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Basically, I came accross this relation

$$\int f(x)g(x)dx=\int f(x)dx\int g(x)dx+c$$

in a "joke" leading me to wonder what are the actual pair of functions that satisfy this relation.

For my initial attempt i assumed the funtions to be continuous and differentiable over $\mathbb{R}$.

Now differentiating both sides w.r.t $x$ we get $$f(x) g(x)= f(x)\int g(x)\ dx +g(x)\int f(x)\ dx$$ Divding by $f(x),g(x)$ both sides we get $$1=\frac{\int g(x) \ dx}{g(x)} + \frac{\int f(x)\ dx}{f(x)}$$

and on solving the integral we get $$1=\frac{1}{\ln(F(x))}+\frac{1}{\ln(G(x))}$$ (ignoring the constant, will add it later, and assuming $F(x),G(x)$ to be the antiderivatives of $f(x)$ and $g(x)$ respectively)

This can be further simplified to

$$\ln(F(x))\ln(G(x))=\ln(F(x)G(x))$$

which is basically what the question was (the constants of int can be written as one constant in the end again to match it)

Lost after this. Not integrating doesn't help either .

Sigma Algebra
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Scypher_Tzu
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1 Answers1

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Let $F(x) = \int f(x) \text dx$ and $G(x) = \int g(x) \text dx$. Then $F'(x) = f(x)$ and $G'(x) = g(x)$, and we may write the equation as:

\begin{align} \int f(x)g(x) \text dx = F(x)G(x) + c \end{align}

which is equivalent to

\begin{align} F'(x)G'(x) = f(x)g(x) = F'(x)G(x) + F(x)G'(x) \end{align} by differentiating both sides of the equation. Then:

$\Rightarrow F'(x)G'(x)-F'(x)G(x)=F(x)G'(x)$

$\Rightarrow F'(x)(G'(x)-G(x)) = F(x)G'(x)$

$\Rightarrow \displaystyle \frac{F'(x)}{F(x)} = \frac{G'(x)}{G'(x)-G(x)}$

Thus, since $\frac{\text d}{\text dx} (\text{ln} |F(x)|) = \frac {F'(x)}{F(x)}$ by the chain rule, and integrating both sides of the equation, we get:

\begin{align} \text {ln} |F(x)| + C = \int \displaystyle \frac{G'(x)}{G'(x)-G(x)} \text dx \end{align}

Then, exponentiating both sides to $e$, and setting $A = e^C$ we get:

\begin{align} A|F(x)|= \text{exp}\displaystyle \left( \int \frac{G'(x)}{G'(x)-G(x)} \text dx \right) \end{align}. Here are two examples of solutions for $f$ and $g$:

\begin{equation} f(x) = ±\frac{k}{(1-x)^2},\text{ }g(x) = 1 \text{ corresponding to } F(x) = ±\frac{k}{1-x} \text{ and } G(x) = x \end{equation} \begin{equation} f(x) = ±kre^{rx},\text{ } g(x) = \frac{r-1}{r} e^{\frac{r-1}{r}x} \text{ for } r\neq1 \text{ corresponding to } F(x) = ±ke^{rx} \text{ and } G(x) = e^{\frac{r}{r-1}x}\end{equation}

Integrable Chessman
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