Laplace of $\int_0^t \frac{sinx}{x}dx$

Question

What is the Laplace transform of

$\int_0^t \frac{\sin x}{x}dx$

I'm thinking about approaching it as a convolution but I am not sure how. Could I define it as the convolution of $1$ and $\frac{\sin x}{x}$?

Answer 1

We need $$F(s) = \int_0^{\infty} e^{-st} \int_0^t \dfrac{\sin(x)}xdx dt = \int_0^{\infty}\int_x^{\infty} e^{-st} \dfrac{\sin(x)}xdt dx = \int_0^{\infty} \dfrac{\sin(x)}x\cdot \dfrac{e^{-sx}}sdx$$ We have $$F'(s) = \int_0^{\infty} \dfrac{\sin(x)}x\cdot \dfrac{-xe^{-sx}}sdx - \int_0^{\infty}\dfrac{\sin(x)}{x}\cdot \dfrac{e^{-sx}}{s^2}dx$$ This gives us \begin{align} sF' + F & = -\int_0^{\infty}e^{-sx}\sin(x)dx = -\dfrac1{s^2+1} \implies \dfrac{d}{ds}\left(sF(s)\right) = -\dfrac1{s^2+1} \end{align} \begin{align} sF(s) - \lim_{y \to 0} yF(y) & = -\int_0^s \dfrac{dt}{t^2+1} = -\arctan(s) \end{align} We have $$\lim_{y \to 0}yF(y) = \lim_{y \to 0}\int_0^{\infty} \dfrac{\sin(x)}xe^{-yx}dx =\int_0^{\infty} \lim_{y \to 0}e^{-yx} \dfrac{\sin(x)}x dx = \int_0^{\infty} \dfrac{\sin(x)}xdx = \dfrac{\pi}2$$ Hence, $$\boxed{\color{blue}{F(s) = \dfrac{\dfrac{\pi}2-\arctan(s)}s = \dfrac1s \arctan\left(\dfrac1s\right)}}$$

Answer 2

Note that the laplace transform of $\sin{c t}{c}$ is $\frac{1}{s^2+c^2}$. Now, integrate this with respect to $c$ from $0$ to $1$ to get the laplace transform of the sine integral, as desired.

(normally, you get the laplace transform of sinc using the rule for derivatives and laplace transforms from this)

Answer 3

Integrate by parts: $$ \int_0^{\infty} e^{-sx} \left( \int_0^x \frac{\sin{t}}{t} \, dt \right) \, dx = \left[ -\frac{e^{-sx}}{s}\left( \int_0^x \frac{\sin{t}}{t} \, dt \right) \right]_{0}^{\infty} + \frac{1}{s}\int_0^{\infty} e^{-sx}\frac{\sin{x}}{x} \, dx $$ The first term is obviously zero since the integral converges. Now consider $$ A(s) = \int_0^{\infty} e^{-sx}\frac{\sin{x}}{x} \, dx. $$ We clearly have $$ A'(s) = -\int_0^{\infty} e^{-sx} \sin{x} \, dx. $$ It is well-known how to do this (parts, complex exponentials...), and we find $$ A'(s) = -\frac{1}{1+s^2} $$ Then $$ A(s) = A(\infty) + \int_s^{\infty} \frac{dt}{1+t^2} = \arctan{\left( \frac{1}{s} \right)}. $$ Hence we have the Laplace Transform as $$ \frac{1}{s} \arctan{\left(\frac{1}{s}\right)}. $$

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To answer your suspicions, you can see here that we have derived the answer as $$ \int_0^{\infty} e^{-sx} \int_0^x \frac{\sin{t}}{t} \, dt \, dx = \int_0^{\infty} \int_t^{\infty} e^{-sx} \, dx \, dt = \frac{1}{s} \int_0^{\infty} e^{-st} \frac{\sin{t}}{t} \, dt. $$ And hence it is the product of the Laplace transforms of $1$ and $\frac{\sin{t}}{t}$.

Answer 4

The Laplace transform is $L(p)= \int \limits _0 ^\infty \mathbb e ^{-pt} \int \limits _0 ^t \frac {\sin x} x \mathbb d x \mathbb d t$. Compute the outer integral by parts (integrating the exponential): $\frac {\mathbb e ^{-pt}} {-p} \int \limits _0 ^t \frac {\sin x} x \mathbb d x \Big| _0 ^\infty - \int \limits _0 ^\infty \frac {\mathbb e ^{-pt}} {-p} \frac {\sin t} t \mathbb d t$. The first term of this difference is $0$ both for $t=\infty$ and $t=0$, so it is $0$. Let us now investigate the integral $f(p)=\int \limits _0 ^\infty e ^{-pt} \frac {\sin t} t \mathbb d t$ that shows up in the second term.

Compute $f'(p)=-\int \limits _0 ^\infty e ^{-pt} \sin t \mathbb d t = -\frac 1 {1+p^2}$ (just integrate by parts twice), therefore $f(p)=-\arctan p +C$, $C$ being some integration constant. Note that, from the definition of $f(p)$, you have that $\lim \limits _{p \to \infty} f(p) = 0$; on the other hand, from the formula of $f$ just found out after integration, you have that $\lim \limits _{p \to \infty} f(p) = -\frac \pi 2 + C$, therefore $C=\frac \pi 2$ and thus $f(p)=-\arctan p +\frac \pi 2$.

Going back in $L(p)$, you get that $L(p)=\frac 1 p (-\arctan p + \frac \pi 2)$. For a nicer result, note that $\frac \pi 2 - \arctan p = \arctan \frac 1 p$, so the end result is $$\frac {\arctan \frac 1 p} p$$