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Compute the value of an integral of a function satisfying a given integral equation

Let f be a function continuous on the positive real axis. Assume that for all positive x, y, the integral

    \[ \int_x^{xy} f(t) \, dt \]

depends only on y (hence, is independent of our choice of x). Compute the value of

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

if we are given that f(2) = 2.


First, since the integral \int_x^{xy} f(t) \, dt is independent of our choice of x, we may let x = 1 to obtain,

    \[ \int_x^{xy} f(t) \, dt = \int_1^y f(t) \, dt. \]

Next, we can use the fact that the integral is additive with respect to the interval of integration to write,

    \[ \int_1^{xy} f(t) \, dt = \int_1^x f(t) \, dt + \int_x^{xy} f(t) \, dt. \]

Substituting in our expression for \int_x^{xy} f(t) \, dt we then have

    \[ \int_1^{xy} f(t) \, dt = \int_1^x f(t) \, dt + \int_1^y f(t) \, dt. \]

This implies,

    \[ A(xy) = A(x) + A(y). \]

Now, holding y fixed and differentiating both sides of this equation with respect to x, we find,

    \begin{align*}  y A'(xy) = A'(x) && \implies && A'(xy) &= \frac{A'(x)}{y} \\   && \implies && \int A'(xy) \, dx &= \int \frac{A'(x)}{y} \, dx \\  && \implies && f(xy) &= \frac{f(x)}{y} \end{align*}

by the fundamental theorem of calculus. Then, since we are given that f(2) = 2 we have,

    \[ f(2y)  = \frac{2}{y} \quad \implies \quad f(1) = 4. \]

Furthermore, since f(xy) = \frac{f(x)}{y}, letting x = 1 gives us

    \[ f(y) = \frac{f(1)}{y} = \frac{4}{y}. \]

Therefore,

    \[ A(x) = \int_1^x f(t) \, dt \quad \implies \quad A(x) = 4 \int_1^x \frac{1}{t} \, dt = 4 \log x. \]

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