# Is The Inverse Laplace Transform of $e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)$ doable using inversion formula?

I'm trying to solve inverse laplace transform using inversion formula and given by this integral: $$\frac{1}{2\pi i}\int_{\gamma-i\infty}^{\gamma+i \infty} e^{st}\operatorname{Log}\left(\frac{s+1}{s}\right)\,\Bbb ds.$$

Here is my contour, since the branch points of $$\operatorname{Log} \left(\frac{s+1}{s}\right)$$ are $$0$$ and $$-1$$

First, i want to show integral on $$L_u\cup L_d$$ is $$0$$ by bounding the integral with ML and then take the limit when $$R$$ goes to $$\infty$$.

By letting $$L_u,\, L_d: s= \xi\pm iR,0\leq \xi\leq \gamma$$, where $$\gamma$$ is the real number that the vertical line of the given contour passed by.

Since the $$L$$ is $$\lvert e^{t(\xi\pm iR)} \rvert$$, then i have ML inequality as below:

$$\lvert F(s)e^{st} \rvert \leq M_R \lvert e^{t(\xi\pm iR)} \rvert = M_R e^{\xi t} \leq M_R e^{at}$$

Next, i need to find $$M_R$$ and take the limit. \begin{align} \left|F(s)\right| &= \left|\operatorname{Log}\left(\frac{s+1}{s}\right)\right|\\ &= \left|\operatorname{Log}\left(\frac{\xi\pm iR+1}{\xi\pm iR}\right)\right| = M_R \end{align}

And by taking the limit of the last expression when $$R$$ goes to infinity yields $$0$$. Meaning the integrals along those lines are $$0$$.

So, from here, am i doing this right? I'm not sure my work is correct. Maybe there are some mistakes there. Help me please!

Edit: Working with my $$L_u$$ with $$ML$$ inequality, i have $$L=\gamma$$. Assuming $$-\pi<\operatorname{arg}{s}\leq \pi$$ and parametrizing $$s=-\xi+iR$$, $$\xi\in [-\gamma,0]$$:

\begin{align} \left|\int_{L_u}\right| &\leq \left|e^{st} \log\left(1 + \frac 1s\right)\right|\\ &\leq \left|e^{-\xi t}\right| \left|e^{iRt}\right|\left|\ln\left|1+ \frac{1}{-\xi+iR}\right| + i\pi\right|\\ &\leq 1\cdot 1 \cdot \ln\left(1+\frac 1R\right) + \pi\\ &\approx \frac 1R + \pi \end{align}

Combining the $$ML$$ i have $$\frac{\gamma}{R} +\gamma\pi$$ Which does NOT approach to $$0$$. Why? Please spot my mistake. I can't think about how to make it goes to $$0$$ since yesterday. Hope you kind to help me.

New Edit: Now my main question is not about ML. I managed to set the big and small arc goes to $$0$$. My main question now is "is it possible to evaluate this by using inversion formula (without differentiate both sides) and use log form instead?"

What i mean by differentiate both sides is: \begin{align} \mathcal{L}^{-1} &= \operatorname{Log}\left(\frac{s+1}{s}\right)\\ (\mathcal{L}^{-1})' &= \frac{1}{1+s} - \frac{1}{s} \end{align}

I don't want this kind of solution. What i really want is evaluating $$\operatorname{Log}\left(\frac{s+1}{s}\right)$$ by inversion formula. That's it. because I keep getting integrals on the contour lines above and below the branch cut cancel each other and make everything 0, which of course I don't want it.

Attempt:


One idea, in this particular case, is to get rid of the $$\ds{\ln}$$-function by introducing an integral representation of it. Namely, \begin{align} &\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic} \int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s} \,\dd s\,\right\vert_{\,\gamma\ >\ 0}} \\[5mm] = &\ \int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ \overbrace{\int_{0}^{1}{\dd x \over x + s}} ^{\ds{\ln\pars{s + 1 \over s}}}\ \,{\dd s \over 2\pi\ic} \\[5mm] = &\ \int_{0}^{1}\int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} {\expo{st} \over s + x}\,{\dd s \over 2\pi\ic}\,\dd x \\[5mm] = &\ \int_{0}^{1}\bracks{t > 0}\expo{-xt}\,\dd x = \bbx{\bracks{t > 0}\,{1 - \expo{-t} \over t}} \\ & \end{align}

$$\ds{\LARGE\left. b\right): {\large Contour Integration}}$$'

