How to evaluate Integrate[Exp[Csch[(x - τ)]^2], {x, 0, x}]

Question

I have been trying to integrate the following and Mathematica couldn't integrate it. Can someone help me on this?

Assuming[{Element[{x, τ}, Reals], x > 0, τ > 0, (x - τ) != 0},
  Integrate[Exp[Csch[(x - τ)]^2], {x, 0, x}]]

Answer 1

The integral doesn't seem to have any closed form solution. You can get an approximation to it by building an interpolating function and integrating it. Like so:

Clear[f, pts, ff, int]
f[τ_][u_] := Exp[Csch[(u - τ)]^2]; 
ff[τ_, x_, dx_] := Interpolation[Table[{u, f[τ][u]}, {u, 0, x, dx}]]; 
int[x_, dx_][τ_] := Function[u, Integrate[ff[τ, x, dx][uu], {uu, 0, u}]]

Then, given τ = 2, x = 1.25, dx = .001, we can use int[x, dx][τ] as function that behaves like the function that is the integral of f. For example:

With[{τ = 2, x = 1.25, dx = .001},
  Plot[{f[τ][u], int[x, dx][τ][u]}, {u, 0, x}, PlotLegends -> "Expressions"]]

Answer 2

As suggested by @JM, you can use ParametricNDSolveValue to obtain a numerical approximation to your integral:

pf = ParametricNDSolveValue[
    {
    int'[u] == Exp[Csch[u-τ]^2],
    int[0] == 0
    },
    int,
    {u, 0, τ-.1}, (* avoid singularity at τ *)
    τ
];

Visualization (using @m_goldbergs settings):

Plot[{Exp[Csch[x-2]^2], pf[2][x]}, {x, 0, 1.25}]