Find all functions of this integral equation, very tricky

[DrummingAtom]

Homework Statement

$$\int{f(x)dx}\int{\frac{1}{f(x)}dx} = -1$$

Homework Equations

The Attempt at a Solution

At first glance, I thought e^x is a solution. But I'm not convinced yet because I didn't formally go through it. When I did it turns into some nasty integral. Here's a couple things I've tried:

From here I took the derivative of each side and then cleaned it up:

$$\int{f(x)dx}\int{\frac{1}{f(x)}dx} = -1$$

Evaluating the left side integral(if this is allowed) and some algebra,.. gives:

$$f(x)\int{\frac{1}{f(x)}dx} + \int{f(x)dx}\frac{1}{f(x)} = 0$$

I then took the derivative of each side again and then some algebra gives:

$$\int{f(x)} = -f(x)^2*ln(f(x))$$$$f(x) = -2f(x)*f ' (x)*ln(f(x)) - f(x)*f ' (x)$$

Cleaning it up and canceling gives finally:

$$f '(x) = \frac{1}{-2}*\frac{1}{ln(f(x))-1}$$

Which then I would just need to integrate both sides and there's my final function. That integral is nasty (wolfram gave me some strange function as the solution). Which makes me think that it must be wrong. Here's a second try at the solution.
The hint was: $$\int{\frac{1}{f(x)}dx} = -\int{f(x)}^{-2}*f(x)}$$ by the chain rule.

If the hint is used then:

$$\int{f(x)dx}\int{\frac{1}{f(x)}dx} = -1$$

After integrating both sides:

$$f(x) = \int{f(x)}dx^{-2}*f(x)$$

Then:

$$\int{f(x)}^{2} = 1$$

From here I can square root and give:

$$\int{f(x)} = 1$$

If this is the case I have no clue as to what integrated function gives 1 as the solution.

Any help would be greatly appreciated. Thank you.

 

[tiny-tim]

Hi DrummingAtom!

Be simple! …

start "if g' = f then (1/g)' = … " ​

 

[DrummingAtom]

Thank tiny-tim,

By that, g' = f then (1/g)' = f '

So then plugging into the original problem would give:

∫g' * ∫(1/g)' = -1

or

∫f * ∫f` = -1

By that it gives:

g = -1/f

From this point do I just guess functions that fit this?

 

[Char. Limit]

Thank tiny-tim,

By that, g' = f then (1/g)' = f '

So then plugging into the original problem would give:

∫g' * ∫(1/g)' = -1

or

∫f * ∫f` = -1

By that it gives:

g = -1/f

From this point do I just guess functions that fit this?

But if g' = f then (1/g)' is 1/g^2 * g', not what you gave.

 

[SammyS]

Why isn't f(x) = e^x a solution?

 

[DrummingAtom]

But if g' = f then (1/g)' is 1/g^2 * g', not what you gave.

Ooopps, I was thinking inverse functions. Thanks for clearing that up.

Ok so using that instead would give:

∫g` = ∫f or g = ∫f

and

∫(1/g)` = (1/g^2 * g')


Which would all give:

g' = -g

Which the only function that would fit that is e^-x, right?

 

[DrummingAtom]

Why isn't f(x) = e^x a solution?

I think it is too, but I don't know how to formally show that it is. I think that e^x and e^-x would have to work.

 

[Char. Limit]

I think it is too, but I don't know how to formally show that it is. I think that e^x and e^-x would have to work.

If 1/e^(x) is a solution, as you proved that it is, then e^(x) must be a solution as well, because the original equation involves multiplying the integrals of f(x) and 1/f(x).

 

[SammyS]

I got that f '(x) = ± f(x) .

 

[tiny-tim]

(just got up :zzz: …)

∫g` = ∫f or g = ∫f

and

∫(1/g)` = (1/g^2 * g')

DrummingAtom, stop using ∫s …

the whole point of introducing g is to simplify by converting everything to differential equations, with no ∫s

Which would all give:

g' = -g

No, I get something with squareds in

 

[DrummingAtom]

No wonder this is giving me trouble, I know next to nothing about differential equations.

I'll try this again:

If g' = f and (1/g)' = 1/g²*g'

then the original equation would say:

g' * (1/g²)*g' = -1

or

(g')² = -g²

At this point I can't take the square root of a negative function, does this mean there are no real solutions?

 

[tiny-tim]

Hi DrummingAtom!

Yes, that's fine except …

If g' = f and (1/g)' = 1/g²*g' …

… it should be (1/g)' = -1/g²*g'

ok, now take the square-root of your last equation (and remember to include a ±).

 

[SammyS]

(just got up :zzz: …)


DrummingAtom, stop using ∫s …

the whole point of introducing g is to simplify by converting everything to differential equations, with no ∫s

You set f = g', which gives (1/g)' = -g'/g² .

I don't see how these help with ∫ (1/f) dx .

 

[SammyS]

What I suggest:

$$\int{f(x)dx}\int{\frac{1}{f(x)}dx} = -1\quad\quad\to\quad\quad\int{\frac{1}{f(x)}dx} = \frac{-1}{\displaystyle \int{f(x)dx}}$$

Take the derivative w.r.t. x of both sides of the equation on the right.

 

[DrummingAtom]

What I suggest:

$$\int{f(x)dx}\int{\frac{1}{f(x)}dx} = -1\quad\quad\to\quad\quad\int{\frac{1}{f(x)}dx} = \frac{-1}{\displaystyle \int{f(x)dx}}$$

Take the derivative w.r.t. x of both sides of the equation on the right.

I can see that working for the left side because it's a set integral. But wouldn't you have to use the quotient rule for the $$\frac{-1}{\displaystyle \int{f(x)dx}}$$, doing so would give:

$$\frac{1}{f(x)} = \frac{f(x)}{\int{f(x)}^{2}dx}$$

 

[SammyS]

I can see that working for the left side because it's a set integral. But wouldn't you have to use the quotient rule for the $$\frac{-1}{\displaystyle \int{f(x)dx}}$$, doing so would give:

$$\frac{1}{f(x)} = \frac{f(x)}{\int{f(x)}^{2}dx}$$

Not quite!

$$\text{Let }u=\int{f(x)\,dx}, \text{ then } \frac{d}{dx}\left(\frac{1}{u}\right)=-\frac{1}{u^2}\,\frac{du}{dx}$$

So you should have: $$\frac{1}{f(x)}=\frac{f(x)}{\displaystyle \left(\int{f(x)}dx}\right)^{2}$$

Solve for f(x) and differentiate, or let g' = f & solve for g.

 

[DrummingAtom]

Haha, yeah, my bad I should have finished that. Would that cause the e^-x solution to disappear though? Or do you have to $$\pm$$ on both sides?

 

[SammyS]

You will have ±. (You only need it on one side!)

 

[tiny-tim]

i stil don't understand why you're using all those integrals …

i got g'² = g² ​