Laplace transform of sine integral

0.1 Derivation of $ℒ\{\mathrm{Si}t\}$

If one performs the change of integration variable

$$u=tx,du=tdx$$

in the defining integral (http://planetmath.org/DefiniteIntegral)

$$\mathrm{Si}t={\int }_0^t\frac{\sin u}{u}𝑑u,$$

of the sine integral function, one obtains

$$\mathrm{Si}t={\int }_0^1\frac{\sin tx}{tx}t𝑑x={\int }_0^1\frac{\sin tx}{x}𝑑x,$$

getting limits (http://planetmath.org/UpperLimit).  We know (see the entry Laplace transform of sine and cosine) that

$$ℒ\left\{\frac{\sin tx}{x}\right\}=\frac{1}{s^2+x^2}.$$

This transformation formula can be integrated with respect to the parametre $x$:

$$ℒ\left\{{\int }_0^1\frac{\sin tx}{x}𝑑x\right\}={\int }_0^1\frac{1}{s^2+x^2}𝑑x=\frac{1}{s}\underset{x=0}{\overset{1}{/}}\arctan \frac{x}{s}=\frac{1}{s}\arctan \frac{1}{s}.$$

Thus we have the transformation formula of the sinus integralis:

$ℒ\left\{\mathrm{Si}t\right\}={\displaystyle \frac{1}{s}}\arctan {\displaystyle \frac{1}{s}}.$(1)

0.2 Laplace transform of sinc function

By the formula  $ℒ\{f^{\prime }\}=sℒ\{f\}-{lim}_{x\to 0+}f(x)$  of the parent (http://planetmath.org/LaplaceTransform) entry,we obtain as consequence of (1), that

$$ℒ\left\{\frac{d}{dt}\mathrm{Si}t\right\}=s\cdot \frac{1}{s}\arctan \frac{1}{s}-\mathrm{Si}\mathrm{\hspace{0.17em}0},$$

i.e.

$ℒ\left\{{\displaystyle \frac{\sin t}{t}}\right\}=\arctan {\displaystyle \frac{1}{s}}.$(2)

The formula (2) may be determined also directly using the definition of Laplace transform.  Take an additional parametre $a$ to the defining integral

$$ℒ\{\frac{\sin t}{t}\}={\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{\sin t}{t}𝑑t$$

by setting

$${\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{\sin at}{t}𝑑t:=\phi (a).$$

Now we have the derivative  ${\phi }^{\prime }(a)={\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt$,  where one can partially integrate twice, getting

$${\phi }^{\prime }(a)={\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt=\frac{1}{s}-\frac{a^2}{s^2}{\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt.$$

Thus we solve

$${\int }_0^{\mathrm{\infty }}{e}^{-st}\cos atdt=\frac{\frac{1}{s}}{1+{\left(\frac{a}{s}\right)}^2}={\phi }^{\prime }(a),$$

and since  $\phi (0)=0$, we obtain  $\phi (a)=\arctan {\displaystyle \frac{a}{s}}$.  This yields

$${\int }_0^{\mathrm{\infty }}{e}^{-st}\frac{\sin t}{t}𝑑t=\phi (1)=\arctan \frac{1}{s},$$

i.e. the formula (2).

Formula (2) is derived here (http://planetmath.org/LaplaceTransformOfFracftt) in a third way.