Inverse Laplace Transform of $\frac{s}{(s-a)^{3/2}}$

Question

Find the inverse laplace of: $\frac{s}{(s-a)^{3/2}}$

I tried working through this using partial fractions and convolution but I can't seem to get a requitible answer. How would I go about solving this? (by the way, we have yet to learn the integral definition of inverse laplace transforms, so we are expected not to use that.)

Answer 1

It is easy to show that the Laplace transform of $x^{c-1}e^{ax}$ is $\Gamma(c) (s-a)^{-c}$

So, the inverse Laplace transform of $\frac{1}{(s-a)^c}$ is $\frac{1}{\Gamma(c)}x^{c-1}e^{ax}$

$$\frac{s}{(s-a)^{3/2}}=\frac{1}{(s-a)^{1/2}}+\frac{a}{(s-a)^{3/2}}$$ With successively $c=1/2$ and $c=3/2$, we obtain the inverse Laplace transform of $\frac{s}{(s-a)^{3/2}}$ : $$\frac{1}{\Gamma(1/2)}x^{-1/2}e^{ax}+a\frac{1}{\Gamma(3/2)}x^{1/2}e^{ax}=\frac{1+2ax}{\sqrt{\pi\:x}}e^{ax}$$

Answer 2

Somehow, I missed the part about not using the definition of the ILT. However, since I invested a lot of effort into the solution and I think it demonstrates some interesting techniques I have not seen used here on M.SE, I will keep the solution posted.

Let's assume that $a \in \mathbb{R}$. The definition of the ILT is as follows:

$$f(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s-a)^{3/2}} e^{s t} $$

where $c \gt a$. We evaluate using Cauchy's theorem by considering the following contour integral:

$$\oint_C dz \frac{z}{(z-a)^{3/2}} e^{z t} $$

where $C$ is a deformed Bromwich contour closed to the left of outer radius $R$ with a small circular contour of radius $\epsilon$ about the pole/branch point on the real axis. As the integral about the outer circular arcs vanish as $R \to \infty$, the contour integral is equal to

$$\int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s-a)^{3/2}} e^{s t} + e^{i \pi} \int_{\infty}^{-a+\epsilon} dx \frac{e^{i \pi} x}{(e^{i \pi} x-a)^{3/2}} e^{-x t} \\ + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{a+\epsilon e^{i \phi}}{\epsilon^{3/2} e^{i 3 \phi/2}} e^{(a+\epsilon e^{i \phi}) t} + e^{-i \pi} \int_{-a+\epsilon}^{\infty} dx \frac{e^{-i \pi} x}{(e^{-i \pi} x-a)^{3/2}} e^{-x t}$$

Now it appears that the expression above diverges as $\epsilon \to 0$, but that is not the case because the divergences cancel. Note that the third integral behaves as

$$\begin{align} i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{a+\epsilon e^{i \phi}}{\epsilon^{3/2} e^{i 3 \phi/2}} e^{(a+\epsilon e^{i \phi}) t} &= i a e^{a t} \epsilon^{-1/2} \int_{\pi}^{-\pi} d\phi\, e^{-i \phi/2} + O \left (\epsilon^{1/2} \right ) \\ &=-i 4 a e^{a t} \epsilon^{-1/2}+ O \left (\epsilon^{1/2} \right ) \end{align}$$

This is of course divergent as $\epsilon \to 0$. However, the second and fourth integrals also have divergences. For example, the second integral may be written as:

$$\begin{align} e^{i \pi} \int_{\infty}^{-a+\epsilon} dx \frac{e^{i \pi} x}{(e^{i \pi} x-a)^{3/2}} e^{-x t} &= e^{a t} \int_{\infty}^{\epsilon} dy \frac{y-a}{e^{i 3 \pi/2} y^{3/2}} e^{-y t} \\ &= -i e^{a t} \int_{\epsilon}^{\infty} dy \, y^{-1/2} e^{-y t} + i a e^{a t} \int_{\epsilon}^{\infty} dy \, y^{-3/2} e^{-y t} \\ &= -i e^{a t} \sqrt{\frac{\pi}{t}} + O \left (\epsilon^{1/2} \right ) + i a e^{a t} \epsilon^{-1/2} \int_{1}^{\infty} du \, u^{-3/2} e^{-\epsilon t u} \end{align} $$

Now, it turns out that the latter integral is related to an incomplete gamma function and has the following behavior as $\epsilon \to 0$:

$$\int_{1}^{\infty} du \, u^{-3/2} e^{-\epsilon t u} = 2-2 \sqrt{\pi t} \epsilon^{1/2} + O(\epsilon) $$

Thus,

$$e^{i \pi} \int_{\infty}^{-a+\epsilon} dx \frac{e^{i \pi} x}{(e^{i \pi} x-a)^{3/2}} e^{-x t} = i 2 a e^{a t} \epsilon^{-1/2} -i e^{a t} \sqrt{\frac{\pi}{t}} - i 2 a \sqrt{\pi t} e^{a t} + O \left (\epsilon^{1/2} \right )$$

Similarly, the fourth integral may be written as

$$e^{-i \pi} \int_{-a+\epsilon}^{\infty} dx \frac{e^{-i \pi} x}{(e^{-i \pi} x-a)^{3/2}} e^{-x t} = i 2 a e^{a t} \epsilon^{-1/2} -i e^{a t} \sqrt{\frac{\pi}{t}} - i 2 a \sqrt{\pi t} e^{a t} + O \left (\epsilon^{1/2} \right )$$

Summing the results for the second, third, and fourth integrals, we see that the terms in $\epsilon^{-1/2}$ cancel, and we are left with

$$\int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s-a)^{3/2}} e^{s t} -i 2 e^{a t} \sqrt{\frac{\pi}{t}} - i 4 a \sqrt{\pi t} e^{a t}$$

By Cauchy's theorem, the contour integral - and therefore the above expression - is equal to zero. The sought-after ILT is then found to be

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \frac{s}{(s-a)^{3/2}} e^{s t} = \frac{1+2 a t}{\sqrt{\pi t}} e^{a t} $$

This of course assumed $t \gt 0$. For $t \lt 0$, the ILT is zero.