Different results of a definite integral $\int_0^{\cosh ^{-1}(a)} \frac{1}{\sqrt{a^2 \text{sech}^2(x)-1}}\, dx$

Question

fixed in 10.0.2

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Update

I have tried like these. I think there is a bug.

Plot[1/Sqrt[-1 + 2^2 Sech[x]^2], {x, 0, ArcCosh[2]},
 Ticks -> {{ArcCosh[2]}, Automatic}]

This is the antiderivative.

primitive = Integrate[1/Sqrt[-1 + 2^2*Sech[x]^2], x]; 

Plot[primitive, {x, 0, ArcCosh[2]},
 Ticks -> {{ArcCosh[2]}, {π/4, π/2}}]

Limit[primitive, x -> 0]

0

So far, that's right. Look at this any situation. Limits of primitive are same regardless of the direction. And the limit value is minus. Is this right?

(version 10)

Limit[primitive, x -> ArcCosh[2], Direction -> -1] // FullSimplify

-π/2

Limit[primitive, x -> ArcCosh[2], Direction -> 1] // FullSimplify

-π/2

But this computation is right at version 9

(version 9)

Limit[primitive, x -> ArcCosh[2], Direction -> 1] // FullSimplify

π/2

And as mentioned earlier origin, the definite integral is an erroneous conclusion at version 9 also.

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Edit

The definite integral is solved using substitution method as Dr. Wolfgang Hintze says like this.

$u$ =$\frac{\cosh ^2(x)-1}{a^2-1}$

$dx$ = $\frac{\left(a^2-1\right)}{2 \sinh (x) \cosh (x)}du$

$\int_0^1 \frac{a^2-1}{2 \sinh (x) \cosh (x) \sqrt{a^2 \text{sech}^2(x)-1}} \, du$

$\frac{1}{2} \int_0^1 \frac{a^2-1}{\sqrt{a^2 \sinh ^2(x)-\sinh ^2(x) \cosh ^2(x)}} \, du$

$\frac{1}{2} \int_0^1 \frac{1}{\sqrt{\frac{\left(\cosh ^2(x)-1\right) \left(a^2-\cosh ^2(x)+1-1\right)}{\left(a^2-1\right) \left(a^2-1\right)}}} \, du$

$\frac{1}{2} \int_0^1 \frac{1}{\sqrt{u (1-u)}} \, du=\frac{\pi }{2}$

It is solved in the real number region. And ArcCos[2]is also real number. But I don't konw why mathematica make $\int_0^{\cosh ^{-1}(2)} \frac{1}{\sqrt{2^2 \text{sech}^2(x)-1}}\, dx$ appear a imaginary term like $\left(\frac{1}{2}-i\right) \pi$ .

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Origin

I have tried two expressions.(version 10)

(1)

 Integrate[1/Sqrt[-1 + 2^2*Sech[x]^2], {x, 0, ArcCosh[2]}]

$\left(\frac{1}{2}-i\right) \pi$

(2)

$Assumptions = {a > 1}; 
Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], {x, 0, ArcCosh[a]}]

$\frac{\pi }{2}$

1. What difference does it make it?

2. The first computing (1) makes the imaginary term. - I π. I don't know why it did such result?

(version 9)

 Integrate[1/Sqrt[-1 + 2^2*Sech[x]^2], {x, 0, ArcCosh[2]}]

$\frac{3 \pi }{2}$

3. If possible, I want to know mathematica's detail process.

Answer 1

It seems to be a branch cut issue. It depends when the substitution is done, that causes the integrator to go one way vs. the other. This below shows difference

ClearAll[x, a];
sol = Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], x]

The integrand at x=0 is always zero, so we can ignore this, The issue is with the upper limit

low = Limit[sol, x -> 0]
(*0*)

Taking the limit using ArcCosh[a] first, then after that replacing a by 2 :

up = Limit[sol, x -> ArcCosh[a]]
Limit[up, a -> 2]
(*  -Pi/2 *)

Now taking the limit using ArcCosh[2] first then replacing a by 2 :

up = Limit[sol, x -> ArcCosh[2]]
Limit[up, a -> 2]
(* Pi/2 *)

So depending when a is set to 2 causes different result. This would indicate a branch cut issue.

Answer 2

This has been fixed in 10.0.2. On windows 7, 64 bit:

Answer 3

$Version

"10.0 for Mac OS X x86 (64-bit) (September 10, 2014)"

$Assumptions = {a > 1};
Integrate[1/Sqrt[-1 + a^2*Sech[x]^2], {x, 0, ArcCosh[a]}]

Pi/2

f[a_?NumericQ] := Module[
  {ar = Rationalize[a, 0]},
  NIntegrate[
    1/Sqrt[-1 + ar^2*Sech[x]^2],
    {x, 0, ArcCosh[ar]},
    WorkingPrecision -> 20] //
   Chop]

RootApproximant[f[2]/Pi]*Pi

Pi/2

Plot[f[a], {a, 1, 5},
 Exclusions -> 1,
 PlotRange -> Pi/2 {.9, 1.1}]

The integral is a constant Pi/2 for a > 1, the integral result with the imaginary component is wrong.