TAOCP 5.2.2 Exercise 46
We evaluate the integral \[ I = \frac{1}{2\pi i} \int_{s-i\infty}^{s+i\infty} \frac{\Gamma(z)\, n^{-z}}{2^{z-1} - 1} \, dz, \] where \(s\) is a positive integer.
Section 5.2.2: Sorting by Exchanging
Exercise 46. [HM20] What is the value of $(1/2\pi i) \int_{s-i\infty}^{s+i\infty} \Gamma(z) n^{-z},dz/(2^{z-1} - 1)$, given that $s$ is a positive integer and $0 < a < s$?
Verified: no
Solve time: 12m49s
Solution
We evaluate the integral
[ I = \frac{1}{2\pi i} \int_{s-i\infty}^{s+i\infty} \frac{\Gamma(z), n^{-z}}{2^{z-1} - 1} , dz, ]
where (s) is a positive integer. The condition (0 < a < s) (with (a) not appearing in the integrand) is interpreted as requiring the contour to lie to the right of the pole at (z=1); i.e. (s > 1). Since (s) is an integer, this means (s \ge 2).
For (\Re(z) > 1) we have the absolutely convergent geometric series
[ \frac{1}{2^{z-1} - 1} = \frac{2^{1-z}}{1 - 2^{1-z}} = \sum_{k=1}^{\infty} 2^{k(1-z)} = \sum_{k=1}^{\infty} 2^k , 2^{-kz}. ]
On the vertical line (\Re(z) = s) (with (s > 1)) this series converges uniformly, and (\Gamma(z)) decays exponentially as (|\Im(z)| \to \infty). Hence we may interchange sum and integral:
[ I = \sum_{k=1}^{\infty} 2^k \left( \frac{1}{2\pi i} \int_{s-i\infty}^{s+i\infty} \Gamma(z) , (2^k n)^{-z} , dz \right). ]
The inner integral is the inverse Mellin transform of (\Gamma(z)). For any (c > 0) and (x > 0),
[ \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \Gamma(z) , x^{-z} , dz = e^{-x}. ]
Since (s \ge 2 > 0), we may apply this with (x = 2^k n) to obtain
[ I = \sum_{k=1}^{\infty} 2^k e^{-2^k n}. ]
The series converges for all (n > 0) and gives the exact value of the integral.
\boxed{\sum_{k=1}^{\infty} 2^k e^{-2^k n}} ∎