I stumbled upon two solutions to the equation L{sin(t)/t} and am having difficult determining which answer is correct, and why.
The two questions and solutions are below:
Let f(t) = L-1{-arctan(s)}
F(s) = -arctan(s)
F'(s) = -1/(s2+1)
Recall L-1{F'(s)} = -t.f(t)
ie. L-1{-1/(s2+1)} = -t.f(t)
-sin(t) = -t.f(t)
f(t) = sin(t)/t
So L-1{-arctan(s)} = sin(t)/t.
2. Find L{sin(t)/t}.
L{f(t)/t} = int(s->inf) F(u)du
f(t) = sin(t)
F(u) = 1/(u2+1)
L{sin(t)/t} = int(s->inf.) [1/(u2+1)]du = [arctan(u)]inf.->s = lim u->inf. arctan(u) - arctan(s) = pi/2 - arctan(s).
--
Both solutions seem correct, but the first suggests L{sin(t)/t} = - arctan(s) and the second suggests L{sin(t)/t} = pi/2 - arctan(s).
If someone could help me out [ie. tell me which one is correct(or incorrect) and why] that would be awesome!! Thanks
The two questions and solutions are below:
- Find L-1{-arctan(s)}.
Let f(t) = L-1{-arctan(s)}
F(s) = -arctan(s)
F'(s) = -1/(s2+1)
Recall L-1{F'(s)} = -t.f(t)
ie. L-1{-1/(s2+1)} = -t.f(t)
-sin(t) = -t.f(t)
f(t) = sin(t)/t
So L-1{-arctan(s)} = sin(t)/t.
2. Find L{sin(t)/t}.
L{f(t)/t} = int(s->inf) F(u)du
f(t) = sin(t)
F(u) = 1/(u2+1)
L{sin(t)/t} = int(s->inf.) [1/(u2+1)]du = [arctan(u)]inf.->s = lim u->inf. arctan(u) - arctan(s) = pi/2 - arctan(s).
--
Both solutions seem correct, but the first suggests L{sin(t)/t} = - arctan(s) and the second suggests L{sin(t)/t} = pi/2 - arctan(s).
If someone could help me out [ie. tell me which one is correct(or incorrect) and why] that would be awesome!! Thanks