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Find a function whose integral satisfies a given equation

Let f(x) be a continuous function on \mathbb{R}_{>0} satisfying

    \[ \int_1^{xy} f(t) \, dt = y \int_1^x f(t) \, dt + x \int_1^y f(t) \, dt \qquad \text{for all } x,y > 0.\]

Find the value of f(x) for all positive real x if f(1) = 3.


First, we use the fundamental theorem of calculus to differentiate both sides of the given equation with respect to y,

    \begin{align*}  && \int_1^{xy} f(t) \, dt &= y \int_1^x f(t) \, dt + x \int_1^y f(t) \, dt \\ \implies && x f(xy) &= \int_1^x f(t) \, dt + x f(y).  \end{align*}

Then, evaluating this at y = 1 we have

    \[ x f(x) = \int_1^x f(t) \,dt + x f(1) \quad \implies \quad xf(x) = \int_1^x f(t) \, dt + 3x. \]

Next, we differentiate this with respect to x,

    \begin{align*}  && xf(x) &= \int_1^x f(t) \, dt + 3x \\ \implies && f(x) + xf'(x) &= f(x) + 3 \\ \implies && f'(x) &= \frac{3}{x} \\ \implies && f(x) &= 3 \log x + C. \end{align*}

Finally, since f(1) = 3 we have

    \[ f(1) = 3 \log 1 + C \quad \implies \quad C = 3. \]

Therefore,

    \[ f(x) = 3 \log x  + 3 \qquad \text{for } x > 0. \]

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