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What is the inverse Laplace transform of L(s)=$exp⁡[-(1/2)sI]$*$I_0[(1/2)sI]$, where $I_0$ is the modified bessel function of first kind.

I am told that the answer is $\frac{1}{π}$ $\left(\frac{1}{(sI-s^2)}\right)^\frac{1}{2}$ but I am struggling to achieve this answer.

Can anyone help?

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  • $\begingroup$ please use mathjax to format your question. Here is a link to a quick tutorial on how to use it . $\endgroup$ May 21, 2018 at 20:24
  • $\begingroup$ Thank you, I will do that ASAP $\endgroup$
    – Mlo27
    May 21, 2018 at 20:24

2 Answers 2

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Well, we are looking at the inverse Laplace transform of:

$$\text{y}_{\space\text{n}}\left(t\right):=\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\cdot\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(t\right)}\tag1$$

Where $\mathcal{I}_{\space\text{p}}\left(\text{z}\right)$ is the modified Bessel function of the first kind.

Now, using the convolution theorem, of the Laplace transform, we can write:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}\space\text{d}\tau\tag2$$

Using the table of selected Laplace transforms, we can write:

$$\mathscr{L}_\text{s}^{-1}\left[\exp\left(-\text{n}\cdot\text{s}\right)\right]_{\left(t-\tau\right)}=\delta\left(t-\tau-\text{n}\right)\tag3$$

Where $\delta\left(x\right)$ is the Dirac Delta function.

So, we can rewrite equation $\left(2\right)$ as follows:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau\tag4$$

Using the definition of the modified Bessel function of the first kind, we can write:

$$\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\tag5$$

Where $\Gamma\left(\text{s}\right)$ is the Gamma function.

So using the table of selected Laplace transforms, we can write:

$$\mathscr{L}_\text{s}^{-1}\left[\mathcal{I}_{\space0}\left(\text{n}\cdot\text{s}\right)\right]_{\left(\tau\right)}=\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\mathscr{L}_\text{s}^{-1}\left[\left(\frac{\text{n}\cdot\text{s}}{2}\right)^{2\text{k}}\right]_{\left(\tau\right)}=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\tag6$$

Now, we can rewrite equation $\left(4\right)$ as follows:

$$\text{y}_{\space\text{n}}\left(t\right)=\int_0^t\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\cdot\frac{1}{\tau^{1+2\text{k}}}\right\}\cdot\delta\left(t-\tau-\text{n}\right)\space\text{d}\tau=$$ $$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{\Gamma\left(1+\text{k}\right)}\cdot\frac{1}{\Gamma\left(-2\text{k}\right)}\cdot\left(\frac{\text{n}}{2}\right)^{2\text{k}}\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau\tag7$$

When $\text{n}>0\space\wedge\space\text{k}\ge0\space\wedge\space t\ge0$, we can write:

$$\int_0^t\frac{\delta\left(t-\tau-\text{n}\right)}{\tau^{1+2\text{k}}}\space\text{d}\tau=\frac{2\cdot\theta\left(t\right)-1}{\left(t-\text{n}\right)^{1+2\text{k}}}\cdot\theta\left(t-\text{n}-t\cdot\theta\left(-t\right)\right)\cdot\theta\left(\text{n}-t+t\cdot\theta\left(t\right)\right)\tag8$$

Where $\theta\left(\text{a}\right)$ is the Heaviside step function.

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This is pretty straightforward if one writes

$$I_0(z) = \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \, e^{z v} $$

Then, letting $a=I/2 \gt 0$,

$$\begin{align} f(t) &= \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, I_0(a s) e^{(t-a) s}\\ &= \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{(t-a+a v)s} \\ &= \frac1{\pi} \int_{-1}^1 dv \, (1-v^2)^{-1/2} \delta(a v+t-a) \\ &= \frac1{\pi a} \left (1- \left (1-\frac{t}{a} \right )^2 \right )^{-1/2} \\ &= \frac1{\pi} (2 a t-t^2)^{-1/2} \end{align} $$

That said, note that for the argument of the delta function inside the $v$-integral to vanish, $t \in [0,2 a]$. Noting that I defined $a=I/2$, the ILT for the given Laplace transform is

$$f(t) = \frac1{\pi} (I t-t^2)^{-1/2} \theta(t) \theta(I-t) $$

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