# Math Help - Inverse Laplace Transform

1. ## Inverse Laplace Transform

Hi, I know how to find the Inverse Laplace Transform for $\ln(1+1/s^2)$ by this method:
Let $F(s)=\ln(1+1/s^2)$ so that $\displaystyle F'(s)=-2\left(\frac{1}{s}-\frac{s}{s^2+1}\right)$.
Since $\displaystyle \mathcal{L}^{-1}\{F'(s)\}=-2(1-\cos t)=-tf(t)$, it follows that $\displaystyle f(t)=\frac{2(1-\cos t)}{t}$

I was wondering if anyone can show me how to find $\displaystyle\mathcal{L}^{-1}\{\ln(1+1/s^2)\}$ by the complex inversion formula (Bromwich's integral formula). Thanks.

2. See contour below.

\begin{aligned}
\lim_{R\to\infty}\frac{1}{2\pi i}\mathop\oint\limits_{\substack{\text{my 3-key} \\ \text{contour}}} e^{st} \text{Log}(1+1/z^2) ds\\
&\hspace{-125pt}=\frac{1}{2\pi i}\mathop\int\limits_{a-i\infty}^{a+i\infty} e^{st}\text{Log}(1+1/z^2)ds+\frac{1}{2\pi i}\lim_{R\to\infty}\mathop\int\limits_{\text{excip ients}}e^{st} \text{Log}(1+1/z^2) ds
\end{aligned}

where $\text{Log}$ is the holomorphic extension of the principal branch to the left helf-plane described below. Therefore
$\mathcal{L}^{-1}\left\{\text{Log}(1+1/z^2)\right\}=-\frac{1}{2\pi i}\lim_{R\to\infty}\mathop\int\limits_{\text{excip ients}}e^{st} \text{Log}(1+1/z^2) ds
$

with
$
\mathop\int\limits_{\text{excipient}}=\mathop\int\ limits_{\text{gray}}+
\mathop\int\limits_{\substack{\text{Red(i)} \\ \text{Blue(i)}}}
+\mathop\int\limits_{\substack{\text{Red(-i)} \\ \text{Blue(-i)}}}
+\mathop\int\limits_{\substack{\text{Black} \\ \text{Green}}}
+\mathop\int\limits_{\text{Yellow}}
$

We now use the following which I believe is correct:

For $p,q$ polynomials, there exists a holomorphic extension of $\log(p/q)$ to the entire complex plane except along branch-cuts extending from each of the zeros and poles of the expression $p/q$. Additionally, the difference across each branch-cut is determined by the order of the branch point such that traveling in the positive sense around the branch point, this difference is $2n\pi i$ for a zero or order n and $-2m\pi i$ for a pole of order m.

And I'll do the red and blue branch-cut integrals at $s=i$. You can do the others just like how I do this one (and what will the difference of the integrand across the branch-cut at the real axis be?) I'll let $f=1+1/z^2$.


\begin{aligned}
\mathop\int\limits_{\substack{\text{Red(i)}\\ \text{Blue(i)}}} e^{st} \text{Log}(1+1/z^2)dz&=\int_{-\infty}^{-\epsilon} e^{st}\left(\ln|f|+i\arg f\right)d\sigma\\
&+\int_{-\epsilon}^{-\infty} e^{st}\left(\ln|f|+i(\arg f-2\pi)\right)d\sigma, s=\sigma+i
\end{aligned}

$
=\int_{-\infty}^{-\epsilon} e^{st}\left(\ln|f|+i\arg f\right)d\sigma-\int_{-\infty}^{-\epsilon} e^{st}\left(\ln|f|+i(\arg f-2\pi)\right)d\sigma
$

$
=2\pi i\int_{-\infty}^0 e^{t(\sigma+i)}d\sigma=2\pi i \frac{e^{it}}{t}
$

And of course you'll get $e^{-it}$ for the one at $-i$ which then combine to give you the cos term and the two at the real axis give you the other term.

This stuff is fun ain't it?
(and for the record, we really need to be addressing how the integral goes to zero on all the other parts of the contour)