Find inverse Laplace transform of : $\ln(\frac{s^2+a^2}{s^2+b^2})$

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

• Correct to me... – Nosrati May 18 at 9:41
• yes all seems good .. actually this is the simplest way of doing it :) – Ahmad Bazzi May 18 at 9:44
• Everything looks good to me. There are a simple missing content, which I am adjusted. – nmasanta May 18 at 14:24

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 $$f\left( t \right)=\frac{2\cos \left( bt \right)-2\cos \left( at \right)}{t}$$