I want to prove/disprove the following equation: $$\text{Log}{\frac{1}{\sqrt{z^2+1}}} = -\frac12 \ln\left(\frac{1}{z^2}+1\right)-\ln\lvert z\rvert - \gamma$$

Which $\gamma$ is the Euler Mascheroni constant.

This problem arise when my textbook said $$\mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} = \text{Log}\left(\frac{1}{\sqrt{s^2+1}}\right)\tag{1}$$ But, Wolfram gave me $$\mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} = -\frac12 \ln\left(\frac{1}{s^2}+1\right)-\ln\lvert z\rvert - \gamma\tag{2}$$

My attempt: Here, my working start is deal with my laplace transform: $$\begin{align} \mathcal{L} \Bigg\{\frac{\cos(t)}{t}\Bigg\} &= \int_0^{\infty}\frac{e^{-st}\cos(t)}{t}\,\Bbb dt\\ &= \int_0^{\infty} \frac{e^{-st}\left(e^{it}+e^{-it}\right)}{2t}\,\Bbb dt\\ &= \int_0^{\infty} \frac{e^{t(s-i)}}{2t}\,\Bbb dt + \int_0^{\infty} \frac{e^{-t(s+i)}}{2t}\,\Bbb dt \end{align}$$

I'm stuck there. If i continue, maybe i would get some exponential integral function. But how to produce equation $(1)$ or $(2)$??

I need a help. Thanks!

P.S. haven't tried to use $\text{Log}(z) = \ln|z| + i\text{Arg}(z)$ since i don't know what $\text{Arg}\left(\dfrac{1}{\sqrt{z^2+1}}\right)$ is. And btw, i'm not sure Wolfram use $\log$ as $\text{Log}$ or $\ln$. Here is the screenshot: enter image description here

  • $\begingroup$ Umm... what is ${\cal L}_t$? $\endgroup$ Jan 14 at 23:52
  • $\begingroup$ A laplace transform w.r.t variable t? $\endgroup$
    – user516076
    Jan 15 at 0:08
  • $\begingroup$ According to the definition of Laplace transform they are both wrong because the integral does not exist due to behavior at $\,t=0.$ Your definition of Laplace transform is wrong because it should be $dt$ and not $ds$. $\endgroup$
    – Somos
    Jan 15 at 1:09
  • $\begingroup$ @Somos edited. Thanks. $\endgroup$
    – user516076
    Jan 15 at 1:18
  • $\begingroup$ Branches of the logarithm differ by $2\pi n i$, the first identity cannot be correct. The issue is how the LT of $\cos(t)/t$ is defined, since the integral $\int_0^\infty e^{-s t} \cos(t)/t \, dt$ doesn't exist. In terms of distributions, $$\mathcal L[\operatorname {Ci}](s) = -\frac {\ln(s^2 + 1)} {2 s}, \\ \operatorname {Ci}'(t) = t_+^{-1} \cos t + \gamma \delta(t), \\ \mathcal L[\operatorname {Ci}'](s) = s \mathcal L[\operatorname {Ci}](s)$$ ($\operatorname {Ci}$ is a regular distribution supported on $[0, \infty)$). $\endgroup$
    – Maxim
    Feb 8 at 14:51

Here's one way the Laplace transform can be derived:

$$ \begin{align} f(t) &= cos(t)/t \\ t f(t) &= cos(t) \\ \mathcal{L} \Bigg\{t f(t)\Bigg\} &= \mathcal{L} \Bigg\{ cos(t) \Bigg\} \\ -\frac{d}{ds} \mathcal{L} \Bigg\{ f(t)\Bigg\} &= \frac {s}{s^2 + 1} \\ \mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \int \frac {s}{s^2 + 1} ds \\ u &= s^2 + 1 \\ \mathcal{L} \Bigg\{ f(t)\Bigg\} &= - \frac{1}{2} \int \frac {1}{u} du \\ &= - \frac{1}{2} ln(u) + C \\ &= - \frac{1}{2} ln(s^2 + 1) + C \\ &= ln(\frac{1}{\sqrt{s^2 + 1}}) + C \end{align} $$

IVT/FVT to find the value of the integration constant.

You'll notice that both of your forms are very close to each other:

$$ -\frac{1}{2} ln (\frac{1}{s^2}+1) - ln(s) \\ -\frac{1}{2} ln (\frac{s^2 + 1}{s^2}) - ln(s) \\ \frac{1}{2} ln (\frac{s^2}{s^2+1}) - ln (s) \\ ln (\frac{s}{\sqrt{s^2+1}}) - ln(s) \\ ln (\frac{1}{\sqrt{s^2+1}}) $$

So the question comes down to the integration constant $-\gamma$


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