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I need to find a continuous function defined for real and positive $x$ such that $f(x) = 1 + \frac1{x} \int_1^x f(t)\ \mathrm{d}t$. What I did is the following:

$$\begin{align*}f(x) &= 1 + \frac1{x} \int_1^x f(t)\ \mathrm{d}t\\ x(f(x) - 1) &= \int_1^x f(t) \ \mathrm{d}t \end{align*}$$

Differentiating both sides:

$$\begin{align*} f(x) - 1 + x f'(x) &= f(x)\\ f'(x) &= \frac1{x}\\ f(x) &= \ln x \end{align*}$$

Which looks alright. But when checking to make sure, I get this:

$$\begin{align*} f(x) &= 1 + \frac1{x} \int_1^x \ln t \ \mathrm{d}t\\ f(x) &= 1 + \frac1{x} (x(\ln x -1)|_1^x)\\ f(x) &= 1 + \frac1{x} (x(\ln x -1) - (\ln 1 - 1))\\ f(x) &= 1 + \ln x - 1 + \frac1{x}\\ f(x) &= \ln x + \frac1{x} \end{align*}$$

which is different from what I got before. What is my mistake here?

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How about the constant when integrating 1/x into lnx? –  gary Nov 25 '11 at 1:51
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You are correct up to here: $ f'(x) = 1/x.$ However, that does not imply immediately $ f(x) = \ln x$, but rather $ f(x) = \ln x + C $ where $C$ is the arbitrary constant of integration. –  Ragib Zaman Nov 25 '11 at 1:52
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We have $f(x)=\ln x +C$. Evaluate $C$ by going back to the original equation and putting $x=1$. The integral dies, and you get $\ln( 1) + C=1$. –  André Nicolas Nov 25 '11 at 1:59
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1 Answer 1

up vote 6 down vote accepted

You left out the constant of integration... the answer is actually $\ln(x) + 1$.

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That was dumb. Thanks. –  Javier Badia Nov 25 '11 at 1:55
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