I want to calculate the inverse Laplace transform of $$ F(s) = \frac{1}{\sqrt{s^2+a^2}} $$ which results in the Bessel function of first kind of order $0$. I will use the Bromwich integral $$ f(t) = \frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty} e^{st} F(s)\,ds $$
I found this thread, where the inverse Laplace transform of $$ F(s) = \frac{1}{\sqrt{s^2-a^2}} $$ is calculated but I have problems to apply the same argument as there. Here, the branch points are on the imaginary axis and I want to use the following contour.
The integrals $c_1$, $c_3$, $c_5$, $c_7$, $c_9$ vanish for the large radius going to $\infty$ and the small circles going to $0$, so the integrals $c_2$, $c_4$, $c_6$, $c_8$ are left. Due to the branch cuts, I get a sign difference between the left and right side of the imaginary axis. I get the following expressions for the integrals $$ \int_{c_2} ds \frac{e^{st}}{\sqrt{s^2+a^2}} = \int_{\infty}^{a} dr \frac{i e^{irt}}{\sqrt{a^2-r^2}}\\ \int_{c_4} ds \frac{e^{st}}{\sqrt{s^2+a^2}} = \int_{a}^{\infty} dr \frac{-i e^{irt}}{\sqrt{a^2-r^2}}\\ \int_{c_6} ds \frac{e^{st}}{\sqrt{s^2+a^2}} = \int_{\infty}^{a} dr \frac{i e^{-irt}}{\sqrt{a^2-r^2}}\\ \int_{c_8} ds \frac{e^{st}}{\sqrt{s^2+a^2}} = \int_{a}^{\infty} dr \frac{-i e^{-irt}}{\sqrt{a^2-r^2}} $$ Summing up these integrals and multipling with the prefactor from the Laplace transform it follows $$ f(t) = -\frac{1}{2\pi i} \int_{a}^{\infty} \frac{-2i(e^{irt} + e^{-irt})}{\sqrt{a^2-r^2}}\,dr = \frac{2}{\pi} \int_{a}^{\infty} \frac{\cos(rt)}{\sqrt{a^2-r^2}}\, dr $$ A substitution of $r = a\sin(\varphi)$ would change the integrand into the correct result, as the integral representation of the Bessel function looks like $$ J_0(t) = \frac{1}{\pi}\int_{0}^{\pi} \cos(at\sin(\varphi)) d\varphi $$
However I do not known how to deal with the upper border of $\infty$. In the case of the other thread, the integration limits were between $-a$ and $a$. In addition, I have a factor of 2 too much. What am I doing wrong?
