Find a continuous function $f$ that satisfies...

Solution 1:

Hint: $xf(x) = x + \int_1^x f(t) \, dt $, except possibly at $x=0$.

Differentiate this to conclude that $f(x) = \ln x + C $.

Evaluate at $x=1$.

Solution 2:

I will also assume that $f$ is differentiable. We are given that $f(x)=1+\frac{1}{x} \int_1^xf(t)dt$. Multiplying by $x$ we see that $$xf(x)=x+\int_1^xf(t)dt$$ Differentiating, $$f(x)+xf'(x)=1+f(x)$$ Subtracting $f(x)$ then dividing through by $x$,

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

Now, integrating we obtain

$$f(x)=\ln(x)+C$$

We must now deal with the initial conditions. from the original condition that $f(1)=1$. So, $$f(1) = 1 =\ln(1)+C \implies C=1$$

Thus the only continuous (+differentiable) function that satisfies the given condition is $$f(x)=\ln(x)+1.$$

Solution 3:

$\displaystyle F(x) = \int_1^xf(t) dt$

So

$\displaystyle \frac{dF}{dx} = 1 + \frac{1}{x} F$

with condition $F(1)=0$

Solution 4:

Given that

$$f(x)=1+\frac{1}{x}\int_1^x f(t)dt$$ Differentiating both sides

$$ f'(x)=\frac{1}{x}f(x)-\frac{1}{x^2}\int_1^xf(t)dt$$ $\implies$

$$ f'(x)=\frac{1}{x}f(x)-\frac{1}{x}\left(\frac{1}{x}\int_1^xf(t)dt\right)$$ But

$$\frac{1}{x}\int_1^x f(t)dt=f(x)-1$$ So $$f'(x)=\frac{f(x}{x}-\frac{1}{x}\left(f(x)-1\right)$$ $\implies$

$$f'(x)=\frac{1}{x} \implies f(x)=Ln(x)+c $$ Finally Use $f(1)=1$ to get the value of $c$