Need help coding with TI-89. Int{sin(x),x,0,inf}

[DyslexicHobo]

I'm trying to code a laplace transform function into my calculator. Right now, I'm doing it by definition:
\int{e^{-st}*f(t)dt} from 0 to \infty

But whenever I try to use a transcendental function, it loops infinitely. I figured this is because the calculator cannot evaluate \int{sin(x)dx} from 0 to \infty.

I'm not exactly sure how to the laplace transforms are derived for sin(x) and cos(x) using the definition, so I can't even code some sort of if statement to use a separate derivation.

Can anyone help? Thanks!

 

[confinement]

You need to write the sines and cosines as imaginary exponentials and then simplify to the familiar results as is done on this page:

http://cnyack.homestead.com/files/alaplace/lapfun1.htm

For your program this means you need to detect cases involving sines and cosines and have them automatically rewritten as exponentials.

 

[HallsofIvy]

Well, certainly, the TI-89 or any other calculator won't do it- that integral doesn't exist! The Laplace transform of sin(x) is, just as you state,
\int_0^\infty e^{-sx}sin(x)dx
and it is that "e^{-sx}" that makes it converge.

 

[DyslexicHobo]

Oh sorry. I should have been more explicit. I'm defining s>0. That makes e^{-s*t} converge, correct?

Or were you talking about sin(x) not converging as x->infinity?

I understand how to resolve imaginary numbers into their real and imaginary parts (we needed to learn this in order to understand some methods for solving differential equations), but I don't understand how to

write the sines and cosines as imaginary exponentials and then simplify to the familiar results

I'm really disappointed how little time we get to spend on the derivations of a lot of the formulas we use. I'm an engineering major so they cram Calc 2, calc 3 (linear algebra), diff eq into 2 semesters.

 

[HallsofIvy]

sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/itex] <br /> so <br /> \int sin(x)e^{-sx} dx= \frac{-i}{2}\int e^{(i-s)x}- e^{(-i-s)x}dx<br /> <br /> Though I think it would be much simpler just to integrate the original form using integration by parts.<br /> <br /> But I do not understand why you are saying that you TI89 will not do that. Certainly what you put in your title, Int{sin(x),x,0,inf}, the TI89 cannot do because it does not exist, but with the exponential, it does exist.

 

[DyslexicHobo]

Using integration by parts, won't the calculator eventually try to evaluate sin(\infty) and return undefined?

But if I have a statement in the code that converts sin(x) to \frac{e^{ix}-e^{-ix}}{2i} then evaluate the limit after I integrate, it will work?

Edit: Oh, I see how it works! Thanks for your help. I completely forgot about that step in the process of resolving sin(x) into its real and imaginary components.

 

[DyslexicHobo]

Well, certainly, the TI-89 or any other calculator won't do it- that integral doesn't exist! The Laplace transform of sin(x) is, just as you state,
\int_0^\infty e^{-sx}sin(x)dx
and it is that "e^{-sx}" that makes it converge.

I don't think I understand you here. We learned that L\{sin(bx)\} = \frac{b}{s^2+b^2}

Is that not a mathematically sound definition? The class is "Math for engineering analysis" so I guess the teacher feels that we only need to know HOW to do it as opposed to how it's derived. It leaves me with a lot of questions, though. :(Edit: Oh, I see where some of the confusion could have been coming from. I meant to say "trigonometric" in the first post, not "transcendental".