Test the following improper integral for convergence:
![\[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-f587d2ff657f2a1eb5b45ce35104ef15_l3.png "Rendered by QuickLaTeX.com")
The integral converges.
Proof. First, we write
![\[ \int_{0^+}^{1^-1} \frac{\log x}{1-x} \,dx = \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx + \int_{\frac{1}{2}}^{1^-} \frac{\log x}{1-x} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-dd94406096e673a94f18807772f7bac5_l3.png "Rendered by QuickLaTeX.com")
For the first integral we know for all 
we have 
; hence,
![\[ \frac{\log x}{1-x} < 2 \log x \qquad \text{for all } x \in \left( 0, \frac{1}{2} \right). \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d8dbd791e2f285c37649283da947d343_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-ce216434319ffd0bf90e17343d555cd3_l3.png "Rendered by QuickLaTeX.com")
(We know 
by L’Hopital’s, writing 
or by Example 2 on page 302 of Apostol.) Hence,
![\[ \int_{0^+}^{\frac{1}{2}} \frac{\log x}{1-x} \, dx \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-14c873cdbba32e2d179d157fb66772a3_l3.png "Rendered by QuickLaTeX.com")
converges by the comparison theorem (Theorem 10.24 on page 418 of Apostol).
For the second integral, we use the expansion of 
about 
,
![\[ \log x = (x-1) - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \cdots. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-eef6e91fcf0845d51af608be81d3b3c1_l3.png "Rendered by QuickLaTeX.com")
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*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-fded1fb5ab5592e4147b7b99b15ed568_l3.png "Rendered by QuickLaTeX.com")
But this integral converges since it has no singularities.
Thus, we have established the convergence of
![\[ \int_{0^+}^{1^-} \frac{\log x}{1-x} \, dx. \qquad \blacksquare \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-79d3c966bc5b8b7862baf0394c57168c_l3.png "Rendered by QuickLaTeX.com")
Can we substitute 1-x=t from the very beginning, then use Taylor expansion for log(1-t), divide it by t, integrate and get ourselves Riemann Zeta function with s=2 which converges by comparison test with 1/(n^2+n) telescoping series somewhere in the book?
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.