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Test the improper integral ∫ log x / x1/2 for convergence

Test the following improper integral for convergence:

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


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

    \begin{align*}  u &= \log x & du &= \frac{1}{x} \\ dv &= x^{-\frac{1}{2}} & v &= 2 \sqrt{x}. \end{align*}

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*}

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*}

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