Find a continuous function $f$ that satisfies

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

Note: I tried differentiating with respect to $x$ to get an ODE but you get one that contains integrals - likely difficult to solve.

  • $\begingroup$ What is the domain of $f$ supposed to be? $\endgroup$ – Potato Jun 11 '13 at 0:17
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    $\begingroup$ I downvoted because this post "does not show any research effort." I would upvote if you explained what motivated you to ask the question, what efforts you have used to try to solve it that didn't work, etc... $\endgroup$ – treble Jun 11 '13 at 0:18
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    $\begingroup$ Also, have you tried differentiating with respect to $x$? $\endgroup$ – Potato Jun 11 '13 at 0:18
  • $\begingroup$ Not specified in the problem statement - I would assume it to be all real numbers $\endgroup$ – Ryan Jun 11 '13 at 0:18
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    $\begingroup$ @GitGud Oh, come on, are we really going to get pedantic about removable singularities? $\endgroup$ – Potato Jun 11 '13 at 0:23

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 $x=1$.


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 $x$,


Now, integrating we obtain


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 $$f(x)=\ln(x)+1.$$

  • $\begingroup$ $ \ln( 0 ) + C = 0 \implies C = 0$? $\endgroup$ – Lucas Jun 11 '13 at 4:23
  • $\begingroup$ @Lucas My mistake, but an easy fix. $\endgroup$ – Gamma Function Jun 11 '13 at 4:32
  • $\begingroup$ I believe you meant to write that $ f(1) = 1 $ (and that $ f(1) = 1 = \ln(0) + C \implies C = 1 $)? $\endgroup$ – ruakh Jun 11 '13 at 6:06
  • $\begingroup$ Good catch! I really should have been more careful and checked to see if the solution I provided worked. You are entirely correct. $\endgroup$ – Gamma Function Jun 11 '13 at 6:42

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


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

with condition $F(1)=0$


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 $c$


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