1
$\begingroup$

After much laboring, sparked by the incredibly helpful comment on my previous question:

Showing that the pmf of a complicated expression sums to 1 (i.e. it converges)

I obtained a closed form for the summation of the pmf (i.e. it was not 1)...

So, based on the context of my problem, I started looking over Integral equations of the first kind, and found that I can use Inverse Laplace transforms.

My question now is something I could not get in Mathematica (again).

How to compute the inverse Laplace transform of $$F(s) = \dfrac{s}{(b+s)(a+b+s)^c}$$ where $a,b > 0$ and $c$ is a whole number?

If I can get this, then I am definitely done with my problem!! Because the pmf will sum to 1!

  • 0
    Did you try fraction $$\dfrac{s}{b+s)(a+b+s)}=\dfrac{k_1}{b+s}+\dfrac{k_2}{a+b+s}$$ ? – 2017-01-25
  • 0
    But $(a+b+s)$ is raised to a power?! – 2017-01-25
  • 0
    Edit the question ? (ok ,I am sorry) – 2017-01-25
  • 0
    MMA 11.0 `Assuming[{a > 0, b > 0, c > 0}, InverseLaplaceTransform[s/((b + s)*(a + b + s)^c), s, t]]=(a^-c E^(-(a + b) t) (-b E^(a t) t (Gamma[c] - Gamma[c, a t]) + (a t)^ c (1 + DiscreteDelta[t] - 2 HeavisideTheta[-t])))/(t Gamma[c])` – 2017-01-25

1 Answers 1

2

Using the convolution theorem:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{b}+\text{s}}\right]_{\left(t\right)}*\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}\tag1$$

Now, use:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\text{b}+\text{s}}\right]_{\left(t\right)}=\delta\left(t\right)-\frac{\text{b}}{e^{\text{b}t}}\tag2$$

And:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{1}{\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\frac{t^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot t}\cdot\Gamma\left(\text{c}\right)}\tag3$$

So:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\left(\delta\left(t\right)-\frac{\text{b}}{e^{\text{b}t}}\right)*\frac{t^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot t}\cdot\Gamma\left(\text{c}\right)}\tag4$$

In order to undo the convolution, solve:

$$\mathcal{L}_\text{s}^{-1}\left[\frac{\text{s}}{\left(\text{b}+\text{s}\right)\cdot\left(\text{a}+\text{b}+\text{s}\right)^\text{c}}\right]_{\left(t\right)}=\int_0^t\left(\delta\left(\tau\right)-\frac{\text{b}}{e^{\text{b}\tau}}\right)\cdot\frac{\left(t-\tau\right)^{\space\text{c}-1}}{e^{\left(\text{a}+\text{b}\right)\cdot\left(t-\tau\right)}\cdot\Gamma\left(\text{c}\right)}\space\text{d}\tau\tag5$$

  • 0
    @MariuszIwaniuk Give me some ground why it wrong? Your way of checking can also be wrong! – 2017-01-26
  • 0
    Now is ok.OP think that this is the multiplication sign ,not a convolution sign.Just I saw not a complete answer :P Good Job. – 2017-01-26
  • 0
    @MariuszIwaniuk Oh no, $*$ is the convolution sign and $\times$ or $\cdot$ are multiplication signs. – 2017-01-26