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If one performs the change of integration variable $$u = tx, \quad du = t\,dx$$ in the defining integral $$\mathrm{Si}\,t = \int_0^{\,t}\frac{\sin{u}}{u}\,du,$$ of the sine integral function, one obtains $$\mathrm{Si}\,t = \int_0^1\frac{\sin{tx}}{tx}t\,dx = \int_0^1\frac{\sin{tx}}{x}\,dx,$$ getting constant limits. We know (see the entry Laplace transform of sine and cosine) that $$\mathcal{L}\left\{\frac{\sin{tx}}{x}\right\} = \frac{1}{s^2+x^2}.$$ This transformation formula can be integrated with respect to the parametre $x$ : $$\mathcal{L}\left\{\int_0^1\frac{\sin{tx}}{x}\,dx\right\} = \int_0^1\!\frac{1}{s^2+x^2}\,dx \,=\, \frac{1}{s}\sijoitus{x=0}{\quad1}\!\arctan\frac{x}{s} \,=\, \frac{1}{s}\arctan\frac{1}{s}.$$ Thus we have the transformation formula of the sinus integralis:
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(1) |
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 $$\mathcal{L}\left\{\frac{d}{dt}\mathrm{Si}\,t\right\} \;=\; s\cdot\frac{1}{s}\arctan\frac{1}{s}-\mathrm{Si}\,0,$$ i.e.
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(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 $\varphi'(a) = \int_0^\infty e^{-st}\cos{at}\,dt$ , where one can partially integrate twice, getting $$\varphi'(a) = \int_0^\infty e^{-st}\cos{at}\,dt = \frac{1}{s}-\frac{a^2}{s^2}\int_0^\infty e^{-st}\cos{at}\,dt.$$ Thus we solve
$$\int_0^\infty e^{-st}\cos{at}\,dt = \frac{\frac{1}{s}}{1+\left(\frac{a}{s}\right)^2} = \varphi'(a),$$ and since $\varphi(0) = 0$ , we obtain $\displaystyle\varphi(a) = \arctan\frac{a}{s}$ . This yields $$\int_0^\infty e^{-st}\,\frac{\sin t}{t}\,dt = \varphi(1) = \arctan\frac{1}{s},$$ i.e. the formula (2).
Formula (2) is derived here in a third way.
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