2
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Can the following integration be performed with an analytical output?

 intq = Integrate[
          Exp[-(Sin[θ]^2/(2*x))]*(((2*x^2)/
                E^((a*Cos[θ])/x) + (2*a*x*Cos[θ])/
                E^((a*Cos[θ])/x) + (-1 - 
                  2*x^2 - (2*x)/Sqrt[1 + a^2*Cos[θ]^2])/
                E^(Sqrt[1 + a^2*Cos[θ]^2]/x) +                      
               a^2*Cos[θ]^2*(E^(-((a*Cos[θ])/x)) + (-1 - (2*x)/
                      Sqrt[1 + a^2*Cos[θ]^2])/
                   E^(Sqrt[1 + a^2*Cos[θ]^2]/x)))/x^2), {θ, 0, 
           Pi/2}, Assumptions -> {a > 0, x > 0}]

The above form was derived from the following:

intp = Integrate[ y^2*Exp[-y], {y, (a*Cos[θ])/x, Sqrt[1 + (a*Cos[θ])^2]/x}, 
                Assumptions -> {θ > 0, x > 0}]

(*     (2 E^(-((a Cos[θ])/x)) x^2 + 
     2 a E^(-((a Cos[θ])/x)) x Cos[θ] + 
     E^(-(Sqrt[1 + a^2 Cos[θ]^2]/
       x)) (-1 - 2 x^2 - (2 x)/Sqrt[1 + a^2 Cos[θ]^2]) + 
     a^2 Cos[θ]^2 (E^(-((a Cos[θ])/x)) + 
        E^(-(Sqrt[1 + a^2 Cos[θ]^2]/
          x)) (-1 - (2 x)/Sqrt[1 + a^2 Cos[θ]^2])))/x^2
 *)
intq = Integrate[Exp[-(Sin[θ]^2/(2*x))]*intp, {θ, 0, Pi/2}]

The second integral stalls, and so the need to use the first original form.

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1
  • 3
    $\begingroup$ Looks unlikely. :( $\endgroup$
    – J. M.'s torpor
    Jul 10 '15 at 8:51
-1
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Integrate[
  y^2 Exp[-y], {y, (a Cos[θ])/x, Sqrt[1 + (a Cos[θ])^2]/x}, 
  Assumptions -> {θ > 0, x > 0}]

$\frac{e^{-\frac{\sqrt{a^2 \cos ^2(\theta )+1}}{x}} \left(-\frac{2 x}{\sqrt{a^2 \cos ^2(\theta )+1}}-2 x^2-1\right)+a^2 \cos ^2(\theta ) \left(e^{-\frac{\sqrt{a^2 \cos ^2(\theta )+1}}{x}} \left(-\frac{2 x}{\sqrt{a^2 \cos ^2(\theta )+1}}-1\right)+e^{-\frac{a \cos (\theta )}{x}}\right)+2 x^2 e^{-\frac{a \cos (\theta )}{x}}+2 a x \cos (\theta ) e^{-\frac{a \cos (\theta )}{x}}}{x^2}$

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1
  • 3
    $\begingroup$ What happen to the second integration over theta? $\endgroup$
    – thils
    Jul 11 '15 at 0:44

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