Evaluate $\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$ by Laplace transform $\require{cancel}$
Note: All logarithmic functions in the post include the absolute value function (this doesn't change the convergence of the variables).
For an entrance exam i got the integral
$$\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt$$
The exercise is to calculate it by the Laplace transform.
By the frequency domain integration theorem (only in unilateral), it holds that:
$$\mathcal{L}\left[\frac{f(t)}{t} \right]= \int_s^\infty F(\omega)\; d\omega = \int_s^\infty \mathcal{L}(f(t))\; d\omega $$
$\omega$ being our dummy variable here
Let our $f(t)$ time-domain function be $e^{-\alpha t}\sin(-t)$
$$\mathcal{L}\left[\frac{f(t)}{t}\right]= \mathcal{L}\left[\frac{e^{-\alpha t}\sin(-t)}{t} \right] =\int_s^\infty \mathcal{L}\left[{e^{-\alpha t}\sin(-t)}\right]\; d\omega $$
$$\mathcal{L}\left[e^{-\alpha t}\sin(\beta t)\right]= \dfrac{\beta}{(\omega+\alpha)^2+\beta^2} \hspace{10mm}\alpha,\beta \in \mathbb{R} $$
If $\beta=-1$
$$\mathcal{L}\left[e^{-\alpha t}\sin(-t)\right]= \dfrac{-1}{(\omega+\alpha)^2+(-1)^2} = \dfrac{-1}{(\omega+\alpha)^2+1} $$
$$\int_s^\infty \dfrac{-1}{(\omega+\alpha)^2+1}\; d\omega = -\int_s^\infty \frac{1}{\omega^2 + 2\alpha\omega + \alpha^2 +1}\; d\omega =
- \int_s^\infty \frac{\frac{-i}{2}}{\omega-\alpha-i}\;d\omega \; - \int_s^\infty \frac{\frac{i}{2}}{\omega-\alpha+i}\;d\omega=$$
$$=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}= -\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty       \bigg)= $$
$$\stackrel{\color{red}{\Delta}}{=}-\frac{i}{2} \bigg(  \left[\cancelto{-\infty}{-\ln(\infty)}+\ln(s-\alpha-i)\right] + \left[\cancelto{\infty}{\ln(\infty)}-\ln(s-\alpha+i)\right] \bigg)= \boxed{-\frac{i}{2}\ln\left(\frac{s-\alpha-i}{s-\alpha+i} \right)} $$
Thus the integral
$$\int_0^\infty \frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt = -\int_0^{\infty}\frac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{t}\; dt$$
is
$$-\left[-\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right) + -\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)\right]= \frac{i}{2}\ln \left(\frac{s-2-i}{s-2+i}\cdot \frac{s-3-i}{s-3+i} \right)=$$
$$= \boxed{\frac{i}{2}\ln\left(\frac{\left(s^2-5s+5\right)+i\left(5-2s\right)}{\left(s^2-5s+5\right)-i\left(5-2s\right)} \right)} $$
The problem here is that i cannot really take the Inverse Laplace Transform of a rational function which is inside a logarithm.
How could i proceed?
If someone is confused by the $\color{red}{\Delta}$ step:
$$\color{red}{\Delta}:=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty  \bigg)=-\frac{i}{2} \left\{ \int_s^\infty -\frac{1}{\omega-\alpha-i}\;d\omega + \int_s^\infty \frac{1}{\omega-\alpha+i}\;d\omega  \right\}=$$
$$=-\frac{i}{2} \bigg( \left[ -\ln(\omega-\alpha-i) \right]_s^\infty + \left[\ln(\omega-\alpha+i)\right]_s^\infty\bigg) = \frac{i}{2}\left(\Bigg[-\ln(\infty-\alpha-i)\underbrace{--}_{+}\ln(s-\alpha-i)\Bigg]+\Bigg[\ln(\infty-\alpha+i)-\ln(s-\alpha+i)\Bigg]\right)=$$
$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\left(-\ln(\sigma-\alpha-i)+\ln(\sigma-\alpha+i)\right)}+\ln(s-\alpha-i)-\ln(s-\alpha+i)\right)=$$
$$=-\frac{i}{2}\left(\color{blue}{\lim_{\sigma\to\infty}\ln\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$
$$=-\frac{i}{2}\left(\color{blue}{\underbrace{\ln\underbrace{\lim_{\sigma\to\infty}\left(\dfrac{\sigma-\alpha-i}{\sigma-\alpha+i}\right)}_{1}}_{0}}+\ln\left(\dfrac{s-\alpha-i}{s-\alpha+i}\right)\right)$$
 A: This is easily done by doing the following.
Note that $$\mathcal{L} \left( \frac{\sin t}{t} \right) = \tan^{-1} \left( \frac{1}{s} \right) $$
(use the formula $\mathcal{L} \left( \frac{f(t)}{t} \right) = \int_s ^\infty \mathcal{L} (f)(w) \ dw$ and an arctan identity), so we can do the following:
\begin{align*}
\int_0^\infty \dfrac{e^{-2t}\sin(-t)+e^{-3t}\sin(-t)}{-t}\; dt &= \int_0^\infty \dfrac{-e^{-2t}\sin(t)-e^{-3t}\sin(t)}{-t}\; dt \\
&= \int_0^\infty \dfrac{e^{-2t}\sin(t)+e^{-3t}\sin(t)}{t}\; dt \\
&= \int_0^\infty \frac{e^{-2t} \sin t}{t} \ dt + \int_0 ^\infty \frac{e^{-3t} \sin t}{t} \ dt \\
&= \mathcal{L} \left( \frac{\sin t}{t} \right) \Big\vert_{s = 2} + \mathcal{L} \left( \frac{\sin t}{t} \right) \Big\vert_{s = 3} \\
&= \tan^{-1} \left( \frac {1}{2} \right) + \tan^{-1} \left( \frac {1}{3} \right) \\
&= \frac{\pi}{2}.
\end{align*}
For the last line, use the tangent sum identity - let $\tan \alpha = 1/2$ and $\tan \beta = 1/3$. Then
\begin{align*}
\tan(\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\
&= \frac{1/2 + 1/3}{1 - (1/2)(1/3) } \\
&= \frac{5/6}{5/6} \\
&= 1
\end{align*}
so $\alpha + \beta = \frac{\pi}{2}.$
TL;DR, You weren't asked to take the Laplace transform of the integrand, rather, you were to recognize this as a Laplace transform evaluated at a point.
A: It turns out that there is a complex logarithmic identity that allows us to turn a logarithm of a complex complex number into an inverse trigonometric function.
$$\color{red}{\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right)} + \color{blue}{\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)}$$
$$\color{red}{\frac{i}{2}\ln\left(\frac{s-2-i}{s-2+i}\right)=-\frac{i}{2}\ln\left(\frac{s-2+i}{s-2-i}\right)=\mathrm{arccot}(s-2)=\arctan\left(\frac{1}{s-2}\right)}$$
$$\color{blue}{\frac{i}{2}\ln\left(\frac{s-3-i}{s-3+i}\right)=-\frac{i}{2}\ln\left(\frac{s-3+i}{s-3-i}\right)=\mathrm{arccot}(s-3)=\arctan\left(\frac{1}{s-3}\right)}$$
Although I'm not really sure how to continue from here now.
From table:
$$\ln\left(\frac{s+a}{s+b}\right)\stackrel{\mathcal{L}^{-1}}{\Rightarrow}\frac{e^{-bt}-e^{-at}}{t}$$
