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Could you please help me to solve this integration problem? $$\int_0^2\frac{1}{2+\sqrt{3\,e^x+3\,e^{-x}-2}}dx$$ Its approximate numeric value is $0.419197813818367...$, but I could not find an exact symbolic expression for it.

Marty Colos
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    What leads you to believe it even has a closed form? :) – apnorton Sep 27 '13 at 20:01
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    @experimentX When I give this integral as a direct input to Mathematica, it returns unevaluated. Did you perform any transformations before feeding it to Mathematica? – Marty Colos Sep 27 '13 at 20:17
  • One thing you could try is to pull one factor of $e^{-x}$ out of the square root and then treat the remaining parts as a quadratic. – abiessu Sep 27 '13 at 20:17
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    @MartyColos $3 e^{x} + 3 e^{-x} = 6 \cosh(x)$ – S L Sep 27 '13 at 20:28
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    When I changed the variable $y=e^x$ and fed the transformed integral $\displaystyle\int_1^{e^2}\frac{dy}{y,\left(2+\sqrt{3,y+\frac{3}{y}-2}\right)}$ to Mathematica, it returned a huge result involving incomplete elliptic functions with complex arguments, that seems to agree with the numerical integration. I will try to simplify it. – Marty Colos Sep 27 '13 at 20:31
  • Mathematica gives

    $ \int f(x)dx = \int \frac{dx}{2+\sqrt{6\cosh x-2}}
    = I(x) = (1/6) \left[(2 - \sqrt{-2 + 6 \cosh x}~)\cdot \coth \frac{x}{2} - 2~ i~ E_2((i~ x/2),~ 3) - 4~i ~E_1((i~x/2),~ 3)\right]. $ Using Santosh Linkha's idea, and taking care to take the $\lim_{a\to 0} \int_a^2$ gives the answer in the OP. $E_1,E_2$ are elliptic integrals.

     – daniel Aug 15 '14 at 20:07
  • May relate to https://math.stackexchange.com/questions/2043030 – Harry Peter Jun 30 '18 at 09:03

1 Answers1

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$\int_0^2\dfrac{1}{2+\sqrt{3e^x+3e^{-x}-2}}dx$

$=\int_1^{e^2}\dfrac{1}{2+\sqrt{3y+\dfrac{3}{y}-2}}d(\ln y)$

$=\int_1^{e^2}\dfrac{1}{y\left(2+\sqrt{3y+\dfrac{3}{y}-2}\right)}dy$

$=\int_1^{e^2}\dfrac{1}{2y+y\sqrt{3y+\dfrac{3}{y}-2}}dy$

$=\int_1^{e^2}\dfrac{1}{\sqrt{3y^3-2y^2+3y}+2y}dy$

$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}-2y}{\left(\sqrt{3y^3-2y^2+3y}+2y\right)\left(\sqrt{3y^3-2y^2+3y}-2y\right)}dy$

$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}-2y}{3y^3-2y^2+3y-4y^2}dy$

$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}}{3y^3-6y^2+3y}dy-\int_1^{e^2}\dfrac{2y}{3y^3-6y^2+3y}dy$

$=\int_1^{e^2}\dfrac{\sqrt{3y^3-2y^2+3y}}{3y(y-1)^2}dy-\int_1^{e^2}\dfrac{2}{3(y-1)^2}dy$

which can express in terms of Lauricella Functions

Harry Peter
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