Test the following improper integral for convergence:
![\[ \int_{0^+}^1 \frac{\log x}{\sqrt{x}} \, dx. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-67e495fd6a3b5173214e651fffa09464_l3.png "Rendered by QuickLaTeX.com")
The integral converges.
Proof. We can compute the value of the improper integral directly. First, we can evaluate the indefinite integral using integration by parts with
Therefore,
![\begin{align*} \int \frac{\log x}{\sqrt{x}} \, dx &= uv - \int v \, du \\[9pt] &= 2 \sqrt{x} \log x - \int \frac{2\sqrt{x}}{x} \, dx \\[9pt] &= 2 \sqrt{x} \log x - 2 \int x^{-\frac{1}{2}} \, dx \\[9pt] &= 2 \sqrt{x} \log x - 4 \sqrt{x}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-41b6b2b1f2f6963d2886529d9813e602_l3.png "Rendered by QuickLaTeX.com")
So, to evaluate the improper integral we take the limit,
![\begin{align*} \int_{0^+}^1 \frac{\log x}{\sqrt{x}} \, dx &= \lim_{a \to 0^+} \int_a^1 \frac{\log x}{\sqrt{x}} \, dx \\[9pt] &= \lim_{a \to 0^+} \left( 2 \sqrt{x} \log x - 4 \sqrt{x} \right)\Bigr \rvert_a^1 \\[9pt] &= \lim_{a \to 0^+} \left( -4 - 2 \sqrt{a} \log a + 4 \sqrt{a} \right) \\[9pt] &= -4. \qquad \blacksquare \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-010fd24ca424ba14201fe204828bad39_l3.png "Rendered by QuickLaTeX.com")
The substitution for
will not give the limit 0, since logarithm is not defined there – it is an indeterminate form 