# Find a continuous function $f$ that satisfies…

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.

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What is the domain of $f$ supposed to be? – Potato Jun 11 '13 at 0:17
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... – treble Jun 11 '13 at 0:18
Also, have you tried differentiating with respect to $x$? – Potato Jun 11 '13 at 0:18
Not specified in the problem statement - I would assume it to be all real numbers – Ryan Jun 11 '13 at 0:18
@GitGud Oh, come on, are we really going to get pedantic about removable singularities? – 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 $x=1$.

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

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

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

So

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

with condition $F(1)=0$

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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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