Finding the inverse Laplace transform of $ \ln \! \left( 1 + \frac{1}{s^{2}} \right) $.

Question

Can someone help me find the inverse Laplace transform of $ \ln \! \left( 1 + \dfrac{1}{s^{2}} \right) $? I have no idea where to start.

Answer 1

The Gamma Function $\Gamma$ is defined as

$$\Gamma(z) = \int_0^{\infty} t^{z-1}e^{-t}dt$$

for $\text{Re}\{z\}>0$.

We can write this integral representation as a Laplace Transform by letting $t \to st$. Then, we have

$$\begin{align} \Gamma(z) &= \int_0^{\infty} (st)^{z-1}e^{-st}sdt\\\\ &=s^z\int_0^{\infty} t^{z-1}e^{-st}dt \end{align}$$

The derivative of the Gamma Function follows directly as

$$\begin{align} \Gamma'(z) &= s^z\log (s)\int_0^{\infty} t^{z-1}e^{-st}dt+s^z\int_0^{\infty} t^{z-1}e^{-st}\log (t) \,dt \end{align}$$

Note that $\Gamma'(1)=\log(s)+s\mathscr{L}\{\log (t)\}(s)$, where

$$\mathscr{L}\{\log (t)\}(s)=\int_0^{\infty}\log(t) e^{-st}dt$$

is the Laplace Transform of $\log (t)$. Solving for $\log (s)$ yields

$$\log(s) = -\gamma -s\,\mathscr{L}\{\log (t)\}(s)$$

where we have noted that $\Gamma'(1)=-\gamma$, is the Euler-Mascheroni constant.

Now, following the decomposition given by Alamos, we have that

$$\begin{align} \log\left(\frac{s^2+1}{s^2}\right)&=\log(s+i)+\log(s-i)-2\log(s)\\\\ &=-(s+i)\int_0^{\infty} \log(t)e^{-(s+i)t}dt-(s-i)\int_0^{\infty} \log(t)e^{-(s-i)t}dt+2s\int_0^{\infty} \log(t)e^{-st}dt\\\\ &=\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-(s+i)t})dt+\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-(s-i)t})dt-2\int_0^{\infty} \log(t)\frac{d}{dt}(e^{-st})dt\\\\ &=2\int_0^{\infty}\log(t)\frac{d}{dt}((\cos(t)-1)e^{-st})dt\\\\ &=2\int_0^{\infty}\frac{1-\cos(t)}{t}e^{-st}dt\\\\ &=2\mathscr{L}\left(\frac{1-\cos(t)}{t}\right)(s) \end{align}$$

Inverting both sides reveals

$$\mathscr{L}^{-1}\left(\log\left(\frac{s^2+1}{s^2}\right)\right)(t)=2\,\,\frac{1-\cos(t)}{t}$$

Answer 2

The idea is to undo the operations you may find in the transform using the properties of the Laplace transform. The Laplace transform of the logarithm times Heaviside is essentially a logarithm divided by $s$. The transform of a derivative is essentially the transform of the function times $s$. And finally the transform of a function multiplied by an exponential is the transform of the function translated.

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We can write

$$\begin{align}\ln\left(1+\frac{1}{s^2}\right)&=\ln(s+i)+\ln(s-i)-2\ln(s)\\&=(s+i)\frac{\ln(s+i)+\gamma}{s+i}+(s-i)\frac{\ln(s-i)+\gamma}{s-i}-2s\frac{\ln(s)+\gamma}{s}\end{align}$$

We know that $$\mathcal{L}(-\ln(t)u(t))=\frac{\ln(s)+\gamma}{s}$$

Therefore

$$\mathcal{L}\left(D\left(-\ln(t)u(t)\right)\right)=s\frac{\ln(s)+\gamma}{s}$$

where $D$ is derivative and

$$\mathcal{L}\left(e^{\pm it}D\left(-\ln(t)u(t)\right)\right)=(s\pm i)\frac{\ln(s\pm i)+\gamma}{s\pm i}$$

Now you can complete it.

Answer 3

Use the following :

$$x(t)\to X(s)$$ $$-tx(t) \to X'(s)$$

Now we write

$$X(s) = \ln (1+1/s^2) = \ln(s^2+1) - 2\ln s$$ $$X'(s) = \frac{2s}{s^2+1}-\frac{2}{s}$$ $$-tx(t) = 2\cos t -2\to x(t) = \frac{2-cos t}{t}$$