i have an integral to solve: $\int_{-\infty}^\infty v^2 \frac{e^{-i\omega\tau}}{\omega^2-k^2v^2}d\omega$
I got solution $\frac{\pi v}{k} \sin(kv\tau)$, but in a book the solution was $\frac{2\pi v}{k} \sin (kv\tau)$, why i got answer only half of the true answer? Here is my work
and this is what i got from some books that the answer was twice from my answer:This
did i make a mistake?
There is a variant of the Residue Theorem that states if a singularity is actually on the contour instead of inside it, it's residue is counted at half value (See, for example, George Arfken's "Mathematical Methods for Physicists", Section 7.2). Setting $$f(z) = v^2\dfrac{e^{-iz\tau}}{z^2-k^2v^2}$$, the residues are $$R(f,kv) = \lim_{z\to kv}(z - kv)f(z) = \lim_{z\to kv}\dfrac{e^{-iz\tau}}{z+kv} = \dfrac{e^{-ikv\tau}}{2kv}$$ and similarly $$R(f,-kv) = -\dfrac{e^{ikv\tau}}{2kv}$$
So by the Residue theorem and noting that the integral around the top arc $\to 0$ as the radius $\to \infty$, the Residue theorem says
$$\begin{align}\int_{-\infty}^\infty v^2f(z)dz &= v^2\pi i(R(f,-kv)+R(f,kv))\\ &=v^2\pi i\left(\dfrac{e^{-ikv\tau}}{2kv}-\dfrac{e^{ikv\tau}}{2kv}\right)\\ &=\frac{\pi v}k\sin{kv\tau}\end{align}$$
So the answer is: you are not wrong, the book is. That quote you included from another book doesn't even make sense. It says $C$ is "chosen as a closed path" to "avoid those poles". But it is claiming an integral on the real line which must pass through those poles. I'm guessing your books are copying from a single source who made the error and the copiers, already being aware of the basic concepts, never bothered to double check it themselves (or in the case of the quote, gave a vague hand-wavy explanation without bothering to think much about it).