Evaluating an improper integral

Question

So I have this function:
$$ int_{-infty}^{infty} mathrm {sech}(x+s)^{2} {rm sech}(x)^{2}dx$$
And when I try to integrate it, I can obtain the indefinite integral:
In[42]:= Integrate[Cosh[x + s]^-2*Cosh[x]^-2, x]

Out[42]= -2 Coth[s] Csch[s]^2 Log[Cosh[x]]+2Coth[s] Csch[s]^2 Log[Cosh[s + x]]-Csch[s]^2 Sech[s] Sech[s+x] Sinh[x]-Csch[s]^2Tanh[x]

But when I evaluate the limits, it cancels to $0$. The solution I was given stated that the answer should be some form of:
$$frac{cosh(s)cdot s}{sinh(s)^3}- frac{1}{sinh(s)^2} $$
None of what I'm doing seems to get me the answer and as you can see, it's not like Mathematica even makes the output easy to parse.

Ulrich Neumann · Answer

Try the definite integral
Integrate[Cosh[x + s]^-2*Cosh[x]^-2, {x, -Infinity,Infinity}]
(*2 Csch[s]^2 (-2 + Coth[s] Log[E^(2 s)]) if ...*)

which Mathematica conditional solves.
Numerical  "confirmation"
int[s_?NumericQ] := 
 NIntegrate[
   Cosh[x + s]^-2*Cosh[x]^-2, 
   {x, -Infinity, Infinity}]

Plot[{int[s], 2 Csch[s]^2 (-2 + Coth[s] Log[E^(2s)])} , {s, -10, 10},PlotStyle -> {Automatic, Dashed}]

Bob Hanlon · Answer

$Version

(* "12.2.0 for Mac OS X x86 (64-bit) (December 12, 2020)" *)

Clear["Global`*"]

expr = Cosh[x + s]^-2*Cosh[x]^-2;

An indefinite integral (anti-derivative) of expr is
ad = Integrate[expr, x] // Simplify

(* -Csch[s]^2 (2 Coth[s] (Log[Cosh[x]] - Log[Cosh[s + x]]) + 
   Sech[s] Sech[s + x] Sinh[x] + Tanh[x]) *)

Verifying that ad is a valid anti-derivative of expr
D[ad, x] == expr // Simplify

(* True *)

The definite integral is then
int1 = Limit[ad, x -> Infinity] - Limit[ad, x -> -Infinity] // Simplify

(* -2 Csch[s]^2 (2 + Coth[s] (Log[E^-s] - Log[E^s])) *)

Calculating the definite integral directly
int2 = Integrate[expr, {x, -Infinity, Infinity}]

The results are equivalent for real s
diff = int1 - int2 // Simplify[#, Element[s, Reals]] &

(* 0 *)

Graphically,
Plot[{int1, int2}, {s, -10, 10},
 PlotStyle -> {Automatic, Dashed}]

Evaluating an improper integral

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