To begin with, we define ϕ as μ+u, where u:=(OverBar[u])>2. We also let J=I=2.
We also let tH denote (OverBar[τ]), i.e. the upper bound of the support.
Finally, we denote (ℓ^0) by L0 and normalize (ℓ^0)=(ℓ^( ** )):=Log[(1+μ)/(γ(ϕ-2))]

Step 1: Write Incentive Compatibility as a Linear Operator

The IC constraint is given by

Using the FGM copula

f[t,s]=f[t]f[s](1+ρ(1-2F[s])(1-2F[t]))

the IC becomes

IC=(E^(-t μ))-(((E^(-t (-2+ϕ))) (1+μ)(1-F[t]))/(-1+ϕ))-((1+μ) (-(E^(-t (-2+ϕ)))F[t](1-(((-1+ϕ))^2))-(1-ϕ)(∫_t^tH)(E^(s+t-s ϕ)) F[s]\[DifferentialD]s-(((-1+ϕ))^2) (∫_0^t)(E^(s+t-t ϕ)) F[s]\[DifferentialD]s))/((-2+ϕ) (-1+ϕ) ϕ)+ρ(1-2F[t])(1+μ)(1/((-2+ϕ) ϕ))(-(∫_t^tH)-(E^(s+t-s ϕ)) (-1+F[s]) F[s]\[DifferentialD]s+(-1+ϕ) (∫_0^t)-(E^(s+t-t ϕ)) (-1+F[s]) F[s]\[DifferentialD]s)

Let the first line of IC define the operator K0 and the second line K1. We obtain

K0=(E^(-t μ))-(((E^(-t (-2+ϕ))) (1+μ) (1-F[t]))/(-1+ϕ))-(1/((-2+ϕ) (-1+ϕ) ϕ))(1+μ) (-(E^(-t (-2+ϕ))) (1-(((-1+ϕ))^2)) F[t]-(1-ϕ) (∫_t^tH)(E^(s+t-s ϕ)) F[s]\[DifferentialD]s-(((-1+ϕ))^2) (∫_0^t)(E^(s+t-t ϕ)) F[s]\[DifferentialD]s);
K1=(1/((-2+ϕ) ϕ))(1+μ)  (1-2 F[t]) (-(∫_t^tH)-(E^(s+t-s ϕ)) (-1+F[s]) F[s]\[DifferentialD]s+(-1+ϕ) (∫_0^t)-(E^(s+t-t ϕ)) (-1+F[s]) F[s]\[DifferentialD]s)

The IC constraint is then given by K0+ρ K1=0.

Step 2: Solving for the distribution (OverTilde[F ])that satisfies K0((OverTilde[F ]))+K1(F0)

Let F0 denote the equilibrium distribution under independence:

F0=1+(((E^(t (-2-μ+ϕ))) ((1+μ-ϕ) (-2+ϕ)+(E^(t μ)) (-1+ϕ) (-2-μ+ϕ)))/μ);

Fs=F0/.t->s;

Let g[t] denote the closed-form expression for K1(F0[t]), up to a multiplicative constant as we shall define below. The IC constraint can then be written in terms of the distribution (OverTilde[F ])(that we denote by F[s] below)

LL=Simplify[(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))-(∫_t^tH)(E^(s+t-s ϕ)) F[s]\[DifferentialD]s+(-1+ϕ) (∫_0^t)(E^(s+t-t ϕ)) F[s]\[DifferentialD]s+g[t],Assumptions->{tH>0,μ>0,ϕ>2}]

(The following quantity is just used to simplify below.)

x0=((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ))+(-1+ϕ) (∫_0^t)(E^(s+t-t ϕ)) F[s]\[DifferentialD]s+g[t];

Because IC must hold identically over an interval, differentiate the above expression

D[LL,t]

Using the quantity x0 defined above, we simplify,

FullSimplify[(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) μ (1+μ)+(E^(t (-2+ϕ))) (-2+ϕ) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (2-μ-ϕ) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+(E^(2 t-t ϕ)) F[t]-x0+(-1+ϕ) ((E^(2 t-t ϕ)) F[t]+(∫_0^t)(E^(s+t-t ϕ)) (1-ϕ) F[s]\[DifferentialD]s)+g'[t]]

and solve d IC / dt = 0 with respect to the last remaining integral of F:

x1=FullSimplify[x/.Solve[(-(E^(t (-2+ϕ))) (-2+ϕ) ϕ+(E^(t μ)) ϕ (-2+ϕ+F[t])+(E^(t (-2+μ+ϕ))) (-g[t]-(-1+ϕ) (ϕ x)+g'[t]))==0,x][[1]]]

