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Compute f(x) + f(1/x) where f satisfies a given integral equation

by RoRi

Define a function by an integral equation as follows:

    \[ f(x) = \int_1^x \frac{\log t}{t+1} \, dx \qquad \text{for} \quad x > 0. \]

Using this formula compute f(x) + f(1/x).

Substituting x and \frac{1}{x} into the formula for f(x) we have

    \[ f(x) + f \left( \frac{1}{x} \right) = \int_1^x \frac{\log t}{t+1} \, dt + \int_1^{\frac{1}{x}} \frac{\log t}{t+1} \, dt. \]

For the second integral we make the substitution t = \frac{1}{u}, dt = -\frac{1}{u^2} \, du. This gives us

    \begin{align*} f(x) + f \left( \frac{1}{x} \right) &= \int_1^x \frac{\log t}{t+1} \, dt + \int_1^x \frac{\log \left( \frac{1}{u} \right)}{\frac{1}{u} + 1} \left(-\frac{1}{u^2} \right) \, du \\[9pt] &= \int_1^x \frac{\log t}{t+1} \, dt + \int_1^x \frac{\log u}{u+u^2} \, du \\[9pt] &= \int_1^x \left( \frac{\log t}{t+1} + \frac{\log t}{t+t^2} \right) \, dt \\[9pt] &= \int_1^x \left( \frac{t \log t + \log t}{t(t+1)} \right) \, dt \\[9pt] &= \int_1^x \frac{\log t}{t} \, dt \\[9pt] &= \left. \frac{(\log t)^2}{t} \right|_1^x \\[9pt] &= \frac{(\log x)^2}{2}. \end{align*}

As a check, we evaluate at x = 2,

    \[ f(2) + f \left( \frac{1}{2} \right) = \frac{1}{2} (\log 2)^2. \]

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