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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

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

One comment

  1. Artem says:

    In the first part, we cannot really use the comparison theorem directly. In fact, the logarithm from 0 to 1/2 is negative, thus your inequality is in the wrong order. However, if it is multiplied by the minus, we arrive at the similar identity, which can now be used with the comparison theorem.

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