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Test the improper integral ∫ log x / (1-x) for convergence

Test the following improper integral for convergence:

    \[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \]


The integral converges.

Proof. First, we write

    \[ \int_{0^+}^{1^-1} \frac{\log x}{1-x} \,dx = \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx  + \int_{\frac{1}{2}}^{1^-} \frac{\log x}{1-x} \, dx. \]

For the first integral we know for all x \in \left( 0 , \frac{1}{2} \right) we have \frac{1}{2} \leq (1-x) < 1; hence,

    \[ \frac{\log x}{1-x} < 2 \log x \qquad \text{for all } x \in \left( 0, \frac{1}{2} \right). \]

Then, the integral

    \begin{align*}  2 \int_{0^+}^{\frac{1}{2}} \log x \, dx &= 2 \cdot \lim_{a \to 0^+} \int_a^{\frac{1}{2}} \log x \, dx \\[9pt]  &= 2 \cdot \lim_{a \to 0^+} \left( x \log x - x \right)\Bigr \rvert_a^{\frac{1}{2}} \\[9pt]  &= 2 \cdot \lim_{a \to 0^+} \left( \frac{1}{2} \log \frac{1}{2} - \frac{1}{2} - a \log a + a \right) \\[9pt]  &= \log \frac{1}{2} - 1 \\[9pt]  &= - \log 2 - 1. \end{align*}

(We know \lim_{a \to 0} x \log x = 0 by L’Hopital’s, writing x \log x = \frac{\log x}{1/x} or by Example 2 on page 302 of Apostol.) Hence,

    \[ \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx \]

converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).

For the second integral, we use the expansion of \log x about x = 1,

    \[ \log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots. \]

Then we have

    \begin{align*}  \int_{\frac{1}{2}}^{1^-} \frac{\log x}{1-x} \, dx &= \int_{\frac{1}{2}}^{1^-} \frac{ (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots }{1-x} \, dx \\[9pt]  &= \int_{\frac{1}{2}}^{1^-} \left( -1 + \frac{x-1}{2} - \frac{(x-1)^2}{3} + \cdots \right) \, dx. \end{align*}

But this integral converges since it has no singularities.

Thus, we have established the convergence of

    \[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \qquad \blacksquare \]

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