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.

Title	Laplace transform of sine integral
Canonical name	LaplaceTransformOfSineIntegral
Date of creation	2014-11-07 15:36:06
Last modified on	2014-11-07 15:36:06
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	15
Author	pahio (2872)
Entry type	Example
Classification	msc 44A10
Synonym	Laplace transform of sinc function
Related topic	SubstitutionNotation
Related topic	SincFunction
Related topic	TableOfLaplaceTransforms
Related topic	SineIntegral