How Do You Solve This Challenging Integral Involving a Modified Bessel Function?

  • Thread starter loloPF
  • Start date
  • Tags
    Integral
In summary, Funky integral is driving HallsofIvy nuts, and HallsofIvy has found that it is the "modified Bessel function of second kind".
  • #1
loloPF
15
0
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).
 
Physics news on Phys.org
  • #2
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
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
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
 
  • #5
I just found the answer: this integral is well known as the "modified Bessel function of second kind", period.
 

FAQ: How Do You Solve This Challenging Integral Involving a Modified Bessel Function?

What is a funky integral?

A funky integral refers to an integral that is difficult to solve using traditional methods, often involving complex or unconventional functions.

How do I know when an integral is considered "funky"?

It can be subjective, but generally an integral is considered funky if it cannot be solved using basic techniques such as substitution, integration by parts, or trigonometric identities.

What are some strategies for solving funky integrals?

Some strategies for solving funky integrals include using special techniques such as u-substitution, partial fractions, or trigonometric substitutions. Another approach is to manipulate the integral algebraically to make it more manageable.

Are there any online resources or tools that can help with solving funky integrals?

Yes, there are many online integral calculators and step-by-step solvers that can help with solving funky integrals. However, it is important to understand the concepts and techniques behind solving the integral rather than just relying on a calculator.

Is there a systematic approach to solving funky integrals?

While there is no one set method for solving funky integrals, there are some general steps that can be followed such as identifying the type of integral, trying different techniques, and checking for errors or mistakes in the solution.

Similar threads

Replies
4
Views
2K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
20
Views
3K
Replies
1
Views
2K
Replies
3
Views
1K
Replies
10
Views
2K
Back
Top