Laplace transform of $\int_{0}^\infty\frac{e^{-t}\sin^2t}{t}dt$

Question

Laplace transform of $\int_{0}^\infty\frac{e^{-t}\sin^2t}{t}dt$.

So far I've calculated that $\frac{e^{-t}\sin^2t}{t}$ transformed equals $\frac{1}{8}(\ln((s+1)^2+4)-2\ln(s+1))$.

My question is what should I do with integral $\int_{0}^\infty$ ?

Answer 1

My free interpretation is: how to use the Laplace transform to compute the integral $$ I=\int_{0}^{+\infty}\frac{e^{-t}\sin^2(t)}{t}\,dt $$ ? Well, since: $$\mathcal{L}\left(e^{-t}\sin^2(t)\right)=\frac{2}{(1+s)(5+2s+s^2)}\tag{1}$$ we have: $$ I = 2\int_{0}^{+\infty}\frac{ds}{(1+s)(5+2s+s^2)}=2\int_{0}^{+\infty}\frac{du}{4+e^{2u}}=\int_{0}^{+\infty}\frac{dv}{4+e^v}\tag{2} $$ or: $$ I = \int_{1}^{+\infty}\frac{dw}{w(4+w)}=\color{red}{\frac{\log 5}{4}}.\tag{3}$$

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{\infty}{\expo{-t}\sin^{2}\pars{t} \over t}\,\dd t} & = \int_{0}^{\infty}\expo{-t}\ \overbrace{{1 - \cos\pars{2t} \over 2}}^{\ds{=\ \sin^{2}\pars{2t}}}\ \overbrace{\int_{0}^{\infty}\expo{-tx}\,\dd x}^{\ds{=\ {1 \over t}}}\ \,\dd t \\[3mm] & = \half\,\Re\int_{0}^{\infty}\int_{0}^{\infty} \bracks{\expo{-\pars{x + 1}t} - \expo{-\pars{x + 1 + 2\ic}t}}\,\dd t\,\dd x \\[3mm] & = \half\,\Re\int_{0}^{\infty} \pars{{1 \over x + 1} - {1 \over x + 1 + 2\ic}}\,\dd x = \left.\half\,\Re\ln\pars{x + 1 \over x + 1 + 2\ic}\right\vert_{\ x\ =\ 0}^{\ x\ \to\ \infty} \\[3mm] & = \half\,\Re\ln\pars{1 + 2\ic} = \half\,\ln\pars{\root{1^{2} + 2^{2}}} = \color{#f00}{{1 \over 4}\,\ln\pars{5}} \end{align}