Billateral laplace transform sin(t) / t u (t)

Question

Hello everyone at my course I have problem solving Laplace transform of

$\frac{\sin(t)}{t}$ $u{(t)}$

I have no idea I tried by definiton but get integral which cant be solved I already took a look at Finding the Laplace Transform of sin(t)/t

But It doesnt help me at all becouse there is used Taylor series expansion becouse I m still begginer is there any easier way to solve it

Thanks in advante

Answer 1

The hint is very useful on that linked problem! Using it, we have

$$\displaystyle f(s) = \int_0^\infty\dfrac{\sin t}{t}e^{-st}~ds$$

hence

$$f'(s) = \int_0^\infty \sin t e^{-st}~ds = -\dfrac{d}{ds} \arctan s$$

Can you now find $F(s)$ and solve for the constant?

Answer 2

In THIS ANSWER, I showed that the inverse Laplace Transform of $\arctan(s)-\pi/2$ is $-\frac{\sin(t)}{t}$ by carrying out the integral

$$\mathscr{L}^{-1}(\arctan(s)-\pi/2)=\int_{\sigma -i\infty}^{\sigma+i\infty}e^{st}(\arctan(s)-\pi/2)\,ds$$

---

Herein, we carry out the forward Laplace Transform of the sinc function. Proceeding, we have

$$\begin{align} F(s)&=\int_0^\infty e^{-st}\frac{\sin(t)}{t}\,dt \end{align}$$

Now, differentiating we have

$$\begin{align} F'(s)&=-\int_0^\infty e^{-st}\sin(t)\,dt\\\\ &=-\frac1{s^2+1} \end{align}$$

whereupon integrating and using $\lim_{s\to \infty }F(s)=0$ yields

$$F(s)=\pi/2-\arctan(s)$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\sin\pars{t} \over t}\,\expo{-st}\,\dd t & = \int_{0}^{\infty}\ \overbrace{\pars{{1 \over 2}\int_{-1}^{1}\expo{-\ic \omega t}\,\dd \omega}}^{\ds{\ds{\sin\pars{t} \over t} \atop}}\ \,\expo{-st}\,\dd t = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{s + \ic\omega}t} \,\,\dd t\,\dd\omega \\[5mm] & = {1 \over 2}\int_{-1}^{1}{\dd\omega \over s + \ic\omega} = {1 \over 2}\int_{-1}^{1}{s \over \omega^{2} + s^{2}}\,\dd\omega = \int_{0}^{1/s}{\dd\omega \over \omega^{2} + 1} = \bbx{\ds{\arctan\pars{1 \over s}}} \end{align}