Anicin' problem

[Victor Adamchik <victor@wolfram.com>]

Here is a brief idea. 

1. apply the Mellin integral transform wrt 'a' to

f(a) = \int_0^\infty g(z)cos(2 a z)dz

2. evaluate the integral

f*(s) = \int_0^\infty g(z) a^(-s) dz

3. represent inverseMellintransform(f*(s), s) in terms of Meijer G-function

4. check Luke's book


	Complete solution with Mathematica V3.0

First note that elliptic integrals in Mathematica are define slightly
differently

	EllipticK[z] == Pi/2 Hypergeometric2F1[1/2,1/2,1,z]

so that a given g(z) is

-2*Sqrt[1 + z^2]*EllipticE[(1 + z^2)^(-1)] + 

 (-(1/Sqrt[1 + z^2]) + 2*Sqrt[1 + z^2])*EllipticK[(1 + z^2)^(-1)]

Let us denote

f[a] = Integrate[g[z] Cos[2 a z], {z, 0, Infinity}]

Applying the Mellin transform to f[a], we obtain

f[a] = 1/(2 Pi I) IMT[ f*[s] a^(-s), contour]

where

f*[s] = Cos[Pi s/2] Gamma[s]/2^s Integrate[g[z] z^(-s),{z,0,Infinity}]

and IMT is the inverse Mellin transform. 

Next  we change the variable of integration in f*[s]

z -> Sqrt[1-x]/Sqrt[x]

Thus

Integrate[g[z] z^(-s),{z,0,Infinity}]
==
1/2 Integrate[(1 - x)^((-1 - s)/2)*x^((s-4)/2)*
(-2*EllipticE[x] + (2 - x)*EllipticK[x]), {x, 0, 1}]

where
-2 < Re[s] < 1

The above integral can be evaluated either by hand (using the Mellin
transform) or by Mathematica V3.0: Finally we have

f*[s] = Sqrt[Pi]/8 Gamma[s/2] Gamma[1+s/2] Gamma[1/2-s/2]/Gamma[2-s/2]


and

f[a] = 1/(2 Pi I) Sqrt[Pi]/4 *
IMT[Gamma[s] Gamma[1+s] Gamma[1/2-s]/Gamma[2-s] a^(-2s), contour]


where the contour is a straight line, lying in the strip 0<Re[s]<1/2

or in terms of Meijer's function


        Sqrt[Pi]   2,1 /    | 1/2      \
f[a] = ---------  G    |a^2 |          |
           4       1,3 \    | 0, 1, -1 /


>From Y. Luke, Mathematical Functions and Their Approximations,
AP, 1975, formula (32), page 309

we have, that

f[a] = Pi/2 BesselI[1, a] BesselK[1, a]
and
f[0] = Pi/4

Victor Adamchik