Fourier Type Integral with Branch Cut

207 Views Asked by At

I'm trying to solve the following integral:

\begin{equation} I = \int_{-\infty}^{\infty}dk e^{ikx} \Big[\frac{-\sqrt{1-k^{2}}+1}{2\sqrt{1-k^{2}}} e^{-\sqrt{1-k^{2}}t} -\frac{\sqrt{1-k^{2}}+1}{2\sqrt{1-k^{2}}} e^{\sqrt{1-k^{2}}t}\Big] \end{equation}

where x and t are positive, and I expect the integral to be zero for x>t.

The function has a branch cut, which I chose to be on the real line between 1 and -1, and choose the square root to be

\begin{equation} \sqrt{1-k^{2}} = i\sqrt{k^{2}-1} \end{equation}

I chose a contour which goes along the real line, circles the two branch points with small circumferences, and closes with a big arc that closes in the upper plane or bottom one (depending on x and t).

The integral on the small arcs of radius epsilon goes to zero for epsilon that goes to zero (the integral goes like the square root of epsilon).

\begin{equation} \text{for } z-1=\epsilon e^{i\theta} \\ I_{small arc}\approx \int_{0}^{\theta}\epsilon e^{i\theta} \frac{1}{\sqrt{(1+\epsilon e^{i\theta})^{2}-1}} \approx \int_{0}^{\theta}\epsilon e^{i\theta} \frac{1}{\sqrt{\epsilon}}(1+\frac{1}{4}\epsilon e^{i\theta}) \end{equation}

that goes to zero for vanishing epsilon (I have ignored the exponential terms as they are always of order 1).

Also the integral along the big arc vanishes for R that goes to infinity, depending on where you close the integral. In fact, for k = +ir or k = -ir (depending on if we close in the upper or bottom half f the plane) we get that asymptotically

\begin{equation} \text{for } k = \pm i r \\ e^{ ikx-i \sqrt{k^{2}-1}t} +e^{ ikx + i\sqrt{k^{2}-1}t} \approx e^{\mp r(x- t)} +e^{\mp r (x + t)} \end{equation}

The second term can always be closed in the bottom half of the plane, while for the first one one has to look at x and t. For x>t the first term can be closed in the upper plane and the integral will exponentially decrease for r that goes to infinity. For x< t then the first term can be closed in the bottom one. So essentially I get that because the function has no poles, the integral is zero. Any ideas of what I am getting wrong?