$PV \int_{ia-\infty}^{ia+\infty}\frac{e^{ikt}}{\sqrt{t^2-b^2}}dt$

Question

How would I calculate

\begin{align} PV \int_{ia-\infty}^{ia+\infty}\frac{e^{ikt}}{\sqrt{t^2-b^2}}dt \end{align}

The square root should be dealt with as is most appropriate, e.g. by taking a branch cut from $t=-b$ to $t=+b$.

Answer 1

You do not need the $PV$. Assume $k, b \gt 0$ and $a \lt 0$. Consider

$$\oint_C dz \frac{e^{i k z}}{\sqrt{z^2-b^2}} $$

where $C$ consists of a number of segments:

$$C_1 = \{ z = i a + t | t \in [-R,R] \} $$ $$C_2 = \{ z = R e^{i \theta} | \theta \in [0,\pi] \}$$ $$C_3 = \{ z = -b+\epsilon e^{i \phi} | \phi \in (0, 2 \pi) \} $$ $$C_4 = \{ z = x + i \delta | x \in [-b+\epsilon,b-\epsilon] \} $$ $$C_5 = \{ z = b+\epsilon e^{i \phi} | \phi \in (-\pi, \pi) \} $$ $$C_6 = \{ z = x - i \delta | x \in [-b+\epsilon,b-\epsilon] \} $$

where $\delta \gt 0$ depends on $\epsilon$ and goes to zero with $\epsilon$.

We are interested in the limits as $R \to \infty$ and $\epsilon \to 0$. In these limits, the integral vanishes over $C_2$, $C_4$, and $C_6$. The result is, for the contour integral,

$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} - i \int_b^{-b} dx \frac{e^{i k x}}{\sqrt{b^2-x^2}} + i \int_{-b}^b dx \frac{e^{i k x}}{\sqrt{b^2-x^2}}$$

A little explanation. On $C_4$, $\arg{(x+b)} = 0$ and $\arg{(x-b)} = \pi$. However, on $C_6$, $\arg{(x+b)} = 0$ and $\arg{(x-b)} = -\pi$. Thus, because the contour integral is zero by Cauchy's theorem,

$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} = -i 2 \int_{-b}^b dx \frac{e^{i k x}}{\sqrt{b^2-x^2}} = -i 2 \pi J_0(b k)$$

The above assumed that $k \gt 0$. When $k \lt 0$, the integral on the left is zero because there are no poles or branch points of the integrand in the region $\operatorname{Im}{z} \lt a$.

When $a \gt 0$, reverse the sign of $k$ in the above results. Putting this altogether, we finally have

$$\int_{i a -\infty}^{i a + \infty} dt \frac{e^{i k t}}{\sqrt{t^2-b^2}} = i 2 \pi \operatorname{sgn}{(a)} J_0(b k) H\left [-k \operatorname{sgn}{(a)} \right ]$$

where $H$ is the Heaviside step function.

N.B. Proof of the Bessel integral may be found here. Also, I realize that $C$ as defined is not exactly closed. The pieces that would connect $C_1$ and $C_2$ vanish as $R \to \infty$, so I did not bother to include them. I will expound on this in more detail in a future post, as I have yet to see a clear demonstration of this simple, yet essential fact, in the textbooks.