Solutions to 18.03 Problem Set 7 Part II

31. (a) L(f(at); s) = ∫_{0}^{∞} f(at)e^{−st}dt. Let u = at, so du = adt. Then L(f(at); s) = ∫_{0}^{∞} f(u)e^{−su/a} \frac{du}{a} = \frac{1}{a}F (\frac{s}{a}).

(b) L(e^{(k+iω)t}; s) = \frac{1}{s−(k+iω)} = \frac{(s−k)+iω}{(s−k)^{2}+ω^{2}} , so L(e^{kt} cos(ωt); s) = Re(L(e^{(k+iω)t}; s)) = \frac{s−k}{(s−k)^{2}+ω^{2}} and L(e^{kt} sin(ωt); s) = Im(L(e^{(k+iω)t}; s)) = \frac{ω}{(s−k)^{2}+ω^{2}} .

32. (a) Using the s-derivative rule, L(t sin(ωt); s) = −\frac{d}{ds} (\frac{ω}{s^{2}+ω^{2}} ) = \frac{2ωs}{(s^{2}+ω^{2})^{2}} and L(t cos(ωt); s) = −\frac{d}{ds} (\frac{s}{s^{2}+ω^{2}} ) = \frac{s^{2}−ω^{2}}{(s^{2}+ω^{2})^{2}} . Hence

L (\frac{t}{2ω} sin(ωt); s) = \frac{s}{(s^{2} + ω^{2})^{2}} .

Also L(\frac{1}{ω} sin(ωt); s) = \frac{s^{2}+ω^{2}}{(s^{2}+ω^{2})^{2}} , so

L (\frac{1}{2ω^{2}} (\frac{1}{ω} sin(ωt) − t cos(ωt)); s) = \frac{1}{(s^{2} + ω^{2})^{2}} .

(b) Write f(t) = \frac{e^{at}−1}{t}, and F (s) = L(f(t); s). Then L(tf(t); s) = L(e^{at} −1; s) = \frac{1}{s−a} − \frac{1}{s}. Also L(tf(t); s) = −\frac{dF (s)}{ds} by the s-derivative rule, so \frac{dF (s)}{ds} = \frac{1}{s} − \frac{1}{s−a}. Hence F (s) = ln(s) − ln(s − a) + c = − ln(1 − a/s) + c for some c. We know that lim_{s→∞} F (s) = 0, which will allow us to find c. As s → ∞, − ln(1 − a/s) → − ln 1 = 0, so c = 0.

For the second half, let f(t) = \frac{sin(ωt)}{t}, and let F (s) = L(f(t); s). Then L(tf(t); s) = L(sin(ωt); s) = \frac{ω}{s^{2}+ω^{2}} . Also L(tf(t); s) = −\frac{dF (s)}{ds}, so \frac{dF (s)}{ds} = \frac{−ω}{s^{2}+ω^{2}} . Integrating: F (s) = c − arctan(s/ω) for some c. As s → ∞, arctan(s/ω) → π/2, so c = π/2.

33. (a)

L(t^{2} sin(ωt); s) = Im (L(t^{2}e^{iωt}; s))
= Im (\frac{2}{(s − iω)^{3}} )

Rationalize the denominator by multiplying top and bottom by (s + iω)^{3}:

L(t^{2} sin(ωt); s) = Im (\frac{2(s + iω)^{3}}{(s^{2} + ω^{2})^{3}} )
= Im (\frac{2(s^{3} − 3sω^{2}) + 2i(3s^{2}ω − ω^{3})}{(s^{2} + ω^{2})^{3}})
= \frac{2(3s^{2}ω − ω^{3})}{(s^{2} + ω^{2})^{3}}.

Similarly,

L(te^{t} cos(ωt); s) = Re (L(te^{(1+iω)t}; s))

= Re (\frac{1}{(s − 1 − iω)^{2}} )
= Re (\frac{(s − 1 + iω)^{2}}{((s − 1)^{2} + ω^{2})^{2}} )
= Re (\frac{(s − 1)^{2} − ω^{2} + 2i(s − 1)ω}{((s − 1)^{2} + ω^{2})^{2}} )
= \frac{(s − 1)^{2} − ω^{2}}{((s − 1)^{2} + ω^{2})^{2}} .

Finally, L(t^{1/2}e^{t}; s) = \frac{Γ(3/2)}{(s−1)^{3/2}} = \frac{\sqrt{π}}{2(s−1)^{3/2}} (as Γ(3/2) = \frac{1}{2}Γ(1/2) = \sqrt{π}/2).

(b) The graph of the “bump” is a rectangle with base b and height 1/b (hence area 1). As b → 0, the bump becomes tall and skinny. L(u_{a}; s) = e^{−as}/s, so

L(d(t); s) = \frac{1}{bs}(e^{−as} − e^{−(a+b)s}) = \frac{e^{−as}}{s} (\frac{1 − e^{−bs}}{b} ) .

The limit lim_{b→0}(\frac{1−e^{−bs}}{b}) is by definition the derivative of −e^{−bs} with respect to b at b = 0, which we know is s. Alternatively, by L’Hopital’s rule, lim_{b→0} \frac{1−e^{−bs}}{b} = s. Thus lim_{b→0} L(d(t); s) = e^{−as}.