Question

Laplace transform of a function f(x) is function (Lf)(s) of the variable s defined by the improper integral (Lf)(s)= (definite integral [0,infinity)) f(x)e^(-sx) dx
"s" is a constant and let "a" be a real number. Find:
1.) the sine function f(x)=sin(ax)
2.)The exponential function f(x)=e^(ax). assume s>a
3.) the cosine function f(x)=cos(ax). assume s>a

Answer 1

Integration by parts for all of these integrals.

Answer 2

Which one are you stuck on? All of them? :o

Answer 3

well, I never did them before so if i can see one I can try to teach from the notes. I did a basic one with a constant time e^-sx and did that fine but these are a little more in depth.

Answer 4

*constant times e^-sx

Answer 5

Were you able to get through the constant * e^-sx without too much trouble?
Because part b will work out almost exactly the same way, but with a minor change.

Answer 6

yes

Answer 7

\[\LARGE \mathscr{L}\left\{e^{ax}\right\}\quad=\quad \int\limits_0^{\infty}e^{ax}\cdot e^{-sx}dx\]

Answer 8

Rules of exponents lets us write it like this:\[\LARGE \mathscr{L}\left\{e^{ax}\right\}\quad=\quad \int\limits\limits_0^{\infty}e^{-sx+ax}dx\]We'll further simplify the exponent to:\[\LARGE =\quad \int\limits\limits\limits_0^{\infty}e^{-(s-a)x}\;dx\]

Answer 9

So when you evaluated,\[\LARGE =\quad \int\limits\limits\limits\limits_0^{\infty}e^{\color{orangered}{-s}x}\;dx\]What did you come up with? :)

Answer 10

-(e^(-sx))/s

Answer 11

bound [0,ininity)

Answer 12

Ok good, and evaluating it at the limits should simplify it down really nicely.

Answer 13

pulled out 1/-s cause constant

Answer 14

for the others i would use IPB?

Answer 15

Others? :o
Wait wait, I'm confused, what did you actually end up with for the laplace of 1

Answer 16

You have to "plug in" the limits and it should simplify down.
It's not -1/s, but that's close D:

Answer 17

oh, i know that. The one problem that i finished before is part 4, I did not list it because I was able to solve it. The problem you just helped me with, i did not think of the exponent law and was trying to IPB. I wanted to know if i would use IPB for f(x)=sin(ax) and fx=cosax.

Answer 18

would my u be e^(-sx) and dv be sin(ax) dx?

Answer 19

It'll be a little tricky, we'll have to do integration by parts `two times` and then do some fancy algebra.

It doesn't matter which you choose to be `u` and `dv`, we'll just have to make a note of it for the next time we do our parts, because we have to do them in the same manner we did the first time.

If that doesn't make sense, that's ok. :)
Yes, let u=the exponential and dv=the trig function.

Answer 20

Are you familiar with Euler's Formula?
There is a way to do this that is a bit easier using the complex exponential.
We would convert our sine or cosine to exponentials using Euler and then the problem would work out similar to part 2.

But if you haven't learned about that stuff yet, np. :)

Answer 21

sorry, I never learnt it yet

Answer 22

So we're working on the sine problem right?
After integrating once, I think we end up with this.\[\Large -\frac{1}{s}\sin(ax)e^{-sx}-\frac{a}{s}\int\limits \cos(ax)e^{-sx}\;dx\]
If I'm moving too fast, you can let me know.
I don't wanna take the fun away if you're doing this on paper, hehe.

Answer 23

yeah, I'm on paper cause I have to write this problem up in a 8 page explanation on how it is done.

Answer 24

na, it's a good speed

Answer 25

Don't worry about the limits for now, we'll take care of the whole mess at the end.

Answer 26

So ummm...

Answer 27

Ahh crap I did u=trig function, dv=exponential :(

So yours will end up looking different.
My bad, lemme fix that.

Answer 28

i was just going to ask about that

Answer 29

ha

Answer 30

So if you do it the way you had suggested, I think we come up with:\[\Large =-\frac{1}{a} \cos(ax)e^{-sx}-\frac{s}{a}\int\limits\limits \cos(ax)e^{-sx}\;dx\]

Answer 31

yes i got that

Answer 32

Ok this is the part where we need to be careful.
We're going to do IBP again on the integral that is remaining.

Remember how we chose to let, `u=exponential` , `dv=trig function` ?
We need to that here again.

