Solution of an integral equation

[neelakash]

Homework Statement

Given [tex]\frac{1}{| \int\ f(\ x)\ g(\ x)\ d\ x\ |}=\int \frac{\ f(\ x)}{\ g(\ x)}\ d\ x[/tex]
Does the above put any condition on f(x) and g(x)?

Homework Equations

The Attempt at a Solution

The | | in the denominator reminds me of Darboux inequality...In fact it looks impossible to solve analytically...Can it be solved numerically?

 

[l'Hôpital]

What if you take the derivative of both sides?

[tex]
\frac{-f(x) g(x) }{| \int\ f(x)\ g(x)\ dx\ |^2}= \frac{\ f(\ x)}{\ g(\ x)}
[/tex]

Assuming f(x) =/= 0 and simplifying with a little algebra

[tex]
(g(x))^2 = - (\int\ f(x)\ g(x)\ dx )^2
[/tex]

Which would imply that g(x) = 0, which is impossible.

So, no such function exists, unless we consider complex functions.

 

[neelakash]

Remembering |x|=x if x>0 and |x|=-x if x<0,there is another option:

[tex]

(g(x))^2 = (\int\ f(x)\ g(x)\ dx )^2

[/tex]
This leads to
[tex]\ g(\ x)=\ e^{\pm\int\ f(\ x)\ d\ x}[/tex]

Do you agree?

 

[l'Hôpital]

Yeah, I guess that makes sense. Though it should be

g(x) = e^int f(x).

Not f(x).

 

[neelakash]

Personally I expected g(x) to have some lower bound;and that looked plausible for Darboux inequality says:| integral |>= Maximum value of integrand*length of the contour...

Can that be a way?

Yea...that was a typo..I am fixing it