How to find out the Inverse Laplace transform of following expression?

Question

I am given with the expression $$\bigg(\frac{a^2}{s+\sqrt{s^2-a^2}}\bigg)^j$$ where $a$ is a constant, and $j = 1, 2, \ldots$. I want to find out the inverse laplace transform of above expression. Answer is given in terms of modified Bessel's Function of First kind.

Answer 1

Re Jack's comment, it isn't terribly hard to find the ILT if you are comfortable with complex integration and Cauchy's theorem. I will discuss this below.

The inverse Laplace transform (ILT) is given by the following expression:

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \left (s-\sqrt{s^2-a^2} \right )^n \, e^{s t}$$

where $c$ is greater than the real part of the right-most singularity of the integrand in the complex plane. Note that I simplified the given Laplace transform to be inverted.

To evaluate this integral, consider the following complex contour integral:

$$\oint_C dz \, \left (z-\sqrt{z^2-a^2} \right )^n \, e^{z t}$$

where $C$ is the following contour:

By Cauchy's theorem, this contour integral is zero because there are no singularities inside of $C$. We will use this fact to determine the ILT in terms of real integrals.

The integral over the vertical line on the right of $C$ becomes the ILT as the extent of the line (and the corresponding circular arcs to the left of the line) increases without bound. In this limit, when $t \gt 0$, the integral over the large circular arcs will vanish and therefore can be neglected.

Note that the square root in the integrand introduces branch point singularities at $z=\pm a$. We address this by defining branch cuts along the real axis, one along $(-\infty,-a]$ corresponding to the branch point $z=-a$, and another along $(-\infty,a]$, corresponding to the branch point $z=a$. Note that the overlap of the two branch cuts along $(-\infty,-a]$ (assuming $a \gt 0$) will cause cancellations later on.

As for the small semicircular detours around the branch points, these will vanish as the radius of these detours goes to zero. Accordingly, we may neglect these integrals. We therefore write the contour integral out as

$$\int_{c-i \infty}^{c+i \infty} ds \, \left (s-\sqrt{s^2-a^2} \right )^n \, e^{s t} + e^{i \pi} \int_{\infty}^a dx \, \left (-x-\sqrt{x^2-a^2} \right )^n \, e^{-x t} \\ + \int_{-a}^a dx \, \left (x-i \sqrt{a^2-x^2} \right )^n \, e^{-x t} + \int_a^{-a} dx \, \left (x+i \sqrt{a^2-x^2} \right )^n \, e^{-x t}\\ + e^{-i \pi} \int_a^{\infty} dx \, \left (-x-\sqrt{x^2-a^2} \right )^n \, e^{-x t} = 0$$

Note that the second and fifth integrals cancel because the value of the square root does not vary with branch cut in that integration region (i.e., the branch cut overlap). Thus, we are left with the first integral - which will become the ILT - and the third and fourth integrals. Rearranging and manipulating the resulting integrand, we have the following expression for the ILT:

$$\begin{align}\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \left (s-\sqrt{s^2-a^2} \right )^n \, e^{s t} &= \frac1{\pi} \operatorname{Im}{\int_{-a}^a dx \, \left (x+i \sqrt{a^2-x^2} \right )^n \, e^{-x t}} \\ &= \frac{a^{n+1}}{\pi} \int_0^\pi d\theta \, \sin{\theta}\, \sin{(n \theta)} \, e^{-a t \cos{\theta}} \\ &= \frac{a^{n+1}}{2 \pi} \int_0^\pi d\theta \, [\cos{((n-1) \theta)}-\cos{((n+1) \theta)}] \, e^{-a t \cos{\theta}} \\ &= \frac{a^{n+1}}{2} \left [I_{n-1}(a t) - I_{n+1}(a t) \right ] \end{align}$$

Finally, using the Bessel function identity $I_{n-1}(y) - I_{n+1}(y) = \frac{2 n}{y} I_n(y)$, we have for the ILT, $t \gt 0$:

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \left (s-\sqrt{s^2-a^2} \right )^n \, e^{s t} = n a^n \frac{I_n(a t)}{t} $$

When $t \lt 0$, we need to close the contour to the right of the vertical line. There, there are no singularities of the integrand, so the integral - and the ILT - is zero by Cauchy's theorem.