Funky integral

  • Thread starter loloPF
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  • #1
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Funky integral!!

This integral is driving me nuts :cry: , anyone got a clue?
Given two real variables X and Y, one defines the function:
f(X,Y)=sqrt(a*(X+cY)^2+b*(X-cY)^2)
where a, b and c are reals and a>0, b>0.
Then the function g is defined as:
g(X,Y)=exp(-f(X,Y))
I am looking for:
1- a primitive for g
2- and/or the value of the following integrals I=integral(-infty,+infty;g(X,Y),dX) and J=integral(-infty,+infty;g(X,Y),dY).
 

Answers and Replies

  • #2
HallsofIvy
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Do you have any reason to believe it has an "elementary" primitive?

If you take a=-1, b= c= 0, you have
[tex]f(x,y)= e^{-x^2}[/tex]
which certainly does NOT have a primitive that can be written in terms of elementary functions. It can of course be written as [itex]2\pi Erf(x)[/tex] where Erf(x) is the error function- but it's not "elementary", it is defined as
[tex]\frac{1}{2\pi}\int e^{-x^2}dx[/tex]
 
  • #3
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You are right HallsofIvy and the answer to your question is: "No I do not.", this is why my second point starts with: "and/or [...]".
I have made some (but little) progress on this and I'll let you know in a comming post where I stand now.
 
  • #4
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HallsofIvy, you might have missed the square root in the definition of f:
[tex]f(X,Y)=\sqrt{a(cY+X)^2+b(cY-X)^2}[/tex]

Definition:
[tex]I(X)=\int_{-\infty}^{\infty} e^{-f(X,Y)}dY[/tex]
Changes of variable:
first [tex]u=cY+X[/tex]
then [tex]v=u\sqrt{a+b}[/tex]
and [tex]w=v-\frac{2bX}{\sqrt{a+b}}[/tex]
leading to:
[tex]I(X)=\frac{1}{c\sqrt{a+b}}\int_{-\infty}^{\infty} e^{-\sqrt{w^2+4X^2\frac{ab}{a+b}}}dw[/tex]

I am now considering a trig transformation:
[tex]w=2|X|\sqrt{\frac{ab}{a+b}}tan(\theta)[/tex]
to get rid of the square root but I am then stuck again :grumpy:
 
  • #5
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I just found the answer: this integral is well known as the "modified Bessel function of second kind", period.
 
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