Find all functions: $\left ( \int \frac{dx}{f(x)} \right )\left ( \int f(x)dx \right )=c$ Find all functions $f(x)$ so that:
$$
 \left ( \int \frac{dx}{f(x)} \right )\left ( \int f(x)dx \right )=c
$$
where c is a constant.
My attempt was to differentiate both sides but that appears to lead me to nowhere. I would assume that the solution involves some sort of differential equations but I am very likely to be wrong.
Thanks in advance!
 A: If you differentiate both sides you get
\begin{equation*}
\frac{1}{f(x)} \Big( \int f(x) dx \Big) + \Big( \int \frac{dx}{f(x)} \Big) f(x) = 0
\end{equation*}
or
\begin{equation*}
\Big( \int f(x) dx \Big) =- \Big( \int \frac{dx}{f(x)} \Big) ( f(x) )^2 
\end{equation*}
Differentiate again:
\begin{equation*}
f(x) =- \frac{1}{f(x)} (f(x))^2 -  \Big( \int \frac{dx}{f(x)} \Big)2 f'(x) f(x)
\end{equation*}
or
\begin{equation*}
f(x) =-  \Big( \int \frac{dx}{f(x)} \Big) f'(x) f(x)
\end{equation*}
Since $f(x) \neq 0$, this leads to
\begin{equation*}
1 =-  \Big( \int \frac{dx}{f(x)} \Big) f'(x)
\end{equation*}
or
\begin{equation*}
\frac{-1}{f'(x)} = \Big( \int \frac{dx}{f(x)} \Big) 
\end{equation*}
Differentiating again,
\begin{equation*}
\frac{f''(x)}{(f'(x))^2} = \frac{1}{f(x)} 
\end{equation*}
By rearranging things,
\begin{equation*}
\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)} 
\end{equation*}
From now on you can continue, just integrate twice and carry on some constants to get the full solution.
A: The expressions $\int f(x)\>dx$ and $\int{1\over f(x)}\>dx$ do not denote functions, but sets of functions. Therefore I'm interpreting your problem as follows: For given $c\in{\mathbb R}$ find functions $F$ and $G$ defined in a common interval $I\subset{\mathbb R}$, such that
$$F(x)\>G(x)= c,\qquad F'(x)\>G'(x)=1\qquad(x\in I)\ .\tag{1}$$
It is easily seen that $c=0$ does not allow any solutions. Therefore we may assume $c=\pm\omega^2$ with $\omega>0$.
Case I: $\ c=\omega^2>0$: The first equation $(1)$ then implies that we may write
$$F(x)=\omega e^{p(x)}, \quad G(x)=\omega e^{-p(x)}$$
for a certain function $x\mapsto p(x)$. From the second equation $(1)$ it then follows that
$$\omega^2 p'(x)^2=-1\ ,$$
or $p'(x)=\pm\, i/\omega$. This leads in turn to to $p(x)=\pm ix/\omega +C$, $\> F(x)=\omega e^{\pm ix/\omega+C}$, and finally
$$f(x)=F'(x)=C'e^{\pm ix/\omega},\qquad C'\ne0\ .$$
One now has to verify that such $f$ satisfy all given conditions.
The Case II: $\ c=-\omega^2<0$ is similar, but allows of real solutions. I leave it to you.
A: Trying $$f(x) = \mathrm{e}^{i\sqrt{1/c}x}$$
We find
$$
\int f(x)dx = -i\sqrt{c}\mathrm{e}^{i\sqrt{1/c}x}\\
\int\frac{1}{f(x)}dx =i\sqrt{c}\mathrm{e}^{-i\sqrt{1/c}x}
$$
A: Given $\displaystyle \int\frac{1}{f(x)}dx\cdot \int f(x)dx =c..............................................(1)$
Now Differentiate both side w. r to $x$, We Get
$\displaystyle \int \frac{1}{f(x)}dx\cdot f(x)+\frac{1}{f(x)}\cdot \int f(x)dx = 0...................................................................(2)$
Now from equation $(1)$, We get $\displaystyle \int\frac{1}{f(x)}dx = \frac{c}{\int f(x)dx}$
and put into equation $(2)$, We get $\displaystyle \frac{c\cdot f(x)}{\int f(x)dx}+\frac{\int f(x)dx}{f(x)} = 0$
So $\displaystyle \left(\int f(x)dx\right)^2 = -c\cdot \left(f(x)\right)^2$
Now Let $-c=k^2\;,$ Then $\displaystyle \left(\int f(x)dx\right)^2 = k^2\cdot \left(f(x)\right)^2$
So $\displaystyle \int f(x)dx = \pm k\cdot f(x)\;,$ Now Differentitae both side w  r to $x\;,$ we get
$\displaystyle \Rightarrow f(x) = \pm k\cdot f'(x)\Rightarrow \frac{f'(x)}{f(x)}=\pm q\;,$ where $\displaystyle q = \pm \frac{1}{k}$
Now Integrate both side w. r to $x\;,$ we get
$\displaystyle \Rightarrow \int \frac{f'(x)}{f(x)} = \pm q\int dx\Rightarrow \ln\left|f(x)\right| = \pm q+\ln \left|c\right|\Rightarrow f(x)=c\cdot e^{\pm qx}$
