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]
  • 1
    $\begingroup$ Maybe try converting to exponentials and see if you can use a substitution to get a rational function under the integral? $\endgroup$
    – Yuriy S
    Mar 26, 2021 at 11:21
  • $\begingroup$ When you write $\cosh(x)^{-2}$ do you mean $\cosh\frac{1}{x^2}$ or $1/\cosh^2(x)$? $\endgroup$ Mar 26, 2021 at 11:37
  • $\begingroup$ $1/cosh^2(x)$ sorry $\endgroup$ Mar 26, 2021 at 11:40
  • $\begingroup$ I dont understand how you would get 0 if the integrand is strictly positive. $\endgroup$
    – zoli
    Mar 26, 2021 at 11:58
  • 1
    $\begingroup$ I used the substitution $t=tanh(x)$, and got an answer of $\frac{4scosh(s)}{sinh(s)^3}-\frac{4}{sinh(s)^2}$ $\endgroup$ Mar 26, 2021 at 12:01

1 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 $$I =\frac{4s\cosh s }{\sinh^3s}-\frac4{\sinh^2s} $$


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