What is $f(x)$?

Question

I have the following question.

When $x > 0$, $f(x)$ is differentiable and satisfies that $$f(x)=1+\frac{1}{x}{\int_{1}^{x}}f(t)dt.$$

Then What is $f(x)$?

Thanks for your help.

Answer 1

Since you have an integral equation, you might want to turn it into a differential equation by rearranging terms and differentiating:

$f(x) = 1 + \dfrac{1}{x}\displaystyle\int_{1}^{x}f(t)\,dt$

$xf(x) - x = \displaystyle\int_{1}^{x}f(t)\,dt$

$xf'(x) + f(x) - 1 = f(x)$

$f'(x) = \dfrac{1}{x}$

The solution is $f(x) = \ln x + C$ for some constant $C$. Plugging $x = 1$ into the original integral equation yields $f(1) = 1$, so $C = 1$, and thus, $f(x) = 1 + \ln x$.

Answer 2

First, notice that $f(1) = 1$. By differentiation, we have: $$f'(x) = -\frac{1}{x^2}\int_{1}^{x} f(t) \mathrm{d}t+ \frac{1}{x}f(x)$$ But, on the other side, $$\int_{1}^x f(t) \mathrm{d}t = x(f(x) - 1)$$ By substituition, we get: \begin{align} f'(x) &= -\frac{1}{x^2}x(f(x) - 1) + \frac{1}{x}f(x) \\ &= -\frac{f(x)}{x} + \frac{1}{x} + \frac{f(x)}{x} \\ &= \frac{1}{x} \end{align} Then, by integration, we have $f(x) = \ln |x| + c$. Since $f(1) = 1$, we obtain $f(x) = \ln |x| + 1$. Ok?

Answer 3

Since $f$ is differentiable,

\begin{align} f'(x) &=0+ \frac{1}{x}f(x)+ \left(\frac{-1}{x^2}\right) \left( \int_1^x f(t) dt\right) \tag{1}\\ &= \frac{1}{x}f(x) -\frac{1}{x} \left(f(x)-1 \right)\\ &=\frac{1}{x} \end{align}

So $f(x)= \ln x+C$ where $C$ is some arbitrary constant. Since $f(1)=1$ it follows that $C=1$. Note that on (1) we use the fundamental theorem of calculus with the product rule.