# How to evaluate this function?

I’m trying to evaluate

y[t] = InverseFunction[ NIntegrate[Sqrt/( -8+Exp[y]+2Exp[-y]), {y,0,t}]]

Some of the problems I’m facing, is that i’d like to get the integral’s result in terms of t as a definite integral.

Any help to get y, y and y’ ?

Assuming the integral is not possible to evaluate symbolically, so that Roman's approach was not available, it is still possible to construct an interpolating function solution of the inverse. First, rewrite your relationship as follows:

eqn = y == Inactive[Integrate][Sqrt/(-8+Exp[s]+2Exp[-s]),{s,0,t[y]}];
eqn //TeXForm


$$y=\int _0^{t(y)}\frac{\sqrt{2}}{2 e^{-s}+e^s-8}ds$$

where I renamed the integration variable to avoid confusion. Then, differentiating with respect to y produces an ODE:

ode = D[eqn, y];
ode //TeXForm


$$1=\frac{\sqrt{2} t'(y)}{2 e^{-t(y)}+e^{t(y)}-8}$$

For initial conditions, clearly t == 0. So, we need to solve:

sol = NDSolveValue[{ode, t==0}, t, {y, -1, 1}];


Visualization:

Plot[sol[y], {y, -1, 1}] Comparison with the exact answer (as provided by Roman):

sol
sol
sol'


0.

-1.33756

-3.53553

vs:

roman[y_] := Log[4-Sqrt Tanh[Sqrt y+ArcTanh[3/Sqrt]]]

roman
roman //N
roman' //N


0

-1.33756

-3.53553

No need to do numerical integration, you can do this one analytically.

Integrate[Sqrt/(-8 + Exp[y] + 2 Exp[-y]), {y, 0, t}]
(* ConditionalExpression[-((ArcTanh[3/Sqrt] + ArcTanh[(-4 + E^t)/Sqrt])/Sqrt),
E^t == 4 || (-4 + E^t)^2 <= 0] *)


It turns out that this result is valid for $$\ln(4-\sqrt{14}):

Plot[-((ArcTanh[3/Sqrt] + ArcTanh[(-4 + E^t)/Sqrt])/Sqrt),
{t, Log[4 - Sqrt], Log[4 + Sqrt]}] Invert it:

Solve[-((ArcTanh[3/Sqrt] + ArcTanh[(-4 + E^t)/Sqrt])/Sqrt) == y, t]


(expression that I simplify by hand to get rid of branches:)

t[y_] = Log[4 - Sqrt Tanh[Sqrt y + ArcTanh[3/Sqrt]]];
Plot[t[y], {y, -1, 1}] From this you get your points of interest:

t
(* 0 *)

t
(* Log[4 - Sqrt Tanh[Sqrt + ArcTanh[3/Sqrt]]] *)

t'
(* -5/Sqrt *)