# Thread: Inverse Laplace Transform with square root of f(s^2) in the denominator

1. ## Inverse Laplace Transform with square root of f(s^2) in the denominator

Hi,
I want to find the inverse laplace transform of $f(s)=\frac{1}{\sqrt{s^{2}+a^{2}}}$. I know that the inverse is $F(t)=J_0(at)$ from the tables, but I need to know how it is found. Does it have anything to do with branch points in complex domain?
I do appreciate it if someone can solve it for me and tell me the inversion process.

2. Originally Posted by Ehsan
Hi,
I want to find the inverse laplace transform of $f(s)=\frac{1}{\sqrt{s^{2}+a^{2}}}$. I know that the inverse is $F(t)=J_0(at)$ from the tables, but I need to know how it is found. Does it have anything to do with branch points in complex domain?
I do appreciate it if someone can solve it for me and tell me the inversion process.
1. Look up the complex inversion formula for the Laplace transform (but you will still need a table of integrals since the integral you will get will not be elementary)

2. That is not the way its done in practice, the usual method is to use a table of Laplace transforms as a reverse look up table.

RonL

3. Originally Posted by Ehsan
Hi,
I want to find the inverse laplace transform of $f(s)=\frac{1}{\sqrt{s^{2}+a^{2}}}$. I know that the inverse is $F(t)=J_0(at)$ from the tables, but I need to know how it is found. Does it have anything to do with branch points in complex domain?
I do appreciate it if someone can solve it for me and tell me the inversion process.
I'm not going to spend time on the details because it will take me years to present it all nice and tidy. Others are more than welcome to add them.

But I'll give the general idea:

You know that $f(t) = \frac{1}{2 \pi i} \int_{\alpha - i \infty}^{\alpha + i \infty} \frac{e^{zt}}{\sqrt{z^2 + a^2}} \, dz$.

Note that $\frac{1}{\sqrt{z^2 + a^2}}$ has a branch cut along the imaginary axis connecting ai with -ai.

You can evaluate this integral using the usual closed contour C ..... Horizontal line to the right of the origin with a big ol' arc joining each end and the radius of the arc goes to infinity. The branch cut is enclosed. A diagram would be useful if I had the time to fiddle a nice one into here.

The principle of deformation of contours can be used to show that the integral around the closed contour C is equal to the integal around an inner contour 'defined' as follows:

Imagine the branch cut inside a 'dog bone' (the semicircle is at each end). The shape of this dog bone is the inner contour. Again, a diagram would be useful if I had the time to fiddle a nice one into here. Anyway ......

Integrating around the inner contour you can use the the usual arguments and a bit of re-arranging to show that $f(t) = \frac{1}{\pi} \int_{-a}^{+a} \frac{e^{iyt}}{\sqrt{a^2 - y^2}} \, dy$.

Make the substitution $y = a \sin \theta$ and with a little bit of re-arranging, Eulers famous identity, properties of odd and even functions, some algebra etc. you'll get one of the standard integral forms of $J_0(at)$, namely $\frac{1}{\pi} \int_0^{\pi} \cos (at \sin \theta) \, d \theta$.