Boersma's solution to Anicin's problem
- To: opsftalk
- Subject: Boersma's solution to Anicin's problem
- From: thk (Tom H. Koornwinder)
- Date: Thu, 22 Jan 1998 21:02:32 +0100 (MET)
- Cc: anicin@EUnet.yu, firstname.lastname@example.org
- Content-Length: 1143
- Content-Type: text
- Sender: owner-opsftalk
>From email@example.com Thu Jan 22 13:06 MET 1998
>Subject: Problem in OP-SF NET 5.1
Dear Dr. Anicin,
Here is a solution of your problem as published by Koornwinder
in the OP-SF NET 5.1, dated January 15, 1998.
Clearly, your function g(z) is related to the Fourier cosine
transform of I_1(a) K_1(a). This transform is listed in
[ 1, form. 1.12(47)]. By inversion of the latter transform one has
2 I_1(a) K_1(a) = (4/pi) \int_0^\infty Q_(1/2)(1+2z^2)cos(2az) dz
and, for a = 0,
1 = (4/pi) \int_0^\infty Q_(1/2)(1+2z^2) dz.
Next, the Legendre function Q_(1/2) (that is, Q with subscript 1/2)
is expressible in terms of elliptic integrals K(k) and E(k) :
Q_(1/2)(1+2z^2) = (2/k - k) K(k) - (2/k) E(k)
where k = (1+z^2)^(-1/2); see e.g. [ 2, form. 8.13.7].
I believe this completely solves your problem.
 A. Erdelyi, W. Magnus, F. Oberhettinger, & F.G. Tricomi, Tables
of Integral Transforms, Vol. 1, McGraw-Hill, New York, 1954.
 M. Abramowitz & I.A. Stegun, Handbook of Mathematical Functions,
Dover Publications, New York, 1965.
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