How to calculate the following Laplace transform: $\mathcal{L}[\frac{1-{J}_{0}(t)}{t}]$? where ${J}_{0}(t)$ is the zeroth Bessel function.

Paul Reilly
2022-09-12
Answered

How to calculate the following Laplace transform: $\mathcal{L}[\frac{1-{J}_{0}(t)}{t}]$? where ${J}_{0}(t)$ is the zeroth Bessel function.

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Mateo Tate

Answered 2022-09-13
Author has **18** answers

If

$${\mathcal{L}}_{t}[F(t)](s)=f(s)$$

then:

$${\mathcal{L}}_{t}\left[\frac{F(t)}{t}\right](s)={\int}_{s}^{\mathrm{\infty}}f(u)\phantom{\rule{thinmathspace}{0ex}}du$$

provided: $\underset{t\to 0}{\text{lim}}\frac{F(t)}{t}$ exist.

Since:

$${\mathcal{L}}_{t}[1-{J}_{0}(t)](s)=\frac{1}{s}-\frac{1}{\sqrt{1+{s}^{2}}}$$

and:

$$\underset{t\to 0}{\text{lim}}\frac{1-{J}_{0}(t)}{t}=0$$

we have:

$${\mathcal{L}}_{t}\left[\frac{1-{J}_{0}(t)}{t}\right](s)={\int}_{s}^{\mathrm{\infty}}(\frac{1}{u}-\frac{1}{\sqrt{1+{u}^{2}}})\phantom{\rule{thinmathspace}{0ex}}du=\mathrm{ln}(\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{{s}^{2}}})$$

$${\mathcal{L}}_{t}[F(t)](s)=f(s)$$

then:

$${\mathcal{L}}_{t}\left[\frac{F(t)}{t}\right](s)={\int}_{s}^{\mathrm{\infty}}f(u)\phantom{\rule{thinmathspace}{0ex}}du$$

provided: $\underset{t\to 0}{\text{lim}}\frac{F(t)}{t}$ exist.

Since:

$${\mathcal{L}}_{t}[1-{J}_{0}(t)](s)=\frac{1}{s}-\frac{1}{\sqrt{1+{s}^{2}}}$$

and:

$$\underset{t\to 0}{\text{lim}}\frac{1-{J}_{0}(t)}{t}=0$$

we have:

$${\mathcal{L}}_{t}\left[\frac{1-{J}_{0}(t)}{t}\right](s)={\int}_{s}^{\mathrm{\infty}}(\frac{1}{u}-\frac{1}{\sqrt{1+{u}^{2}}})\phantom{\rule{thinmathspace}{0ex}}du=\mathrm{ln}(\frac{1}{2}+\frac{1}{2}\sqrt{1+\frac{1}{{s}^{2}}})$$

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