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[parent] Laplace transform of sine integral (Example)

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

If one performs the change of integration variable

$\displaystyle u = tx, \quad du = t\,dx$
in the defining integral
$\displaystyle \mathrm{Si}\,t = \int_0^{\,t}\frac{\sin{u}}{u}\,du,$
of the sine integral function, one obtains
$\displaystyle \mathrm{Si}\,t = \int_0^1\frac{\sin{tx}}{tx}t\,dx = \int_0^1\frac{\sin{tx}}{x}\,dx,$
getting constant limits. We know (see the entry Laplace transform of sine and cosine) that
$\displaystyle \mathcal{L}\left\{\frac{\sin{tx}}{x}\right\} = \frac{1}{s^2+x^2}.$
This transformation formula can be integrated with respect to the parametre $ x$:
$\displaystyle \mathcal{L}\left\{\int_0^1\frac{\sin{tx}}{x}\,dx\right\} = \int_0... ...{\!\!\!x=0}^{\,\quad1}\!\arctan\frac{x}{s} \,=\, \frac{1}{s}\arctan\frac{1}{s}.$
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)

Laplace transform of sinc function

By the formula $ \mathcal{L}\{f'\} = s \mathcal{L}\{f\}-\lim_{x \to 0+}f(x)$ of the parent entry, we obtain as consequence of (1), that

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

$\displaystyle \mathcal{L}\{\frac{\sin{t}}{t}\} = \int_0^\infty e^{-st}\,\frac{\sin t}{t}\,dt$
by setting
$\displaystyle \int_0^\infty e^{-st}\,\frac{\sin{at}}{t}\,dt := \varphi(a).$
Now we have the derivative $ \varphi'(a) = \int_0^\infty e^{-st}\cos{at}\,dt$, where one can partially integrate twice, getting
$\displaystyle \varphi'(a) = \int_0^\infty e^{-st}\cos{at}\,dt = \frac{1}{s}-\frac{a^2}{s^2}\int_0^\infty e^{-st}\cos{at}\,dt.$
Thus we solve
$\displaystyle \int_0^\infty e^{-st}\cos{at}\,dt = \frac{\frac{1}{s}}{1+\left(\frac{a}{s}\right)^2} = \varphi'(a),$
and since $ \varphi(0) = 0$, we obtain $ \displaystyle\varphi(a) = \arctan\frac{a}{s}$. This yields
$\displaystyle \int_0^\infty e^{-st}\,\frac{\sin t}{t}\,dt = \varphi(1) = \arctan\frac{1}{s},$
i.e. the formula (2).

Formula (2) is derived here in a third way.



"Laplace transform of sine integral" is owned by pahio.
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See Also: substitution notation, sinc function, table of Laplace transforms, sine integral

Other names:  Laplace transform of sinc function

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Cross-references: derivative, Laplace transform, consequence, sinus integralis, parametre, Laplace transform of sine and cosine, function, sine integral, variable
There are 2 references to this entry.

This is version 10 of Laplace transform of sine integral, born on 2008-05-14, modified 2008-08-10.
Object id is 10587, canonical name is LaplaceTransformOfIntegralSine.
Accessed 782 times total.

Classification:
AMS MSC: 44A10 (Integral transforms, operational calculus :: Laplace transform)
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