Integrate exp(-1/2( ax^2 - b/x^2)

  • Thread starter Raincloud
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Homework Statement



Integrate:
[itex] I(a,b) =
\int^\infty_\infty exp(-1/2(ax^2+b/x^2)) dx [/itex]
given
[itex]\int^\infty_\infty exp(1x^2/2) dx = \sqrt{2\pi} [/itex]


Homework Equations



The suggested substitution is [itex] y = (\sqrt{a}x - \sqrt{b}/x)/2 [/itex]

The Attempt at a Solution



The substitution gives
[itex]\int^\infty_\infty exp(-(2y^2-2\sqrt{ab}) dx [/itex]
and [itex] dy/dx = 1/2(\sqrt{a} + \sqrt{b}/x^2) [/itex]
but I can't seem to rearrange the dy/dx to do anything helpful. I've tried integrating by parts before plugging in the substitution, but it didn't seem to help.

FWIW, i've been told the numerical answer is
[itex] \sqrt{2\pi/a}exp(-ab)[/itex]
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



Integrate:
[itex] I(a,b) =
\int^\infty_\infty exp(-1/2(ax^2+b/x^2)) dx [/itex]
given
[itex]\int^\infty_\infty exp(1x^2/2) dx = \sqrt{2\pi} [/itex]


Homework Equations



The suggested substitution is [itex] y = (\sqrt{a}x - \sqrt{b}/x)/2 [/itex]

The Attempt at a Solution



The substitution gives
[itex]\int^\infty_\infty exp(-(2y^2-2\sqrt{ab}) dx [/itex]
and [itex] dy/dx = 1/2(\sqrt{a} + \sqrt{b}/x^2) [/itex]
but I can't seem to rearrange the dy/dx to do anything helpful. I've tried integrating by parts before plugging in the substitution, but it didn't seem to help.

FWIW, i've been told the numerical answer is
[itex] \sqrt{2\pi/a}exp(-ab)[/itex]
Maple gets [tex]\sqrt{\frac{2\pi}{a}} \;e^{-\sqrt{ab}},[/tex] whiich is not what you wrote.

RGV
 
  • #3
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Maple gets [tex]\sqrt{\frac{2\pi}{a}} \;e^{-\sqrt{ab}},[/tex] whiich is not what you wrote.

RGV


Yes, you're right- looking at my notes I have lost a [tex]\sqrt[/tex] in there.
 

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