Calculus functional equation: Find all positive $f(x)$ such that $\int f\left(x\right)dx\cdot \int \frac{1}{f\left(x\right)}dx=-1$

Question

Is this problem possible to solve, other than through inspection? Through a basic argument, it can be shown that $\int \frac{1}{f(x)}\,ndx$ cannot be described with elementary functions, and this is the main issue I'm facing when trying to analytically solve it.

Answer 1

Here is a hint on how to solve this rigorously: differentiate your equation to obtain $$f \int \frac{1}{f} + \frac{1}{f}\int f = 0 \tag{1}$$ and move terms to obtain \begin{equation}f^2 \int \frac{1}{f} + \int f = 0 \tag{2}.\end{equation} Continue to differentiate (after moving terms appropriately) to get rid of the antiderivatives. You should end up with the second order ODE $$(f')^2 = f\cdot f''$$ which you should be able to solve.

Edit: I'll just fill in the rest. Differentiate equation (2) to see $$2ff'\int \frac{1}{f} + f + f = 0$$ By positivity of $f$ we can factor $f$ to see $$f' \int \frac{1}{f} = -1 \iff \int \frac{1}{f} = \frac{-1}{f'} \tag{3}.$$ If we assume both of these are differentiable, we differentiate both sides to conclude $$\frac{1}{f} = \frac{f''}{(f')^2}\tag{4}.$$ To solve this ODE, rewrite equation $(4)$ as $$\frac{f'}{f} = \frac{f''}{f'} \implies \ln(f) + C = \ln(f') \implies \frac{f}{f'} = C \implies f(x) = De^{Cx}$$ for some positive constant $D$ and constant $C$.

Now we need to check which of these satisfies our equation, because we reduced the amount of information we had by taking derivatives. It's clear that we can set $D = 1$ WLOG. Likewise, if $C = 0$, our functional equation doesn't work (in particular, $f' = 0,$ so equations (3,4) won't hold.)

So, set $C \neq 0, f(x)= e^{Cx}$ and check: $$\int f \cdot \int \frac{1}{f} = \frac{e^{Cx}}{C} \cdot \frac{e^{-Cx}}{-C} = -1.$$

Our final answer is $f(x) = De^{Cx}$ for $D > 0, C = \pm 1.$.

Answer 2

Well, here my try, I'll put every note if I think can be something wrong:

First, this problem does not specify how to do the integral or what we can do, for example with the constants or if they must be evaluated on a interval or are indefinite, I'll assume I can play with them in order to find $f(x)$ with some constants that works.

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

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

Replace with eq.1:

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

After here, will be there any a $f(x)$ with can kill every limit to get 0?

Reorder:

$$\frac{f(x)}{\int f(x) \,dx}*(-1 + (\frac{\int f(x) \,dx}{f(x)})^2)=0$$

Two options:

One:

if $\frac{f(x)}{\int f(x) \,dx}=0$ means $(\frac{\int f(x) \,dx}{f(x)})^2$ will be infinite, in that case the equation can not be $0$.

Two

$$(-1 + (\frac{\int f(x) \,dx}{f(x)})^2)=0$$

This will only happens if

$$\int f(x) \,dx=f(x)$$

Solution:

$$c*e^{x}$$

Or

$$\int f(x) \,dx=-f(x)$$

Solution:

$$c*e^{-x}$$

Tests:

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

One:

$$(c_1*e^{x}+c_2)*(\frac{e^{-x}}{-c_1}+c_3)=-1$$

$$-1+c_3*c_1*e^{x}+\frac{e^{-x}*c_2}{-c_1}+c_2*c_3 = -1$$

How we want a function level to be equal, $c_2=0$ and $c_3=0$

Two:

$$(-c_1*e^{-x}+c_2)*(\frac{e^{x}}{c_1}+c_3)=-1$$

$$-1+c_3*-c_1*e^{-x}+\frac{c_2*e^{x}}{c_1}+c_2*c_3==-1$$

How we want a function level to be equal, $c_2=0$ and $c_3=0$

Now, just check where $f(x)>=0$ which is in all places.

That would be my answer, but no idea if there is a complex solution (?