We differentiate further with respect to time, and impose d IC / dt =0:

D[((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) μ (1+μ)+(E^(t (-2+ϕ))) (-2+ϕ) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ))+((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (2-μ-ϕ) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ))+(E^(2 t-t ϕ)) F[t]-x0+(-1+ϕ) ((E^(2 t-t ϕ)) F[t]+(∫_0^t)(E^(s+t-t ϕ)) (1-ϕ) F[s]\[DifferentialD]s)+g'[t],t]

We obtain the following differential equation for (OverTilde[F])

ic=Simplify[-(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) μ (1+μ)+(E^(t (-2+ϕ))) (-2+ϕ) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (μ^2) (1+μ)+(E^(t (-2+ϕ))) (((-2+ϕ))^2) (-1+ϕ)) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))-(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (2-μ-ϕ) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+((2 (E^(-t (-2+μ+ϕ))) (-(E^(t μ)) μ (1+μ)+(E^(t (-2+ϕ))) (-2+ϕ) (-1+ϕ)) (2-μ-ϕ) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+(((E^(-t (-2+μ+ϕ))) (-(E^(t μ)) (1+μ)+(E^(t (-2+ϕ))) (-1+ϕ)) (((2-μ-ϕ))^2) (-2+ϕ) ϕ)/((1+μ) (-1+ϕ)))+(E^(2 t-t ϕ)) (2-ϕ) F[t]-(-1+ϕ) ((E^(2 t-t ϕ)) F[t]+x1(1-ϕ))+(E^(2 t-t ϕ)) F'[t]+(-1+ϕ) ((E^(2 t-t ϕ)) (1-ϕ) F[t]+(E^(2 t-t ϕ)) (2-ϕ) F[t]+(((1-ϕ))^2) x1+(E^(2 t-t ϕ)) F'[t])-g'[t]+g''[t],Assumptions->{tH>0,μ>0,ϕ>2}]

Solving the differential equation, we obtain

Ftilde=Simplify[F[t]/.DSolve[{inc==0,F[0]==0},F[t],t][[1]]]/.K[1]->s

Simplifying, we can write as

Ftilde=F0+(E^(t (-2+ϕ))) (∫_0^t)(((-1+ϕ) g[s]-(-2+ϕ) g'[s]-g''[s])/ϕ)\[DifferentialD]s;

(One can also verify that solving the ODE with g[t] identically equal to zero yields the equilibrium distribution F0[t] as expected.)

We can integrate g’’[s] explicitly, and we obtain

Ftilde=F0+(E^(t (-2+ϕ))) ((∫_0^t)(((-1+ϕ) g[s])/ϕ)\[DifferentialD]s-(((-2+ϕ)( g[t]-g[0]))/ϕ)-(1/ϕ)(g'[t]-g'[0]))

where g[t] corresponds to the function K1(F0[t]), which we denote as

A=(1-2 F0) ((∫_t^tH)(E^(s+t-s ϕ)) (1-Fs) Fs\[DifferentialD]s+(1-ϕ) (∫_0^t)(E^(s+t-t ϕ)) (1-Fs) Fs\[DifferentialD]s);

Letting G0 denote the second term in (OverTilde[F]), we have

G0=(E^(t (-2+ϕ))) ((∫_0^t)(((-1+ϕ) g[s])/ϕ)\[DifferentialD]s-(((-2+ϕ)( g[t]-g[0]))/ϕ)-(1/ϕ)(g'[t]-g'[0]));

And finally

Ftilde=F0+G0;

Thus the distribution with ρ-correlation is given by

F1(ρ)=F0+ρ (OverTilde[F])=F0(1+ρ)+ρ G0

Step 3: The Objective as a function of ρ

Evaluate G0 around ρ=0 (plugging in G0 and g defined above)

ga0=(E^(t (-2+ϕ))) (-(((-2+ϕ) (-(A/.{t->0})+A))/ϕ)+(((-1+ϕ))/ϕ)(Integrate[A,t]-(Integrate[A,t]/.t->0))+(D[(A/ϕ),t]/.t->0)-D[(A/ϕ),t])

