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One integral (square root and gaussian peak)

by jostpuur
Tags: gaussian, integral, peak, root, square
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jostpuur
#1
Jul19-07, 10:23 AM
P: 2,066
An integral
[tex]
\int\limits_{-\infty}^{\infty} \sqrt{1+x^2} e^{-Ax^2+Bx} dx
[/tex]
is the problem. Anyone knowing the solution?
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AiRAVATA
#2
Jul19-07, 12:40 PM
P: 173
Try completing squares...
jostpuur
#3
Jul19-07, 01:03 PM
P: 2,066
A variable change that simplifies the exponent only makes the expression under the square root more complicated, like this
[tex]
\int\limits_{-\infty}^{\infty} \sqrt{x^2+Ax+B}e^{-x^2} dx
[/tex]
which doesn't seem any easier. (A and B are different constants than in the previous equation). Or was this what you meant by completing the squares?

CompuChip
#4
Jul19-07, 01:08 PM
Sci Advisor
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One integral (square root and gaussian peak)

I think he meant this (also check out the example on that page), although I must admit I don't immediately see how that helps you... think about it :)
jostpuur
#5
Jul19-07, 01:37 PM
P: 2,066
An equation
[tex]
-Ax^2 + Bx = -\big(\sqrt{A}x - \frac{B}{2\sqrt{A}}\big)^2 + \frac{B^2}{4A}
[/tex]
would naturally have preceded my variable change
[tex]
x=\frac{1}{\sqrt{A}}y + \frac{B}{2A}
[/tex]
and renaming [itex]y\mapsto x[/itex], in simplifying the exponent.
Gib Z
#6
Jul19-07, 11:46 PM
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Sorry to tell you but nothing you've done helps you very much. In fact the only thing that might help in the slightest was post 3, where one can at least conclude that the integral even exists.

The integrand has no elementary anti derivative, or even one in known non-elementary functions. You have to use more powerful tools.

Define:

[tex]I(a)= \int^{a}_{-a} \sqrt{1+x^2} e^{-Ax^2 + Bx} dx = \int^{a}_{-a} \sqrt{1+y^2} e^{-Ay^2+By} dy[/tex]

[tex]I^2(a) = \int^{a}_{-a} \int^{a}_{-a} \sqrt{(1+x^2)(1+y^2)} e^{ -A(x^2+y^2) +B(x+y)} dx dy[/tex].

Now applying Fubini's Theorem we can see this double integral is the same as the integral:

[tex]\int_C \sqrt{(1+x^2)(1+y^2)} e^{ -A(x^2+y^2) +B(x+y)} d(x,y)[/tex], where C is the square with vertices {(−a, a), (a, a), (a, −a), (−a, −a)} on the xy-plane.

Now switch to polar co-ordinates, so[tex]x=r\,\cos \theta, y= r\,\sin\theta, d(x,y) = r\, d(r,\theta)[/tex].

Hopefully it will work out.
tookan
#7
Aug5-07, 04:31 PM
P: 2
Hi - I don't suppose that anyone figured this out yet? Been giving me a hard time for a while now - I can't see how the transformation to polar helps - still left with a gaussian and square root product in the integrand.
jostpuur
#8
Aug6-07, 12:13 AM
P: 2,066
Quote Quote by tookan View Post
Hi - I don't suppose that anyone figured this out yet? Been giving me a hard time for a while now - I can't see how the transformation to polar helps - still left with a gaussian and square root product in the integrand.
Do you mean that you took this as a challenge now when you saw this post, or that you had been trying to solve this same integral earlier, and found this post when you were trying to find solution?

The GibZ's hint doesn't work, I checked it. Looks like the circles (in shperical coordinates) should be replaced with some other shapes. In fact, with those shapes, that satisfy

[tex]
(1+x^2)(1+y^2)=c^2
[/tex]

for some constant c, but I couldn't make that stuff work.

Other idea I had, was to define

[tex]
I(A,B) = \int\limits_{-\infty}^{\infty} dx\;\sqrt{1+x^2}e^{-Ax^2+Bx}
[/tex]

Accept [itex]I(A_0,0)[/itex] as some special constant, where [itex]A_0[/itex] is some constant, and try Taylor series

[tex]
I(A,B)=I(A_0,0) + (A-A_0,B)\cdot(\partial_A, \partial_B) I(A_0,0) + \frac{1}{2}\big( (A-A_0,B)\cdot(\partial_A,\partial_B)\big)^2 I(A_0,0) +\cdots
[/tex]

but the partial derivatives produced just more difficult integrals, that I couldn't make return to the original one.

I haven't been trying much after these attempts.
CompuChip
#9
Aug6-07, 02:41 AM
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Is this integral even analytically solvable?
jostpuur
#10
Aug6-07, 04:14 AM
P: 2,066
Quote Quote by CompuChip View Post
Is this integral even analytically solvable?
I don't know how to prove such things, but the Wolfram Integrator at least doesn't give integral function for that. That of course doesn't mean, that it would be impossible to carry out the integration over some given domain.
Gib Z
#11
Aug6-07, 04:41 AM
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When one runs out of ideas, crazy things come to mind. Perhaps the squeeze theorem on the exponent will turn out fine... Give me some time on this ill actually bother to do this now.
jostpuur
#12
Aug6-07, 04:51 AM
P: 2,066
Quote Quote by Gib Z View Post
When one runs out of ideas, crazy things come to mind. Perhaps the squeeze theorem on the exponent will turn out fine... Give me some time on this ill actually bother to do this now.
GibZ, are you interested in physics? We had some interesting discussion about quantum theory in threads

http://www.physicsforums.com/showthread.php?t=175155

http://www.physicsforums.com/showthr...=176514&page=3

and about the operator [itex]\sqrt{-\nabla^2+m^2}[/itex]. I though, that being able to calculate this integral, would bring some understanding to how this operator behaves. That was where I got the original motivation.
tookan
#13
Aug6-07, 04:53 PM
P: 2
Hey people - I actually came across this integral whilst doing a functional integral calculation for the truncated (in real lattice space) Hubbard model. If you don't know then the Hubbard model is a lattice model for stongly correlated electron systems in condensed matter. If I can get this integral then some serious physics should drop out analytically. The problem may be very similar in form to some different field theories. jostpuur - nice one for your help - I had come to the conclusion as well that some sort of expansion is the way to proceed but the details of my specific problem (where x is a complicated function of multivalued auzilliary fields) suggest that the square root remains in the answer! maybe can expand integral, do individual integrals, the gaussian integrals with higher powers of x as a product in the integrand are gamma integrals and can be done easily, and then regroup...?
jostpuur
#14
Aug6-07, 10:37 PM
P: 2,066
One way would be to write [itex]\sqrt{1+x^2}[/itex] as Taylor series. Unfortunately those series don't converge everywhere. An ugly way would be to press on by integrating small pieces somewhere near the origo always using some series that converge there. This would not give precise solution, but an approximation at least.

One might also ask, that can it be written as a different kind of series, that converge everywhere, and whose terms can be integrated.


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