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Test the improper integral ∫ 1 / (x1/2 log x) for convergence

Test the following improper integral for convergence:

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


The integral diverges.

Proof. First, we show that \log x < \sqrt{x} for all x> 0. To do this let

    \[ f(x) = \sqrt{x} - \log x \quad \implies \quad f'(x) = \frac{1}{2 \sqrt{x}} - \frac{1}{x} = \frac{\sqrt{x} - 2}{2x}. \]

This derivative is 0 at x = 4 and is less than 0 for x < 4 and greater than 0 for x > 4. Hence, f(x) has a minimum at x = 4. But, f(4) = \sqrt{4} - \log 4 = 2 - 2 \log 2 > 0 (since 2 < e implies \log 2 < 1). So, f(x) has a minimum at x = 4 and is positive there; thus, it is positive for all x > 0, or

    \[ f(x) = \sqrt{x} - \log x > 0 \quad \implies \quad \sqrt{x} > \log x \qquad \text{for all } x > 0. \]

So, since \log x < \sqrt{x} we know

    \[ \frac{1}{\sqrt{x} \log x} > \frac{1}{\sqrt{x} \cdot \sqrt{x}} = \frac{1}{x}. \]

But then, consider the limit

    \begin{align*}  \lim_{x \to 0^+} \frac{ \frac{1}{x} }{ \frac{1}{\sqrt{x} \log x}} &= \lim_{x \to 0^+} \frac{\log x}{\sqrt{x}} \\  &= \lim_{x \to 0^+} \frac{\log x}{x^{\frac{1}{2}}} \\[9pt]  &= 0 &(\text{by Theorem 7.11}). \end{align*}

Therefore, by the limit comparison test (Theorem 10.25), the convergence of \frac{1}{\sqrt{x} \log x} would imply the convergence of \int_{0^+}^a \frac{1}{x} (for any 0 < a < 1), but we know by Example 5 that this integral diverges. Hence, we must also have the divergence of

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

One comment

  1. Rafael Deiga says:

    You are wrong about theorem 7.11. The hypothesis of theorem 7.11 is with x is going to infinity, not to zero.

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