Let

f0=D[F0,t];

f1a=f0+D[ga0,t];

With the FGM copula, the objective becomes

K-γ ((Exp[L0])/(ϕ-1))(E^(t (2-ϕ)))(∫_0^t) F'[s](1+ρ(1-2F[s])(1-2F[t]))\[DifferentialD]s+γ ((Exp[L0])/ϕ)((1/(1-ϕ))((∫_t^tH)(E^(t+s (1-ϕ)))f[s](1+ρ(1-2F[s])(1-2F[t]))\[DifferentialD]s)+(∫_0^t)(E^(s+t (1-ϕ))) f[s](1+ρ(1-2F[s])(1-2F[t]))\[DifferentialD]s)

We therefore define the following quantities:

Obj=-((E^(-t μ))/μ)+(((E^(t (2-ϕ))) (1+μ))/((-2+ϕ) (-1+ϕ)))-γ ((Exp[L0])/(ϕ-1))(E^(t (2-ϕ)))(F[t]+ρ(F[t]-((F[t])^2))(1-2F[t]))+γ ((Exp[L0])/ϕ)((1/(1-ϕ))((∫_t^tH)(E^(t+s (1-ϕ)))f[s](1+ρ(1-2F[s])(1-2F[t]))\[DifferentialD]s)+(∫_0^t)(E^(s+t (1-ϕ))) f[s](1+ρ(1-2F[s])(1-2F[t]))\[DifferentialD]s);

doa=D[Obj,ρ]/.{F[t]->F0,(f[s]->f0/.t->s),(F[s]->F0/.t->s)};

ooa=Obj/.{ρ->0,F[t]->F0,f[s]->f0/.t->s};

And the change in the mass point at tH=(OverBar[τ])

doth=Simplify[D[Obj f0,tH]/.{ρ->0,f[tH]->f0/.t->tH}];

ju=Simplify[D[Obj/.{t->tH,ρ->0},tH](1-F0/.t->tH)//.{f[tH]->f0/.t->tH,F'[tH]->f[tH],f[s]->f0/.t->s,F[tH]->F0/.t->tH}];

Finally, how does tH vary with ρ (around ρ=0)?

dth=Simplify[-(((∫_0^t)(E^s) (-1+( F0/.t->s)) ( F0/.t->s) (-1+2 F0)\[DifferentialD]s)/((E^t) (-2+u)+(E^t) F0-(-1+u) (∫_0^t)(E^s) ( F0/.t->s)\[DifferentialD]s))//.{t->tH,J->2,m->μ,u->ϕ-μ,tH->((Log[(((J-m-u) (-1+u))/((J-u) (-1+(m^2)+u+m u)))])/m)},Assumptions->ϕ>μ>0];

Adding up terms, the derivative of the expected cost is given by

(∫_0^tH)(C(t)f1(t)+(((dC(t))/(dρ))+((dC(t))/dtH)(dtH/(dρ)))f0(t))dt+(1-F0(tH))((dC(tH))/dtH)(dtH/(dρ))

We therefore separate the first integrand

I1=ooa f1a+doa f0+doth dth;

and the mass point (jump)

ju2=Simplify[ju dth];

Now we plug in expressions for some, including two players (J=2), and the value of (OverBar[τ]) for the independent case

I2=I1//.{J->2,m->μ,u->ϕ-μ,tH->((Log[(((J-m-u) (-1+u))/((J-u) (-1+(m^2)+u+m u)))])/m)}

The expression can be simplified (but this step can be quite slow):

Simplify[I2,Assumptions->{ϕ>μ>0&&t>0}]

Finally, choose parameter values

Clear[ϕ,μ]

ϕ=2.1+μ;
μ=1/4;

Compute the upper bound (OverBar[τ]) under independence, and plug into the derivative of the mass point

t2=N[((Log[(((u-J+m) (-1+u))/((u-J) (-1+(m^2)+u+m u)))])/m)//.{J->2,m->μ,u->ϕ-μ}]

N[dth//.{tH->t2}]

Finally, evaluate the integral:

n2=N[I2];

NIntegrate[n2,{t,0,t2}]+ju2//.{tH->t2}

A negative number indicates that costs decrease with ρ, hence positive correlation is beneficial.