1. I think this looks really cool. At first glance I have no idea how one would solve it. And no, this isn't homework. I don't have an answer anywhere. It would be cool if someone felt like working it out.

2.

3. Originally Posted by Chemboy
I think this looks really cool. At first glance I have no idea how one would solve it. And no, this isn't homework. I don't have an answer anywhere. It would be cool if someone felt like working it out.

Give me some time, I actually think I can do this one. If I remember correctly I can just apply the limit at the end. I treat n as a constant through the integration.

I could be remembering my Calculus II wrong though. I'll attempt it tomorrow though it's later here.

4. Oh, I need a half angle identity by the way. And the notation in the problem is wrong the ellipses in the integrations are suppose to be between the first and second not the second and third. I know it really doesn't matter but I'm just pointing it out because my calculus professors freak out over little stuff like that.

5. I've reduced it to

Any ideas I think I know the next step but I'm not sure how to do it. I could integrate from there I believe. Not sure exactly how I would do it.

6. I think I just solved it. It's infinity.

I think that makes sense to. look at the graph of all the area is above the x axis. This integral just gives us the area underneath and above the x axis. That value tends towards infinity. were only looking at the space between 0 and 1 that area is finite but summing it up infinitely many times gives you infinity. The math affirms this I will write out the rest of the steps if anyone is interested.

I would like to get one of the good mathematicians of the forum to check the answer first.

Here is my final limit though.

7. The answer is , right?

8. I don't have an answer to it... That's why I posted it, so one of you smart people could take a crack at it.

GenerationE... I think you're getting infinitely iterated integrals mixed up with integrals whose limits go to infinity. And I'm going to trust that it's written correctly, it came from an 8th ed. calculus book...

9. Originally Posted by JaneBennet
Um I don't know I thought it was . Do the integral and see what you come up with. my first guess was actually like 1 or something.

Here is the critical step I may have messed up on. When I finally did the integral of the portion I had reduced it to I said that the integral of

was equal to

then the integral of the 1 became

The reason there is no n term in the sin function at the limit is because when you evaluate the integral the n's cancel.

10. Oh wait I'm so stupid. I said that when evaluated equals n. that actually equals 1 so the limit is different. I'm going to go fix it in the old post.

I just did one step wrong and my answer went from 1 to .

I don't know that notation still doesn't seem right to me. the closest integral to the integrand is evaluated by the first differential. The second closes integral sign to the integrand is evaluated by the second differential from the integrand.

Let me see if I can find a picture of what I am saying. I know it's useless arguing with a book though. It's probably right. But it goes against how I have always treated differentials in integrals.

11. Originally Posted by JaneBennet
The that I have multiplied outside is from the half angle identity I used to simplify the integrand.

12. Originally Posted by Chemboy
I don't have an answer to it... That's why I posted it, so one of you smart people could take a crack at it.

GenerationE... I think you're getting infinitely iterated integrals mixed up with integrals whose limits go to infinity. And I'm going to trust that it's written correctly, it came from an 8th ed. calculus book...

See the first integral sign closest to the 5 in that picture. It is the one that goes with the differential of x. That is what I am saying. it should be the same way for all integration techniques. It doesn't make sense to change the notation.

13. Originally Posted by GenerationE
Originally Posted by JaneBennet
Um I don't know I thought it was .

14. Originally Posted by JaneBennet
Originally Posted by GenerationE
Originally Posted by JaneBennet
Um I don't know I thought it was .
I guess you didn't read any of my other post after that one. The answer is I already determined what I had done wrong, I evaluated one of my terms wrong so it left a n in my limit.

So did you do the integral too?

15. Let First

where

using

where

Next

where

where

So we have where and

The last one will turn out to be

Hence

16. Confirmed my result although using a much more concise method. Oh, well I can't be perfect. I have only been doing calculus a year and a half anyways.

So that was fun, anymore problems for us to look at ChemBoy?

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