math

[antstastegood]

Ok, so I just got out of a math midterm (line integrals, Green's theorem), which would have been nearly perfect, except for one part of one problem.

One part of my work came down to this integral:


I tried Integration by Parts, and all the other tricks in the book, and got nowhere. It just looks too darn simple to be impossible, though. It would really put my mind at rest if I knew whether or not that integral can be found, so I'll know whether it was a flaw in my setup to get to that point.

Thanks, and sorry about posting OT,

ants

[elendil]

Thanks, and sorry about posting OT

What's that about???? Hey, watch this! I'm gonna pull an OT thread right outa my...bag of tricks.

[NicoMoreno]

at first glance, it looks like you are missing part of it. You definitely can't do it by parts though... there is only one term.... although... come to think of it

try integration by parts twice, I have a sneeking suspition that you eventually get the same thing in which case you can equate the integral to something...

Let me see, where did I put my calculus brain...

oh that's right... I usd it up this morning.

What year/stuff of calculus is it? I am having the time of my life with ODEs and DEs and DEs that use series... oh boy....

[Ridseard]

The substitution u=exp(t) changes it to the integral from 1 to e of (1/u)cos(u)du. The cosine integral Ci(x) is the integral from infinity to x of (1/u)cos(u)du, so the value of the original integral would be Ci(e)-Ci(1), which is approximately -0.11732.

There might be some trick way of evaluating it directly, but I don't see it right now.

[Ridseard]

I don't think parts will works because it gets nasty. Taking u=cos(exp(t)) and dv=dt,

du=-sin(exp(t))exp(t)dt
v=t

So integral[cos(exp(t))dt]= tcos(exp(t)) - integral[texp(t)(-sin(exp(t)))dt
= tcos(exp(t)) + integral[texp(t)sin(exp(t))dt]

Now then, the thing is to do parts again on that last integral and hope we can bootstrap. Hmmmm, not sure how to break it up. Lemme think on it awhile.

No. It's hopeless. I think that unless you use the Ci function, it has no closed form evaluation. I used to have the full Maple on this machine, but I foolishly deleted it to get more drive space. Damn!

[antstastegood]

Yeah, I just found out before I saw your updated post that it aint gonna work. Sure looked for a while like it would, taking for that last integral u=t and dv= e^t sin e^t dt.

But when I wrote out all the terms, everything cancelled out.

Anyway, the backstory is, that it was a question on line integrals, that called for four separate integrals. (The class has covered line integrals and Green's theorem so far.) Parametrizing in terms of t gave me that integral for one of them. Must have been a screw-up in the parametrizing.

Thanks for all your help, folks.

[Ridseard]

Anyway, the backstory is, that it was a question on line integrals, that called for four separate integrals. (The class has covered line integrals and Green's theorem so far.) Parametrizing in terms of t gave me that integral for one of them. Must have been a screw-up in the parametrizing.

I don't suppose you recall the original problem, do you? I love line integrals, Green's theorem, and all that stuff.

[antstastegood]

Anyway, the backstory is, that it was a question on line integrals, that called for four separate integrals. (The class has covered line integrals and Green's theorem so far.) Parametrizing in terms of t gave me that integral for one of them. Must have been a screw-up in the parametrizing.

I don't suppose you recall the original problem, do you? I love line integrals, Green's theorem, and all that stuff.

I can't remember the given function, but the given info was as though you had used Green's theorem, and were going to turn it into a quick double integral and be done with it, but the question asked us to turn it back into a line integral, and evaluate. The curve to integrate over had to be broken into four integrals (it was a square region), and setting up one of them apparently went sour. Remind me, sometime, and I'll dig it up after I get the test back.

For your enjoyment, straight from the textbook:

evaluate the line integral of

2xsin(y)dx + [(x^2)(cos y) - 3y^2]dy

where C is any path from (-1, 0) to (5, 1)

[Tyghress]

And I thought the shop tool talk was hot. . .oh my. . .integrals and theorems, calculus.

That term just rolls off the tongue. . .cal que luss. . .assonant with delicious.

"Parameterizing". . .mmmm...how it uses lips and jaws. . .tongue and palate....all those wonderful textures. Pah ram met tr eyes sing. . . .

Oh, do go on! <sprawling out and waiting for more>

[emmline]

While I love theoretical physics...this is exactly why I couldn't be a physicist. But you know, Einstein got somebody else to do the really hard math, too.

[Ridseard]

For your enjoyment, straight from the textbook:

evaluate the line integral of

2xsin(y)dx + [(x^2)(cos y) - 3y^2]dy

where C is any path from (-1, 0) to (5, 1)

Well, it's path independent (the integrand is an exact differential). I did it first going from (-1,0) to (5,0) then (5,0) to (5,1) and got 25sin(1)-1. Then I checked it on the path from (-1,0) to (-1,1) then (-1,1) to (5,1) and got the same answer.

[cowtime]

I've gotta show this to my geophysicist daughter. She thinks math is "fun" (shudder). How I produced such a child is beyond me. Math=horror/torture as far as I'm concerned.

[TelegramSam]

I used to know how to do calculus last year. But now I've forgotten it all. How depressing...

[littlejohngael]

So integral[cos(exp(t))dt]= tcos(exp(t)) - integral[texp(t)(-sin(exp(t)))dt= tcos(exp(t)) + integral[texp(t)sin(exp(t))dt]

Now try playing it with a Generation Eb!

Little John

[Jerry Freeman]

Taking u=cos(exp(t)) and dv=dt,

du=-sin(exp(t))exp(t)dt
v=t

So integral[cos(exp(t))dt]= tcos(exp(t)) - integral[texp(t)(-sin(exp(t)))dt
= tcos(exp(t)) + integral[texp(t)sin(exp(t))dt]

Ah, cool!

So this is the new Irish traditional music notation, complete with all the ornaments?

Ridseard, you're a genius. The ITM world will forever be in your debt.

Best wishes,
Jerry