk \\ &= \frac{1}{2k_0} \int_{-\infty}^{\infty} \frac{\cos (k_0 + \xi)x - \cos k_0 x}{\xi} \, d\xi \quad (\xi = k - k_0) \\ (2) \cdots \quad &= -\frac{1}{2k_0} \int_{-\infty}^{\infty} \frac{\cos k_0 x (1 - \cos \xi x) + \sin k_0 x \sin \xi x}{\xi} \, d\xi \\ (3) \cdots \quad &= -\frac{\sin k_0 x}{2k_0} \int_{-\infty}^{\infty} \frac{\sin \xi x}{\xi} \, d\xi \\ (4) \cdots \quad &= -\frac{\pi}{2k_0} \sin k_0 x. \end{align*}

Throughout the calculation, we utilized the following obesrvations:

• In (1), we utilized the substitution $k \mapsto -k$ to obtain $$\int_{0}^{\infty} \frac{\cos kx - \cos k_0 x}{(-k) - k_0} \, dk = \int_{-\infty}^{0} \frac{\cos kx - \cos k_0 x}{k - k_0} \, dk.$$

• In (2), we utilized the addition formula for cosine.

• The step (3) follows from $$\int_{-\infty}^{\infty} \frac{\cos k_0 x (1 - \cos \xi x)}{\xi} \, d\xi = 0,$$ which is true because we are integrating odd function.

• The final step (4) follows from the Dirichlet integral: $$\int_{-\infty}^{\infty} \frac{\sin x k}{k} \, dk = \pi \operatorname{sign}(x).$$

Let $I(x) = I_1(x) -I_2(x)$ where $I_1(x) = \int_{0}^{\infty}\{\frac{cos(kx)}{k^2-k_0^2}\}dk$ and $I_2(x) = \int_{0}^{\infty}\{\frac{cos(k_0x)}{k^2-k_0^2}\}dk$.

Using complex analysis & some algebra

$I_1(x) = \frac{isin(k_0x)}{2k_0}\{\int_{-\infty}^{\infty}\frac{exp(ik)}{k}dk\}$

while

$I_2(x) = \frac{cos(k_0x)}{2}\{\int_{-\infty}^{\infty}\frac{1}{k^2-k_0^2}dk\}$

Expanding the integrand of $I_2$ as an expanded partial fraction, one gets $I_2(x)=0.$

By Cauchy integral around a indented contour & applying Jordan's lemma(See Complex Variables and Applications,Churchill & Brown, 7th ed, Chap 7, Pgs 267-270 )

we find $\int_{-\infty}^{\infty}\frac{exp(ik)}{k}dk = i\pi$.

Thus $I(x)= \frac{-\pi sin(k_0x)}{2k_0}$