How to calculate the following Laplace transform: $\mathcal{L}[te^{-3t}J_0(2t)]$?

Question

I'm trying to calculate the Laplace transform of this function.

$$ \mathcal{L}[te^{-3t}J_0(2t)] $$

where $J_0(t)$ is the zeroth-order Bessel function.

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Solution Attempt

The p-Bessel function is defined as: $$J_p(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+p+1)}\left(\frac{x}{2}\right)^{2m+p}$$

Therefore for $p=0$ the zeroth-order Bessel is: $$J_p(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+1)}\left(\frac{x}{2}\right)^{2m+p}$$

It is trivial to prove that: $$\mathcal{L}\left[J_0(t)\right](s)= \frac{1}{\sqrt{s^2 + 1}}$$

Using transform rule: $ \mathcal{L}[f(t)]= F(s) \to \mathcal{L}[t^kf(t)] = (-1)^k \frac{d}{ds^k}(F(s))$

I calculated that $ \mathcal{L}[te^{-3t}] = \frac{1}{(s+3)^2} $

In any case I'm stuck at this point. $$ \mathcal{L}[te^{-3t}J_0(2t)] $$I don't know how to calculate $J_0(2t)$ neither what rule to use to combine it with the previous finding.

Any ideas?

Answer 1

Well, we can solve a way more general formula. We want to find the following Laplace transform:

$$\text{F}_\text{n}\left(\alpha,\beta,\text{k},\text{s}\right):=\mathcal{L}_x\left[x^\text{n}\exp\left(\alpha x\right)\mathcal{J}_\text{k}\left(\beta x\right)\right]_{\left(\text{s}\right)}\tag1$$

Where $\text{n}\in\mathbb{N}$.

Using the 'frequency-domain general derivative' property of the Laplace transform, we can write:

$$\text{F}_\text{n}\left(\alpha,\beta,\text{k},\text{s}\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\exp\left(\alpha x\right)\mathcal{J}_\text{k}\left(\beta x\right)\right]_{\left(\text{s}\right)}\right)\tag2$$

Using the 'frequency shifting' property of the Laplace transform, we can write:

$$\text{F}_\text{n}\left(\alpha,\beta,\text{k},\text{s}\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\mathcal{J}_\text{k}\left(\beta x\right)\right]_{\left(\text{s}-\alpha\right)}\right)\tag3$$

Now, we can use the definition of the Bessel functions of the first kind in order to write:

$$\text{F}_\text{n}\left(\alpha,\beta,\text{k},\text{s}\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}}{\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot\left(\frac{\beta x}{2}\right)^{2\text{m}+\text{k}}\right]_{\left(\text{s}-\alpha\right)}\right)=$$ $$\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}}{\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot\frac{\beta^{2\text{m}+\text{k}}\cdot x^{2\text{m}+\text{k}}}{2^{2\text{m}+\text{k}}}\right]_{\left(\text{s}-\alpha\right)}\right)=$$ $$\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\mathcal{L}_x\left[\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}\beta^{2\text{m}+\text{k}}2^{-\left(2\text{m}+\text{k}\right)}}{\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot x^{2\text{m}+\text{k}}\right]_{\left(\text{s}-\alpha\right)}\right)=$$ $$\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}\beta^{2\text{m}+\text{k}}2^{-\left(2\text{m}+\text{k}\right)}}{\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot \mathcal{L}_x\left[x^{2\text{m}+\text{k}}\right]_{\left(\text{s}-\alpha\right)}\right)\tag4$$

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

$$\text{F}_\text{n}\left(\alpha,\beta,\text{k},\text{s}\right)=\left(-1\right)^\text{n}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}\beta^{2\text{m}+\text{k}}2^{-\left(2\text{m}+\text{k}\right)}}{\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot \frac{\Gamma\left(1+2\text{m}+\text{k}\right)}{\left(\text{s}-\alpha\right)^{1+2\text{m}+\text{k}}}\right)=$$ $$\left(-1\right)^\text{n}\cdot\sum_{\text{m}\ge0}\frac{\left(-1\right)^\text{m}\Gamma\left(1+2\text{m}+\text{k}\right)\beta^{2\text{m}+\text{k}}}{2^{2\text{m}+\text{k}}\left(\text{m}!\right)\cdot\Gamma\left(1+\text{m}+\text{k}\right)}\cdot\frac{\partial^\text{n}}{\partial\text{s}^\text{n}}\left(\frac{1}{\left(\text{s}-\alpha\right)^{1+2\text{m}+\text{k}}}\right)\tag5$$

Answer 2

We know the basic Bessel integral:

$$ \int_0^\infty e^{ax}J_0(bx)dx = \frac{1}{\sqrt{a^2 + b^2}}$$

Therefore let $g(t) = e^{-3t}J_0(2t)$

$$ \mathcal{L}[g(t)] = \mathcal{L}[e^{-3t}J_0(2t)] = G(s) = \int_0^\infty e^{-(s+3)t}J_0(2t)dt = \frac{1}{\sqrt{(s+3)^2 + 2^2}}$$

Now using the basic Laplace Transform property $ \mathcal{L}[t^nf(t)] = (-1)^nF^{(n)}(s) $ we have

$$ \mathcal{L}[tg(t)] = (-1)^1 * (\frac{1}{\sqrt{(s+3)^2 + 2^2}})' = \frac{s+3}{\left(s^2+6s+13\right)^{\frac{3}{2}}} $$