Glasser's solutions to Anicin problems

["Martin Muldoon" <muldoon@mathstat.yorku.ca>]

Bozidar Anicin (anicin@EUnet.yu) posted the following problems in January 1998
and Larry Glasser (laryg@sun.mcs.clarkson.edu) has provided the solutions which
follow.
Martin Muldoon

Problem A.
To prove, disprove or improve the following relation:

2 I_1(a) K_1(a) = \int_0^\infty g(z)cos(2az)dz / \int_0^\infty g(z) dz,

with g(z) = g(k) = (2/k -k) K(k) - (2/k) E(k) where K(k) and E(k) are
complete elliptic integrals of the first and second kind, respectively,
with the modulus k^2 = 1/(1+z^2). I_1 and K_1 are modified Bessel
functions of the first order. The variable z is real, "a" is also real,
but the possibility of complex "a" could also be considered.

Problem B
With the notation introduced in Problem A, to prove, disprove or improve
the following relation
    
        \int_0^\infty g(z)dz = \pi/4.


From: M.L. GLASSER  <laryg@sun.mcs.clarkson.edu>
To: anicin@EUnet.yu
Subject: Problem A
 
Dear Dr. Anicin,
    Your problem is actually equivalent to a well known integral.
Since
    $$\int_0^{\infty}cos(xz)I_1(x/2)K_1(x/2)dx=g(k)$$
(This follows for example from Prudnikov et al. Tables of Integrals
and Series, Vol. 2 (2.16.30.3)). Hence, by Fourier inversion,
$$\int_0^{\infty}cos(2az)g(z)dz=(\pi/2)I_1(a)K_1(a).$$
In particular, for a-->0 you get the solution to problem B.
 
Sincerely,
Larry Glasser