# Evaluating an improper integral

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.

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}]


\$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}]
`