If we chose them in the reverse order, we would simply undo our last IBP, which is no good :p

Answer 33

got it

Answer 34

\[\Large =-\frac{1}{a} \cos(ax)e^{-sx}\color{orangered}{-\frac{s}{a}\int\limits\limits\limits \cos(ax)e^{-sx}\;dx}\]So I guess we'll end up with:\[\large =-\frac{1}{a} \cos(ax)e^{-sx}\color{orangered}{-\frac{s}{a}\left[\frac{1}{a}\sin(ax)e^{-sx}+\frac{s}{a}\int\limits \sin(ax)e^{-sx}\right]}\]

Answer 35

good. is this one of the problem were we set the original integral equal to a variable and since integral that is in red bottom is the same we sub in that variable and solve?

Answer 36

Yes, it's one of those things :)

Answer 37

like w= bla bla bla +s/a w

Answer 38

nice

Answer 39

exactly.
But don't forget to `distribute` that -s/a in front of the square brackets.
So we really have -(s^2/a^2) W on the right there.

Answer 40

okay

Answer 41

So as you mentioned, we end up with:\[\Large W\quad=\quad=-\frac{1}{a} \cos(ax)e^{-sx}-\frac{s}{a^2}\sin(ax)e^{-sx}-\frac{s^2}{a^2}W\]

Answer 42

\[\Large W+\frac{s^2}{a^2}W\quad=\quad W\left(1+\frac{s^2}{a^2}\right)\]Then we'll just divide by that coefficient on W, right? :)

Answer 43

that big mess 2 post above with out the end term of W divided by (1+s^2/a^2).

Answer 44

Oh I guess getting a common denominator would be good :P heh\[\Large W\left(\frac{s^2+a^2}{a^2}\right)\]

Answer 45

what is the purpose of the common denominator?

Answer 46

It makes it MUCH easier to divide it over to the other side.
Dividing the fraction is the same as multiplying the other side by:\(\Large \left(\cfrac{a^2}{s^2+a^2}\right)\)

Answer 47

okay

Answer 48

See how the a's on cosine and sine will cancel out nicely when we do that? :o

Answer 49

yes

Answer 50

\[\Large W\quad=\quad -\frac{a}{s^2+a^2}\cos(ax)e^{-sx}-\frac{1}{s^2+a^2}\sin(ax)e^{-sx}\]

Answer 51

Mmmmm k, giving us something like that.

Answer 52

Make sure you understand up to that point.
Our next step is to FINALLY "plug in" our limits of integration.

Answer 53

I fine up to now

Answer 54

b-a

Answer 55

It looks like the upper limit of integration will be producing 0 for both terms because:\[\LARGE \lim_{x\to\infty}e^{-sx}\quad=\quad 0\]
And we have that exponential appearing in both terms.

Answer 56

Hmm you might want to be a little bit more rigorous though in that part.
Like maybe mention something about the fact that sine and cosine are bounded by -1 and 1 so we don't need to worry about an indeterminate form.

I dunno, maybe something like that D:

Answer 57

We `subtract` the function evaluated at the lower limit.
See how each term already has a negative in front?
So we're subtracting each of those negative terms, so they'll change to positive, leaving us with:\[\Large \frac{a}{s^2+a^2}\cos(0)e^{0}+\frac{1}{s^2+a^2}\sin(0)e^{0}\]

Answer 58

so a/s^2 + a^2

Answer 59

Yayyy good job team! \c:/

Answer 60

thanks. And how would the (assume s>a) come into play for the #2 when you showed me?

Answer 61

It applies the same way it does to integrating constants, so make sure you understand that little detail.\[\Large \mathscr{L}\{1\}\quad=\quad \int\limits\limits\limits_0^{\infty}e^{-sx}\;dx\quad=\quad -\frac{1}{s}e^{-sx}|_0^{\infty}\]\[\Large =\quad -\frac{1}{s}\left[\color{red}{\lim_{b\to\infty}e^{-sb}}-e^{0}\right]\]

Answer 62

Oh so for the constant case, we say that s > 0.
If s is allowed to be negative, then this red part, our upper limit, would not converge to 0.

Because we would have e^{+infinity} which blows up.

Answer 63

The same thing happens with the s>a problem.
We need the coefficient in front of b to remain negative.
If it switches to positive we have a problem.

Answer 64

okay. And for the f(x)= cosax it is basically what we just did right just working with cos

Answer 65

Yah it'll be the exact same process.
We'll just end up different coefficients in front of each term along the way.

Answer 66

okay. Thanks for all of the help. You really saved me with this.

Answer 67

np \c:/

Answer 68

oops ha repost

Answer 69

thanks again and good night. I have a ton of writing to do ha