Evaluating $ \int_{-\infty}^{\infty} \cosh(x+s)^{-2}\cosh(x)^{-2}dx$

Question

So I have this function:

$$ \int_{-\infty}^{\infty} \cosh(x+s)^{-2}\cosh(x)^{-2}dx$$

And when I try to integrate it, I can obtain $0$. And, when I evaluate the limits, it also 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. This is my Mathematica expression:

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]

Answer 1

Let $x= \frac{t-s}2$ to reexpress the integral as $$I=\int_{-\infty}^{\infty} \frac1{\cosh^2(x+s)\cosh^2(x)}dx = \int_{0}^{\infty} \frac4{(\cosh t+ \cosh s)^2}dt $$ Note that

$$\left( \frac{\sinh t}{\cosh t+ \cosh s} \right)’ = \frac{\cosh s}{\cosh t+ \cosh s} - \frac{\sinh^2s}{(\cosh t+ \cosh s)^2} $$ Integrate both sides $$I =\frac{4 }{\sinh^2s}\left(-1+\cosh s \int_0^\infty \frac{1}{\cosh t+\cosh s}dt\right) $$ where $$\int_0^\infty \frac{1}{\cosh t+\cosh s}dt = \frac{2\tanh^{-1}(\tanh\frac t2\tanh\frac s2)}{\sinh s}\bigg|_0^\infty = \frac s{\sinh s}$$ Thus $$I =\frac{4s\cosh s }{\sinh^3s}-\frac4{\sinh^2s} $$