Hi CRGreatHouse,
In your post 1673, the summation on LHS runs from n=2 to infinity, (n=2,3,4,5,6,7,8,...)
But the summation on RHS runs over all primes.(p=2,3,5,7,...)
From the definition of von Mangoldt function,when n=6,10,12,14,15,18,... , the summand
became (-1/n) whenever n...
when n is a prime or prime power, the summation is okay.
but suppose when n=6, then the sum is (von Mangoldt(6) -1)/6 , which is = -1/6,
as n runs from 2 to infinity,can we settle the problem of convergency or divergency?
-Ng
For the first integral ,
it can be shown to be = Gamma''(1) and
- Euler's Constant = Gamma'(1)
where Gamma(x)=Gamma Integral
I just don't know how to use the above facts.
I have tried many hours on the following integrals and would appreciate any help from you.
1. Int{x=0 to infinity}(exp(-x)*Ln(x)*Ln(x)dx)
2. Int{x=0 to Infinity}(exp(-x*x)*Ln(x)dx)
Any idea guys?
The beautiful point about this calculation is that it is applicable for all positive integers.
Define I(n)=Int((log(sin(x)))^n, {x=0 to pi/2}) then it can be shown that
I(0) = pi/2
I(1) = -(pi/2)*log(2)
I(2) = (pi^3)/(24) + (pi/2)*(log(2))^2
and I(3) is a function of...
when the angle changed from 0 to pi/2 ,you DON'T get the whole curve!
Try drawing out the curve.Note, cos(5*angle) must be > 0 to give a real point on the curve.
when angle=0, r=3
when angle=18 degree, r=0
hope this is helpful
Reading through "Ramanujan's notebook Part 2" and "Collected Papers of Ramanujan ",
I chanced upon an entry which solved my problem beautifully.
I just wish that much more can be learned from Ramanujan's work.
-mathslover
I would like to find the definite integral of (log(sin(x)))^2 under the interval (pi/2,0)
Numerical integration can only give a numerical answer
I would like to find the above integral in terms of well-known constants
-Mathslover