Integration of tan(x) from 0 to pi/2, can the one use Jordan-contour to calculate it?
Yes, residue theorem can calculate it in most cases, but the answer will not be in real plane. Since, the solution is most cases will be in complex plane, so expect "i" = sqrt(-1) in your solutions.
Thanks a lot Vikash Pandey, how could you integrate (cot(x))^(n) between 0 and pi/2 by the means of residue or contour?
For n between -1 and 1, you can find the answer in this link. http://mathworld.wolfram.com/Cotangent.html
For other values, it can be tedious to find, since before applying the residue theorem, one needs to have the Laurent series expansion of the given function!
But the answer in that link has a catch, be careful! The answer provided in the link is real and is valid for -1
By the way, when we say R[n] is between -1 and 1, we are supposing that n must be Complex. what if n is real, i.e. the imaginary part is zero, is that equation still correct, i do believe that imaginary part can tend to zero but never equal to it. do you agree?
By the way, when we say R[n] is between -1 and 1, we are supposing that n must be Complex. what if n is real, i.e. the imaginary part is zero, is that equation still correct, i do believe that imaginary part can tend to zero but never equal to it. do you agree?
Nope, you misunderstood it slightly. It implies that the relationship holds even when n is complex with the condition that real part of n is between -1 and 1! It is independent of the value of the imaginary part. The relation is valid for
n = -0.1 + 2.0 i, 0.001+ 100 i, 0.5 - 20 i, 0.2 + 100000 i, 0.001 - 10000000i, 0 + 0.3i,
But the relation is NOT valid for
n = 1.1 + 2.0 i, -1.5, 2.5, 3.5 + 10 i, -2.8 - 100000i.
Hope it clarifies.
In addition to above, one last thing. The answer mentioned in the link is real and valid for -1
in addition to what you have mentioned, we are integrating say (cotx)^0.99 but by suitable contour we are avoiding, x = 0 by a circle of radius e then letting e go to zero. so practically, the real integral remains divergent while the complex integral is valid with an approximation...is it right?
in addition to what you have mentioned, we are integrating say (cotx)^0.99 but by suitable contour we are avoiding, x = 0 by a circle of radius e then letting e go to zero. so practically, the real integral remains divergent while the complex integral is valid with an approximation...is it right?
In the case of (cot(x))^0.99, you are not avoiding anything as per the solution mentioned in the earlier link, since divergence comes into the picture only when |n| >= 1! The answer is true!
The integral with n = 0, 1, 2... is essentially divergent in real plane and so there is none solution. You can find a convergent solution only in complex plane, but then how do you interpret imaginary part of the solution is more or less an open problem!
Alternatively if you are rigid for a real solution, then you have to escape the singularity through a branch-cut. See, https://en.wikipedia.org/wiki/Methods_of_contour_integration