# 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, boersma@win.tue.nl
• Content-Length: 1143
• Content-Type: text
• Sender: owner-opsftalk

>From boersma@win.tue.nl Thu Jan 22 13:06 MET 1998
>To: anicin@EUnet.yu
>Subject: Problem in OP-SF NET 5.1

Dear Dr. Anicin,

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.

References
[1] A. Erdelyi, W. Magnus, F. Oberhettinger, & F.G. Tricomi, Tables
of Integral Transforms, Vol. 1, McGraw-Hill, New York, 1954.
[2] M. Abramowitz & I.A. Stegun, Handbook of Mathematical Functions,
Dover Publications, New York, 1965.

Sincerely yours,
John Boersma



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