Note that $$\ds{{s + 1 \over s} < 0}$$ whenever $$\ds{s \in \pars{-1,0}}$$: It indicates that the $$\mbox{\ds{\ln}-branch-cut}$$ lies along $$\ds{\bracks{-1,0}}$$. Also, $$\ds{{s \pm \ic\epsilon + 1 \over s \pm \ic\epsilon} \,\,\,\stackrel{{\rm as}\ \epsilon\ \to\ 0^{+}}{\sim} \,\,\, \pars{1 + s \over s} \mp {\epsilon \over s^{2}}\,\ic}$$ such that the $$\ds{\ln}$$-$$\ds{arg}$$ is $$\pars{-\pi}$$ when $$\ds{s}$$ is above the real axis and $$\ds{\pi}$$ when it's below the real axis. \begin{align} &\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic} \int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s} \,\dd s\,\right\vert_{\,\gamma\ >\ 0}} \\[5mm] = &\ -\int_{-\infty}^{-1}\ln\pars{s + 1 \over s}\expo{ts} {\dd s \over 2\pi\ic}\label{1}\tag{1} \\[2mm] - &\ \int_{-1}^{0}\bracks{\ln\pars{-s - 1 \over s} - \ic\pi}\expo{ts} {\dd s \over 2\pi\ic} \\[2mm] - &\ \int_{0}^{-1}\bracks{\ln\pars{-s - 1 \over s} + \ic\pi}\expo{ts} {\dd s \over 2\pi\ic} \\[2mm] - & \int_{-1}^{-\infty}\ln\pars{s + 1 \over s}\expo{ts} {\dd s \over 2\pi\ic}\label{2}\tag{2} \end{align} I included (\ref{1}) and (\ref{2}) "by completeness" but, indeed, it's not necessary because the $$\ds{\ln}$$-branch-cut lies along $$\ds{\bracks{-1,0}}$$ albeit it guarantees the vanishing out of the integration along the"big-arc".

Then, \begin{align} &\bbox[5px,#ffd]{\left.{1 \over 2\pi\ic} \int_{\gamma\ -\ \ic\infty}^{\gamma\ +\ \ic\infty} \expo{st}\ln\pars{s + 1 \over s} \,\dd s\,\right\vert_{\,\gamma\ >\ 0}} \\[5mm] = &\ -\int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} - \ic\pi}\expo{-ts} {\dd s \over 2\pi\ic} \\[2mm] &\ \,\, + \int_{0}^{1}\bracks{\ln\pars{-s + 1 \over s} + \ic\pi}\expo{-ts} {\dd s \over 2\pi\ic} \\[5mm] = &\ \int_{0}^{1}\expo{-ts}\,\dd s = \bbx{1 - \expo{-t} \over t} \\ & \end{align}

• (+1) Thanks. But i'm afraid it still doesn't answer my question about using my contour to perform the ILT. Because the new integrand inside the integral w.r.t $s$ has different contour with mine. Maybe if you can explain why i can't use my contour i'll accept this answer, though i already upvote your answer. Still many thanks for the answer! Jan 20 '21 at 4:25
• @user516076 I just included the contour integration. Jan 20 '21 at 5:53
• Thank you for your wonderful answer! Jan 20 '21 at 7:02
• @user516076 Thanks. Glad to see it's useful. Jan 20 '21 at 21:28
• Hi, Felix. It's been a while since you posted your answer, hope I don't interrupt you. Anyway, the term $\ln$ on the last paragraph about contour integration, can I replace the bracket with absolute value? Just in case when I'm evaluating and do u-subs I wouldn't get negative inside $\ln$? Jan 23 '21 at 0:24

Let's pick a sufficiently large $$\gamma$$ such that $$|s|>1$$, so that we can plug in

$$\log\left(1+\frac1s\right)=\sum_{k=1}^\infty{(-1)^{k+1}\over ks^k}$$

and get

\begin{aligned} f(t) &={1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\sum_{k=1}^\infty{(-1)^{k+1}e^{st}\over s^k}\mathrm ds \\ &=\sum_{k=1}^\infty{(-1)^{k+1}\over k}{1\over2\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}{e^{st}\over s^k}\mathrm ds \end{aligned}

Provided that

$$\mathcal L\{t^{k-1}\}(s)={(k-1)!\over s^k}$$

we can deduce

\begin{aligned} f(t) &=\sum_{k=1}^\infty{(-t)^{k-1}\over k!} \\ &=-\frac1t\left[\sum_{k=0}^\infty{(-t)^k\over k!}-1\right] \\ &={1-e^{-t}\over t} \end{aligned}

where the final step follows from the fact that

$$e^z=\sum_{n=0}^\infty{z^k\over n!}$$

• Ok. Nice and thanks . Actually i know the ILT already, But sorry, my question is why did I fail to perform complex inversion formula to logarithm directly? Without splitting the log term, is that "doable"? Jan 20 '21 at 2:14
• @user516076 I am unsure about that since most of the texts I have found often differentiate to eliminate logarithms in the integrand (perhaps to avoid dealing with logarithmic branch point) Jan 20 '21 at 2:28
• Ok. Still seeking my answer. Btw i still (+1) for your kindly giving another approach. Jan 20 '21 at 2:32