Question

i was given \[\mathcal L \lbrace 3e^{4t} - e^{-2t} \rbrace\]

i got the right answer \[\frac{2s + 10}{(s-4)(s+2)}\]

but it lacked the limits that was supposed to be s>4

so my question is why s>4? how do i find that?

Answer 1

well, you see why not s=4 hopefully
any by definition of the laplace transform s>0
so the question is why is 4>s>0 also not admitted, right?

Answer 2

wait...the laplace transform is ALWAYS s>0?

Answer 3

yes\[\mathcal L\{f(x)\}=\int_0^\infty e^{-st}f(x)dt\]provided that \(s>0\)
if somehow the exponent on the integral is not negative it will not converge
therefor whatever the exponent is, the bounds are such that it must remain negative
for the general case with an unknown function the condition \(s>0) is enough to ensure that the integral converges, but....

Answer 4

ohhh so why isnt it 0<s<4

Answer 5

you have a specific function of the form \(f(x)=e^{at}\) so our transform becomes\[\int_0^\infty e^{at}\cdot e^{-st}dt=\int_0^\infty e^{(a-s)t}dt\]now again, this integral will only converge if the exponent is negative, hence we have that\[a-s<0\implies s>a\]so you have to look to the integral to find the bounds

Answer 6

so how do i figure out what a is?

Answer 7

plug in your function when it is of the form e^a

Answer 8

should be anything ... greater than zero. else your integral is not going to converge.

Answer 9

what do you mean plug in your function when it is of the form e^a? o.O

and @experimentX how would i know htat it's 4 and not -2?

Answer 10

i mean how would i know that it's 4?

Answer 11

whenever you have an exponential function it is easy to check
you have two separate ones on yours:\[\int_0^\infty(3e^{4t})(e^{-st})dt+\int_0^\infty(e^{-2t})(e^{-st})dt\]\[3\int_0^\infty e^{(4-s)t}dt+\int_0^\infty e^{-(2+s)t}dt\]so far do you agree?

Answer 12

hmm yeah...

Answer 13

you can view integration just as sums!!
if what would be sum up to infinity if 4-s is negative
what would be sum if 4-s is positive?

Answer 14

we cannot allow either integral to diverge. we achieve that my making sure the exponents in each integral are always negative

Answer 15

so it becomes s > 4 and s>-2

but since it has to be s> 0 it's just s4?

Answer 16

i mean s>4

Answer 17

... s is assumed to be positive.

Answer 18

just look at the second integral's exponent: \(-(2+s)t\)
since \(s>0\) we know that \(-(2+s)<0\) hence that integral will always converge
we need only worry about the other...

Answer 19

hmm i think im beginning to get the idea

Answer 20

so we say
"ok we want the other integral to converge too, so we need the exponent on that guy to be negative"
that means
4-s<0
s>4

Answer 21

i still lack insight on Laplace transform.

Answer 22

me too, I only have a pdf's worth of understanding
though it makes solving some DE's a breeze!

Answer 23

from analyst point of view ... these things are quite complicated.
but somewhere i remember reading Laplace transforms gives you the frequency view.
I;ve also seen Laplace transforms used to solve definite integral apart from solving IVP DE

Answer 24

\[ \Large \int_0^\infty {\sin x \over x}dx\]