# 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.

• The problem is ill-defined unless you specify lower bounds $a$ and $b$ or if you are looking for solutions $f(x)$, $a$, and $b$ satisfying the above. – abnry Dec 10 '14 at 17:38
• I am not sure if it's problem with my maybe informal wording, but basically you are given the equation and you try to solve for all functions $f(x)$ so that the equation can be satisfied. Klaramun has given a very nice solution to the question, so thanks anyways! – B1ll Dec 10 '14 at 19:56

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.

• The last one reads $(\ln f'(x))' = (\ln f(x))'$, so $f' = C f$ for some $C$, and, same trick, this implies $(\ln f)' = C$ giving the solution $f = \exp(C_1 x + C_2)$. (Of course next step is checking for what constants it satisfies the equation.) – Myself Dec 10 '14 at 19:53

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.

Trying $$f(x) = \mathrm{e}^{i\sqrt{1/c}x}$$

We find $$\int f(x)dx = -i\sqrt{c}\mathrm{e}^{i\sqrt{1/c}x}\\ \int\frac{1}{f(x)}dx =i\sqrt{c}\mathrm{e}^{-i\sqrt{1/c}x}$$

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

$\displaystyle \Rightarrow \int \frac{f'(x)}{f(x)} = \pm q\int dx\Rightarrow \ln\left|f(x)\right| = \pm q+\ln \left|c\right|\Rightarrow f(x)=c\cdot e^{\pm qx}$