[MOVED] improper integral: int[2pi, infty][sin(x)]dx

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The problem-

. . .integral {2Pi to infinity} [sin(x)] dx

Here is what I did-

. . .integral [sin(x)] dx = -cos(x) + C

I substituted "infinity" with "b", so I have-

. . .-cos(x), from 2Pi to b

. . .[-cos(b)] - [-cos(2Pi)]

. . .[infinity] + 1 = infinity

Which means it is diverging.

is this correct?
The answer is the book says it is diverging, but i'm not sure if i worked it out correctly.

Thank You
 
Hint: Since when does cosine take on values outside the interval -1 to 1?

Eliz.
 
almostacasper said:
I substituted "infinity" with "b", so I have-
Click to expand...
Never say that again. You did NOT substitute for "infinity". You created a proper integral to some finite point, 'b'.

[-cos(b)] - [-cos(2Pi)]
[infinity] + 1 = infinity
Click to expand...
This has me truly baffled. You SAID you substituted 'b' for infinity, but the "infinity" magically reappears. What happened?

[infinity] + 1
Click to expand...
Don't ever write that again.
 
ok,
so since it would be [-cos(b)] and cosine can only be -1 to 1 wouldn't it cancel out and i would be left with -[-cos(2Pi)] wich would be 1. But that would be converging.

ps:
when i said substitute i was just trying to explain what i did. It seemed like the easiest way to explain.
 
ps. I know what you were trying to say. You were not actually saying it. Part of mathematics is being clear. Confusion never is the goal. You do not have to simplify anything for us. Give it your best shot.

Think about the cos(b) just a bit. Does it settle down to a single value as 'b' increases without bound?

cos(10) = -0.839
cos(100) = 0.862
cos(1000) = 0.562
cos(10000) = -0.952

Are we getting anywhere?
 
almostacasper said:
cosine can only be -1 to 1
Click to expand...
No; cosine takes on all values between -1 and 1, not "only" -1 and 1.

The point, by the way, was that, no matter what the argument, the cosine can not evaluate to "infinity", or indeed any value outside the interval between -1 and 1.

almostacasper said:
wouldn't it cancel out
Click to expand...
Wouldn't what "cancel out" with which?

Please clarify. Thank you.

Eliz.
 
so cos(b) would have a limit of -1.
so
-1 + -[-cos(2Pi)] = -1 + 1 = 0
 
almostacasper said:
so cos(b) would have a limit of -1.
Click to expand...
Why? As b tends toward infinity, what does the graph of the cosine look like? Does it really approach -1 and then stay there...?

Eliz.
 
You didn't answer my questions.

Does it settle down to a single value as 'b' increases without bound?

cos(10) = -0.839
cos(100) = 0.862
cos(1000) = 0.562
cos(10000) = -0.952

Are we getting anywhere?
 
I think you answered this question: "Are we getting anywhere?"

However, if you were answering this question: "Does it settle down to a single value as 'b' increases without bound?" we would be getting somewhere.

If it is to have a limit, it must stop jumping around! The cosine function (cos(x))continues to bounce about BETWEEN +1 and -1. It does NOT have a limit as x increases wihtout bound.

The limit does not exist. What does this suggest for the value of your integral?
 
so if the limit does not exist as a finite number, then that means it diverges. If that is the case, then what about the extra 1 i would still have in the equation?
 
There is no "equation"; there is only an integral. There is no "extra 1"; there is only a non-existant limit. Even if you could add 1 to "not a number", you're still going to get "not a number".

Please consider having a serious heart-to-heart with your instructor and/or your academic counsellor, to see about getting caught up on the missing background concepts and content. Good luck!

Eliz.
 
There are at least two criteria you need to consider. You seem to be familiar with one of them.

1) Gets really big and never stops getting big. -- No limit -- You get this one.

2) Won't hold still. -- No limit -- You weren't getting this one.
 
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