I have a question about how this substitution in the integral works. Especially with the dt.
Theorem 6.3.1 If \(\displaystyle F(s)\, =\, \mathcal{L} \left\{f(t)\right\}\) exists for \(\displaystyle s\, >\, a\, \geq\, 0\), and if \(\displaystyle c\) is a positive constant, then:
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\}\, =\, e^{-cs}\, \mathcal{L} \left\{f(t)\right\}\, =\, e^{-cs}\, F(s),\, s\, >\, a.\)
Conversely, if \(\displaystyle f(t)\, =\, \mathcal{L}^{-1} \left\{F(s)\right\}\), then:
. . . . .\(\displaystyle u_c (t) f(t\, -\, c)\, =\, \mathcal{L}^{-1} \left\{e^{-cs} F(s) \right\}.\)
Discussion Theorem 6.3.1 simply states that the transformation of \(\displaystyle f(t)\) a distance \(\displaystyle c\) in the positive \(\displaystyle t\) direction corresponds to a multiplication of \(\displaystyle F(s)\) by \(\displaystyle e^{-cs}.\) To prove Theorem 6.3.1, it is sufficient to compute the transform of \(\displaystyle u_c f(t\, -\, c)\):
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\} \, =\, \displaystyle{\int_0^{\infty}}\, e^{-st} u_c (t) f(t\, -\, c)\, dt\)
. . . . . . . . . . . . . . .. .\(\displaystyle =\, \displaystyle{\int_c^{\infty}}\, e^{-st} f(t\, -\, c)\, dt\)
Introducing a new integration variable \(\displaystyle \xi\, =\, t\, -\, c\), we have:
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\}\, =\, \displaystyle{\int_0^{\infty}}\, e^{-(\xi\, +\, c)s}\, f(\xi)\, d\xi\, =\, e^{-cs}\, \displaystyle{\int_0^{\infty}}\, e^{-s \xi}\, f(\xi)\, d\xi\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\, e^{-cs} F(s)\)
Theorem 6.3.1 If \(\displaystyle F(s)\, =\, \mathcal{L} \left\{f(t)\right\}\) exists for \(\displaystyle s\, >\, a\, \geq\, 0\), and if \(\displaystyle c\) is a positive constant, then:
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\}\, =\, e^{-cs}\, \mathcal{L} \left\{f(t)\right\}\, =\, e^{-cs}\, F(s),\, s\, >\, a.\)
Conversely, if \(\displaystyle f(t)\, =\, \mathcal{L}^{-1} \left\{F(s)\right\}\), then:
. . . . .\(\displaystyle u_c (t) f(t\, -\, c)\, =\, \mathcal{L}^{-1} \left\{e^{-cs} F(s) \right\}.\)
Discussion Theorem 6.3.1 simply states that the transformation of \(\displaystyle f(t)\) a distance \(\displaystyle c\) in the positive \(\displaystyle t\) direction corresponds to a multiplication of \(\displaystyle F(s)\) by \(\displaystyle e^{-cs}.\) To prove Theorem 6.3.1, it is sufficient to compute the transform of \(\displaystyle u_c f(t\, -\, c)\):
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\} \, =\, \displaystyle{\int_0^{\infty}}\, e^{-st} u_c (t) f(t\, -\, c)\, dt\)
. . . . . . . . . . . . . . .. .\(\displaystyle =\, \displaystyle{\int_c^{\infty}}\, e^{-st} f(t\, -\, c)\, dt\)
Introducing a new integration variable \(\displaystyle \xi\, =\, t\, -\, c\), we have:
. . . . .\(\displaystyle \mathcal{L} \left\{u_c (t) f(t\, -\, c) \right\}\, =\, \displaystyle{\int_0^{\infty}}\, e^{-(\xi\, +\, c)s}\, f(\xi)\, d\xi\, =\, e^{-cs}\, \displaystyle{\int_0^{\infty}}\, e^{-s \xi}\, f(\xi)\, d\xi\)
. . . . . . . . . . . . . . . . .\(\displaystyle =\, e^{-cs} F(s)\)
Last edited by a moderator: