Inverse Laplace Transform of $\frac{1}{{e}^{s}-1}$ and $\frac{s}{{e}^{s}-1}$

ymochelows
2022-09-11
Answered

Inverse Laplace Transform of $\frac{1}{{e}^{s}-1}$ and $\frac{s}{{e}^{s}-1}$

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asked 2021-12-08

How do I find the inverse Laplace transform of $\frac{4s}{{s}^{2}+4}}^{2$ ?

asked 2022-09-10

What is the closed-form of the following integral

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-x}\mathrm{cos}(x)}{{x}^{2}+1}\mathrm{d}x$$

If we replaced $\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{{x}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}}$ by its integral representation, $\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\displaystyle {\int}_{0}^{\mathrm{\infty}}{e}^{-xt}\mathrm{sin}(t)\text{d}t,\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}}$, we get that

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}{e}^{-x}\mathrm{cos}(x){\textstyle (}{\int}_{0}^{\mathrm{\infty}}{e}^{-xt}\mathrm{sin}(t)\text{d}t{\textstyle )}\mathrm{d}x$$

$$\{\text{reverse the order of integration}\}$$

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\mathrm{sin}(t){\textstyle (}{\int}_{0}^{\mathrm{\infty}}{e}^{-x(t+1)}\mathrm{cos}(x)\text{d}x{\textstyle )}\mathrm{d}t$$

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\frac{(t+1)\mathrm{sin}(t)}{(t+1{)}^{2}+1}\mathrm{d}t$$

$$\{\text{make the change of variable}t+1=u{\textstyle \}}$$

$$={\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{sin}(u-1)}{{u}^{2}+1}\mathrm{d}t$$

$$=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}(1){\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{sin}(u)}{{u}^{2}+1}\mathrm{d}t\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}(1){\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{cos}(u)}{{u}^{2}+1}\mathrm{d}t$$

Have anyone an idea to finish the remaining integrals ?

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\frac{{e}^{-x}\mathrm{cos}(x)}{{x}^{2}+1}\mathrm{d}x$$

If we replaced $\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\displaystyle \frac{1}{{x}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}}$ by its integral representation, $\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}{\displaystyle {\int}_{0}^{\mathrm{\infty}}{e}^{-xt}\mathrm{sin}(t)\text{d}t,\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}}$, we get that

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}{e}^{-x}\mathrm{cos}(x){\textstyle (}{\int}_{0}^{\mathrm{\infty}}{e}^{-xt}\mathrm{sin}(t)\text{d}t{\textstyle )}\mathrm{d}x$$

$$\{\text{reverse the order of integration}\}$$

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\mathrm{sin}(t){\textstyle (}{\int}_{0}^{\mathrm{\infty}}{e}^{-x(t+1)}\mathrm{cos}(x)\text{d}x{\textstyle )}\mathrm{d}t$$

$$I\phantom{\rule{thickmathspace}{0ex}}=\phantom{\rule{thickmathspace}{0ex}}{\int}_{0}^{\mathrm{\infty}}\frac{(t+1)\mathrm{sin}(t)}{(t+1{)}^{2}+1}\mathrm{d}t$$

$$\{\text{make the change of variable}t+1=u{\textstyle \}}$$

$$={\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{sin}(u-1)}{{u}^{2}+1}\mathrm{d}t$$

$$=\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}(1){\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{sin}(u)}{{u}^{2}+1}\mathrm{d}t\phantom{\rule{thickmathspace}{0ex}}-\phantom{\rule{thickmathspace}{0ex}}\mathrm{sin}(1){\int}_{1}^{\mathrm{\infty}}\frac{u\mathrm{cos}(u)}{{u}^{2}+1}\mathrm{d}t$$

Have anyone an idea to finish the remaining integrals ?

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$$F(s)={L}^{-1}(\frac{\Gamma (-\frac{s}{a}+b+\frac{1}{4})}{\Gamma (\frac{1}{4}-\frac{s}{a})})$$

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Is there another method that doesn't include differentiating the Laplace transform of $\mathrm{cosh}t$ ten times?

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Is there another method that doesn't include differentiating the Laplace transform of $\mathrm{cosh}t$ ten times?

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