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Question :

Find inverse Laplace transform of :

$$\ln \left(\frac{s^2+a^2}{s^2+b^2}\right)$$

My try :

I'm trying use this identity :

$f(t)=-\frac{\mathcal{L}^{-1}(\frac{dF(s)}{ds})}{t}$

Let $F(s)=\ln(\frac{s^2+a^2}{s^2+b^2})$

Then :

$\frac{dF(s)}{ds}=\frac{2s}{s^2+a^2}-\frac{2s}{s^2+b^2}$

So : $-tf(t)=\mathcal{L}^{-1}(\frac{2s}{s^2+a^2}-\frac{2s}{s^2+b^2})$ $=2\cos (at)-2\cos (bt)$

Is my try correct ?

If any one have simple method plz help me

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  • $\begingroup$ Correct to me... $\endgroup$ – Nosrati May 18 at 9:41
  • $\begingroup$ yes all seems good .. actually this is the simplest way of doing it :) $\endgroup$ – Ahmad Bazzi May 18 at 9:44
  • $\begingroup$ Everything looks good to me. There are a simple missing content, which I am adjusted. $\endgroup$ – nmasanta May 18 at 14:24
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Using the change of variable $s=\frac{1}{x}$ $$F\left( s \right)=\ln \left( \frac{{{s}^{2}}+{{a}^{2}}}{{{s}^{2}}+{{b}^{2}}} \right)=\ln \left( \frac{1+{{a}^{2}}{{x}^{2}}}{1+{{b}^{2}}{{x}^{2}}} \right)=\ln \left( 1+{{a}^{2}}{{x}^{2}} \right)-\ln \left( 1+{{b}^{2}}{{x}^{2}} \right)$$ and the power series expansion: $$\ln \left( 1+u \right)=u-\frac{{{u}^{2}}}{2}+\frac{{{u}^{3}}}{3}-\cdots =\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{u}^{n}}}$$ hence $$F\left( s \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{\left( \frac{a}{s} \right)}^{2n}}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}}{n}{{\left( \frac{b}{s} \right)}^{2n}}}$$ or $$f\left( t \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{a}^{2n}}}{n}{{\mathcal{L}}^{-1}}{{\left( \frac{1}{s} \right)}^{2n}}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{b}^{2n}}}{n}{{\mathcal{L}}^{-1}}{{\left( \frac{1}{s} \right)}^{2n}}}$$ so $$f\left( t \right)=\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{a}^{2n}}}{n}\frac{{{t}^{2n-1}}}{\left( 2n-1 \right)!}-}\sum\nolimits_{n=1}^{\infty }{\frac{{{\left( -1 \right)}^{n+1}}{{b}^{2n}}}{n}\frac{{{t}^{2n-1}}}{\left( 2n-1 \right)!}}$$ at last $$\begin{align} & \frac{1}{2}tf\left( t \right)=\sum\nolimits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{\left( at \right)}^{2n}}}{\left( 2n \right)!}-}\sum\nolimits_{n=1}^{\infty }{{{\left( -1 \right)}^{n+1}}\frac{{{\left( bt \right)}^{2n}}}{\left( 2n \right)!}} \\ & \quad \quad \quad =\left( 1-\cos \left( at \right) \right)-\left( 1-\cos \left( bt \right) \right) \\ \end{align}$$ and you get $$f\left( t \right)=\frac{2\cos \left( bt \right)-2\cos \left( at \right)}{t}$$

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  • $\begingroup$ Very nice work sir , thank you very much $\endgroup$ – Kînan Jœd May 18 at 22:03

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