# Laplace transform of sine integral

## 0.1 Derivation of $\mathcal{L}\{\mathrm{Si}\,t\}$

If one performs the change of integration variable

 $u\;=\;tx,\qquad du\;=\;t\,dx$

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

 $\mathrm{Si}\,t\;=\;\int_{0}^{\,t}\!\frac{\sin{u}}{u}\,du,$
 $\mathrm{Si}\,t\;=\;\int_{0}^{1}\frac{\sin{tx}}{tx}t\,dx\;=\;\int_{0}^{1}\frac{% \sin{tx}}{x}\,dx,$

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

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

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

 $\mathcal{L}\left\{\int_{0}^{1}\frac{\sin{tx}}{x}\,dx\right\}\;=\;\int_{0}^{1}% \!\frac{1}{s^{2}+x^{2}}\,dx\;=\;\frac{1}{s}\operatornamewithlimits{\Big{/}}_{% \!\!\!x=0}^{\,\quad 1}\;\arctan\frac{x}{s}\;=\;\frac{1}{s}\arctan\frac{1}{s}.$

Thus we have the transformation formula of the sinus integralis:

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

## 0.2 Laplace transform of sinc function

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

 $\mathcal{L}\left\{\frac{d}{dt}\mathrm{Si}\,t\right\}\;=\;s\cdot\frac{1}{s}% \arctan\frac{1}{s}-\mathrm{Si}\,0,$

i.e.

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

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

 $\mathcal{L}\{\frac{\sin{t}}{t}\}\;=\;\int_{0}^{\infty}e^{-st}\,\frac{\sin t}{t% }\,dt$

by setting

 $\int_{0}^{\infty}e^{-st}\,\frac{\sin{at}}{t}\,dt\;:=\;\varphi(a).$

Now we have the derivative  $\varphi^{\prime}(a)=\int_{0}^{\infty}e^{-st}\cos{at}\,dt$,  where one can partially integrate twice, getting

 $\varphi^{\prime}(a)\;=\;\int_{0}^{\infty}e^{-st}\cos{at}\,dt\;=\;\frac{1}{s}-% \frac{a^{2}}{s^{2}}\int_{0}^{\infty}e^{-st}\cos{at}\,dt.$

Thus we solve

 $\int_{0}^{\infty}e^{-st}\cos{at}\,dt\;=\;\frac{\frac{1}{s}}{1+\left(\frac{a}{s% }\right)^{2}}\;=\;\varphi^{\prime}(a),$

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

 $\int_{0}^{\infty}e^{-st}\,\frac{\sin t}{t}\,dt\;=\;\varphi(1)\;=\;\arctan\frac% {1}{s},$

i.e. the formula (2).

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

Title Laplace transform of sine integral LaplaceTransformOfSineIntegral 2014-11-07 15:36:06 2014-11-07 15:36:06 pahio (2872) pahio (2872) 15 pahio (2872) Example msc 44A10 Laplace transform of sinc function SubstitutionNotation SincFunction TableOfLaplaceTransforms SineIntegral