The denominator D factorizes in D=x1*x2*x3*x4, where
x1=sqrt(2+i) (first quadrant)
x2=-sqrt(2-i) (second quadrant)
x3=-sqrt(2+i) (third quadrant)
x4=sqrt(2-i) (forth quadrant)
The integral is given by the residue theorem
int=(2*pi*i)[Res(x1)+Res(x2)]
and Res(x1)=-(i/4)*sqrt(2+1)...
Substitute trig. objects with parametric formulae:
sinx=2t/(1+t^2), cosx=(1-t^2)/(1+t^2), dx=2dt/(1+t^2)
and your trig. integrale becomes the rational integral
2*Integ[-1,1](1/(t^2+2*t+5)dt = pi/4
Plotting the function (remebering x!=Gamma(x+1))
G(x)=Gamma(x + 1) + Gamma(x - 2) - 16x + 24
you can see that for x>0 there are only four zeros.
For x<0 there are infinite solutions near 0,-1,-2,-3...none exacly integer.
There Gamma function is singular (has a pole) and change sign.
If x can be a real number (using Gamma function) you can find 4 solutions betweeen 1 and 4.5 (by plotting method or std numeric method).
x1=1.1837
x2=1.3134
x3=2.1222
x4=4.4099
There are also 5 negative solution... None is integer.
With the substitution u=y/x equation becomes
u''+au=0
which has solution
u=c1*cos(kx)+c2*sin(kx) [k=sqrt(a)]
so that
y=xu=c1*x*cos(kx)+c2*x*sin(kx)
So doing y'(0)=c1=0 y(1)=c1*cos(k)+c2*sin(k)=0
i.e. y=0, onlythe trivial solution ?
The series is convergent by Cauchy criterium since C(n+1)/C(n)->0 for n>infinity,
as can be easily verified using Striling formula : log(n!) => n*log(n)-n
If M1=Qd is the dipole moment (Q :charge, d:separation as a vector) the potential field is
V(r)=1/(4*pi*e) M1.grad(1/r)
Having a ground plane, this can be eliminated if you consider an image dipole (symmetrically placed under the ground plane). of moment M2=-M1.
Then the interaction energy...
The separability of Laplace and Helmoltz equations in a coordinate system is possible if and only if the cofactor of the Staekel matrix and metric tensor satisfy some appropriate relations. Look at the introductory pages of
FIELD THEORY HANDBOOK - P.Moon-D.E.Spencer
Springer 1961
It is the Laplace equation in cylindrical coordinates with symmetry about y-axe.
You can solve it by variable separation, once given the boundary condition:
phi(x,y)=X(x)Y(y)
X''+X'/x+cX=0 (in x, 0-order Bessel equation)
Y''-cY=0 (in y)
c=arbitrary positive/negative real constant