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37m
revised What is probability that out of the first half on N objects, none will be matched with their own label?
order relation
53m
revised What is probability that out of the first half on N objects, none will be matched with their own label?
code
58m
comment What is probability that out of the first half on N objects, none will be matched with their own label?
It is a useful exercise and might help the OP in understanding to draw the poset and the Mobius function for the case of set inclusion (case A, size zero is wanted, case B, size $m$ is wanted) and in the case of the divisor poset, where we get the Mobius function from number theory.
1h
comment What is probability that out of the first half on N objects, none will be matched with their own label?
Do you have an extra factor here or am I misreading things?
1h
answered What is probability that out of the first half on N objects, none will be matched with their own label?
1h
comment A sum of Stirling numbers of the second kind
@joriki Generating functions to the rescue.
1h
answered A sum of Stirling numbers of the second kind
2h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
Nice to know that we have a quality answer for the OP now, My answer was written so as to be first.
2h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
I see it now, I understand what you're doing, thanks and sorry for the trouble. Looks fine.
2h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
Just a quick question. On the negative imaginary axis as asked for by the OP we get $$\frac{1}{\exp(-i\pi/2)^2 x^2+1} = \frac{1}{1-x^2}$$ which, when substituted, is not a multiple of the target integral. It's possible that I misinterpreted the question and that OP means to choose the branch of the logarithm only and keep the slot on the positive real axis, which looks difficult as well.
3h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
(+1). Nice work. Let me respectfully point out that the question does look like it asks to test the skill level with complex logarithms. One advantage of your approach is that it introduces the keyhole contour and the residues at the two poles.
20h
revised Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
remark
20h
revised Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
multiparse by maple
23h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
Well for starters one would need to account for the contribution from the singularity at $-i$ which would end up on the branch cut. Let me remind you that there are several experts on the subject at this site who will see your question I hope. I have omitted some of the complex number algebra as it does not add to an understanding of the subject. You need to implement your own logarithm if you are using a CAS or else you will most likely get the values from the classic $[-\pi,\pi)$ branch.
23h
revised Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
wording
23h
comment Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
The quality above is rather middling (written so as to be quick). Yes you do use a different branch, namely the one with the cut on the positive real axis and argument zero to $2\pi.$
23h
revised Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
clarify II
23h
revised Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
ML
23h
answered Evaluation of $ \int_0^\infty\frac{x^{1/3}\log x}{x^2+1}\ dx $
1d
revised How many distinct patterns exist for a 5x5 grid by filling 3 colors?
Burnside