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

1. Mar 6, 2009

### 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!

2. Mar 6, 2009

### 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:

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

3. Mar 6, 2009

### 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.

4. Mar 7, 2009

### 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
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.

5. Mar 7, 2009

### HallsofIvy

$$sin(x)= \frac{e^{ix}- e^{-ix}}{2i}[/itex] so [tex]\int sin(x)e^{-sx} dx= \frac{-i}{2}\int e^{(i-s)x}- e^{(-i-s)x}dx$$

Though I think it would be much simpler just to integrate the original form using integration by parts.

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.

6. Mar 7, 2009

### 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.

7. Mar 7, 2009

### DyslexicHobo

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".

Last edited: Mar 7, 2009