Question

@TuringTest i need some help...

Answer 1

\[\huge \mathcal L \left\{ \frac{\sin kt}{t} \right \}\]

Answer 2

you wanna derive it from the definition?

Answer 3

if possible yes. or maybe a shortcut formula or something... i know i cant use gamma functions and make it into \[\mathcal L \{t^{-1} \sin kt \}\]

Answer 4

Help plz?

Answer 5

i also have no idea how to do \[\huge \int \limits_0^\infty e^{-st} \left(\frac{\sin kt}{t} \right) dt\]

Answer 6

I just found the following:\[\mathcal L\left\{{f(t)\over t}\right\}=\int_s^\infty F(u)du\]now how to implement that... I gotta think
it's a new formula for me

Answer 7

wonder what F(u) means...

Answer 8

yeah, specifically the u part
F(s) would be the Laplace of f(t)

Answer 9

but what would we put for u I wonder...

Answer 10

for some odd reason wolfram gave me \[\tan^{-1} \left(\frac ks \right)\]

Answer 11

\[\mathcal L\{\sin kx\}={k\over s^2+k^2}=F(s)\]so I guess\[\mathcal L\left\{{f(t)\over t}\right\}=\int_s^\infty {k\over s^2+k^2}du\]means \[u=kx\]???
I'm sorry, I need to investigate that

Answer 12

hmm so \[\mathcal L \left \{\frac{f(t)}{t} \right \} = \int \limits _0^\infty (\mathcal L \{f(t)\})dt\]

Answer 13

so it's \[\mathcal L \{\mathcal L \{f(t) \} \} \]

Answer 14

yeah, but \[\mathcal L\{f(t)\}=F(s)\]and the form is\[\mathcal L \left \{\frac{f(t)}{t} \right \} = \int \limits _0^\infty F(u)du\]and so why the heck is it u not s...?

Answer 15

u is a dummy variable, to replace the s (so you don't confuse the limit)

Answer 16

so F(u) is just like F(s)?

Answer 17

I would replace s with u in your k/(s^2+k^2)

Answer 18

wolfram didnt agree though :C

Answer 19

normally you would use s, but here, the lower limit is s, so to avoid confusion, people rename s to u

Answer 20

even if i remove the limits it says \[\int \frac{k}{s^2 + k^2} ds = \tan^{-1} (\frac sk)\]
but it should be \[\tan^{-1} (\frac ks)\]

Answer 21

maybe we raise s/k to -1 because that's the power of t? lol

Answer 22

that integral is right
u=(ks)tan(theta)

Answer 23

\[\mathcal L \left \{ \frac{\sin kt}{t} \right \}\] http://www.wolframalpha.com/input/?i=laplace+transform&a=*C.laplace+transform-_*Calculator.dflt-&f2=sinkt%2Ft&f=LaplaceTransformCalculator.transformfunction_sinkt%2Ft&f3=t&f=LaplaceTransformCalculator.variable1_t&f4=s&f=LaplaceTransformCalculator.variable2_s
\[\int \frac{k}{s^2+k^2}ds\] http://www.wolframalpha.com/input/?i=int+k%2F%28s%5E2+%2B+k%5E2%29ds

Answer 24

they show different :C

Answer 25

If you put in the limit, sage gives

1/2*pi - arctan(s/k)
which I think is the same as
arctan(k/s)

Answer 26

can we just say\[u=ks\implies du=kds\implies\frac{du}k=dx\]so\[\int_s^\infty F(u)du=\frac1k\int{\over}{ds\over{(\frac sk)^2}+1}\]

Answer 27

wolf agrees with this^

Answer 28

hmm im lost...

Answer 29

can we start from the beginning again?

Answer 30

If you put in the limits, it works

\[ \int \frac{k}{u^2 + k^2} ds = \tan^{-1} (\frac uk) \]
for u= s to oo
you get
atan(inf)- atan(s/k)
= atan(k/s)

Answer 31

why ds?

Answer 32

cut and paste problem. du

Answer 33

so you're putting u as s right?

Answer 34

u is a dummy variable of integration
(s or u does not matter)
but when you plug in the limit s, you may lose track of which s you are looking at.

Answer 35

okay... i get tan^-1 (u/k) what happens after that?

Answer 36

okay... i get tan^-1 (u/k)
don't you mean
tan^-1(k/s)

Answer 37

no..i meant i get how you got tan^-1 (u/k) what did you do next?

Answer 38

put in the limits u goes from s to infinity

Answer 39

what does that mean?

Answer 40

just like integral x dx from x=0 to 1: x^2/2 from x=0 to 1: 1/2 -0= 1/2

Answer 41

well, we got\[\mathcal L\{\sin kt\}={k\over s^2+k^2}=F(s)\]which implies that if we want to call this \(F(u)\) then\[u=s\implies du=ds\]\[\mathcal L\left\{{f(t)\over t}\right\}=\int_s^\infty F(u)du=\int_s^\infty {k\over u^2+k^2}du\]\[=\left.\tan^{-1}(\frac uk)\right|_s^\infty=\frac\pi2-\tan^{-1}(\frac sk)\]ok, what did I do wrong?

Answer 42

@TuringTest looks good
but 90 - atan(s/k) = atan(k/s)

Answer 43

oh shnap, I totally forgot that!
thanks phi :)

Answer 44

wiat... is that \[\Large \int \limits_s^\infty\] or \[\Large \int \limits_0^\infty\]

Answer 45

\[\int_s^\infty\]

Answer 46

...still not seeing

Answer 47

starting from s is a result if the /t part...

Answer 48

oh

Answer 49

I don't know how to derive that :/

Answer 50

no wonder im getting wrong answers from wolfram

Answer 51

well it looks like between us we have wolf's answer, so I think we got it
did you enter the integral from 0 ?

Answer 52

so how come \[90 - \tan^{-1} (\frac sk) = \tan^{-1} (\frac ks)\]
and yes @TuringTest

Answer 53

erm, it's an identity, let me look at a unit circle for a minute
brb

Answer 54

one way is notice that if one angle of a right angle is A, the other is 90-A
and tan(A)= 1/tan(90-A)

Answer 55

so in generalization... \[\Large \mathcal L \left \{ \frac{f(t)}{t} \right \} = \int \limits_s^\infty F(u) du\]
then F(u) is just F(s) but replace s with u

Answer 56

did i say that right?

Answer 57

tan A= y/x --> A = atan(y/x)
tan (90-A)= x/y --> 90-A= atan(x/y)

Answer 58

yes

Answer 59

cool. i'll get some practice for now then