This may be too easy, but I find this one surprising.
Compute
To stop this thread from being moved to the second page of the subforum . I would like to
give my solutions to these three integrals I posted previously . Hope more and more wonderful identities will be created here !
Problem :
Let
Show that :
Solution to the fisrt integral :Spoiler:
Solution to the second integral :Spoiler:
Solution to the third integral :Spoiler:
Sure , my idea is similar to yours :
As the period of the integrand is ( here is the same as ) I first partitioned the interval into . For we have which yields so we find that the values of the first summands are the same ( ) . For the last summand , , we know it is less than . Therefore , . Then i used squeeze principle to evaluate the limit .
For the limit, use L'hospital's rule and see that this is equal
For the sum, we first evaluate:
Consider the following polynomial in :
Then we have:
and therefore
Divide both sides by
to get the following identity:
from which we obtain:
where is the 'th harmonic number (setting ).
Substituting and using
together with
we get
Therefore in the proposed problem:
We don't use L'hopital's around these parts! That said, it was so easy that might as well accept that. >
Also, I like your method! It was totally different than what I did! Which was:
Spoiler:
I propose three more problems in increasing difficulty:
This first one is super-easy but try to do it in a non-obvious way. Anyways, it's an interesting result!
Problem 1: Prove that for
Problem 2: For such that compute
Have fun with this one (you won't, unless you 'see it')
Problem 3: Compute
I like your method very much! Although I liked my solution for being elementary, yours is far more elegant.
What is an obvious way for this? The following?
and then substituting and using Newton's binom, gives the result.
For problem 2, it's the same method:
Was this your method? If not, I'm interested to see how you solved it.
I'd be glad to post some problems, but none have come to my mind thus far, and in the last days I haven't encoutered with any challenge problems which would fit this thread. I will, however, keep looking for nice problems to post in here.
Thank you!
I had a mental lapse. There is another problem which is super easy to do using analysis/calculus but has other interesting ways to do it. It's (go ahead and try it) computeWas this your method? If not, I'm interested to see how you solved it.
Although I just realized you can do this by a little trickery similar to what you did for your proof of the previous problem using Retkes's identities.
Here is quite a beautiful one (in my opinion).I'd be glad to post some problems, but none have come to my mind thus far, and in the last days I haven't encoutered with any challenge problems which would fit this thread. I will, however, keep looking for nice problems to post in here.
Problem: Let be the set of all such that and . Compute
where .
I looked up Retkes's identities, but don't understand how you meant to apply them. Care to clarify?
Also, you had a little typo there - the binomial should have been upside down.
Besides the usual proof for this identity, using the derivative of
I've just found a much more elegant (in my opinion) method: let in the summation to get:
Now using
gives the result.
Edit: an easy problem - prove the following generalization:
Edit 2: there was a little mistake here. Fixed it now.
where are the Stirling numbers of the second kind, .
I'd try the other problem you proposed, but I haven't yet covered multivariable integrals .
Try applying Retkes' third identity (the one with the sum of the reciporcals) and recalling that
Of course! The first proof is the one I had in mind and the second one is nice indeed! I did it by double counting.Also, you had a little typo there - the binomial should have been upside down.
Haha, oh well!I'd try the other problem you proposed, but I haven't yet covered multivariable integrals .