Let ℒ{𝑓(𝑡)} = 𝐹(𝑠). Show that 𝑓(𝑡) = − 1 𝑡 ℒ −1 {𝐹 ′ (𝑠)} Thus, if we know how to invert 𝐹 ′ (𝑠) then we know how to invert 𝐹(𝑠). Use this information to find the Laplace inverse transform of (i) arctan ( 𝑎 𝑠 ), (ii) ln ( 𝑠+𝑎 𝑠−𝑎 ).
F(s)=∫0∞e−tsf(t)dtF(s)=\int^{\infin}_0e^{-ts}f(t)dtF(s)=∫0∞e−tsf(t)dt
F′(s)=e−tsf(t)F'(s)=e^{-ts}f(t)F′(s)=e−tsf(t)
L−1(F(s))=f(t)L^{-1}(F(s))=f(t)L−1(F(s))=f(t)
L−1(F′(s))=f(t)/(−t)L^{-1}(F'(s))=f(t)/(-t)L−1(F′(s))=f(t)/(−t)
f(t)=−tL−1(F′(s))f(t)=-tL^{-1}(F'(s))f(t)=−tL−1(F′(s))
i)
F′(s)=a1+s2F'(s)=\frac{a}{1+s^2}F′(s)=1+s2a
f(t)=−atsintf(t)=-atsintf(t)=−atsint
ii)
F′(s)=1s+a−1s−aF'(s)=\frac{1}{s+a}-\frac{1}{s-a}F′(s)=s+a1−s−a1
f(t)=−t(e−at−eat)f(t)=-t(e^{-at}-e^{at})f(t)=−t(e−at−eat)
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