First, note that if f(0) is 0, then by substituting u=ax in the integral of f(x)/x, our integral is the difference of two equal integrals and so is 0 (the integrals are finite because f is 0 at 0 and differentiable there. Note I make no requirement of continuity).

Second, note that if f is the characteristic function of the interval [0, 1? --- i.e.

1, 0<=x<=1

f (x) =

0 otherwise

then a little arithmetic reduces our integral to that of 1/x from 1/a to 1 (assuming a>1; if a <= 1 the reasoning is similar), which is ln(a) = f(0)ln(a) as required. Call this function g.

Finally, note that the operator which takes the function f to the value of our integral is linear, and that every function meeting the hypotheses (incidentally, I should have said `differentiable from the right', or else replaced the characteristic function of [0,1? above by that of (-infinity, 1]; but it really doesn't matter) is a linear combination of one which is 0 at 0 and g, to wit

f(x) = f(0)g(x) + (f(x) - g(x)f(0)).

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