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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.

Harry Peter
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Ryan
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  • What is the domain of $f$ supposed to be? – Potato Jun 11 '13 at 00:17
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    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 00:18
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    Also, have you tried differentiating with respect to $x$? – Potato Jun 11 '13 at 00:18
  • Not specified in the problem statement - I would assume it to be all real numbers – Ryan Jun 11 '13 at 00:18
  • @Ryan $x=0{}{}$? – Git Gud Jun 11 '13 at 00:19
  • If you differentiate with respect to $x$, which was what I tried first, you get a differential equation that contains integrals so solving it could prove to be quite tricky. – Ryan Jun 11 '13 at 00:19
  • @GitGud I was looking at that too, but I suppose the problem is nice enough that the integral goes to zero there and cancels it. – Potato Jun 11 '13 at 00:20
  • @Potato It could be that the limit is well-behaved, but it still is undefined at $x=0$ – Git Gud Jun 11 '13 at 00:21
  • @RyanPeden The integral is the one that already appears in the problem, so you can substitute to remove it. There are cleverer solutions below, though. – Potato Jun 11 '13 at 00:23
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    @GitGud Oh, come on, are we really going to get pedantic about removable singularities? – Potato Jun 11 '13 at 00:23

4 Answers4

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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 $x=1$.

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

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

So

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

with condition $F(1)=0$

bob.sacamento
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1

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 $c$

Ekaveera Gouribhatla
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