I saw this one and found a cool solution, whats yours?
Let's start by the typical solution:
We also have:
Since each of those integrals exists separately:
We have: where the Remainder is (see the Remainder of Taylor's Polynomials)
Since : thus:
Therefore the extra term disappears as n tends to infinity:
I may as well give my two cents. I will use the double integral we're all so fond of. I looked around to see if the Kriz may have done this at some time or another, but I couldn't find anything, so I assume not.
An identity we can use is:
So, we can use our double integral thing like so:
Now switch em':
You know, there is a little something that is bothering me. Please straighten me out if need be.
While ding this I noticed something. Now, assuming the limit above is 0 we are in good shape, but isn't it undefined?. I supose as long as . Maybe I am wrong, but something about that is bothering me an little.
As a check I ran it through my 92 and it said undefined. Yet we should get .
I am just a little iffy about this. I reckon if we want it to be 0 we can. That's what I say., yet
Then we have
Now if we had that
Where that would be indeterminate but we have
Now this limit is still a number...a non zero number albeit almost zero so we know that no matter how small it gets it never actually reaches zero...therefore