Home » Convergence » Page 3

Tag: Convergence

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

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

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

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

Test the improper integral ∫ 1 / (x3 + 1)1/2 for convergence

Test the following improper integral for convergence:

    \[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \]


The integral converges.

Proof. We know the integral \int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx converges (example #1 on page 417 of Apostol). Applying the limit comparison test (by the note to Theorem 10.25 on page 418, which says that if \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 then the convergence of \int g(x) implies the convergence of \int f(x).) we have

    \begin{align*}  \lim_{x \to +\infty} \frac{f(x)}{g(x)} &= \lim_{x \to +\infty} \frac{ \frac{1}{\sqrt{x^3+1}}}{\frac{1}{x}} \\[9pt]  &= \lim_{x \to +\infty} \frac{x}{\sqrt{x^3+1}} \\[9pt]  &= \lim_{x \to +\infty} \frac{1}{x^{\frac{1}{2}} \sqrt{1 + \frac{1}{x}}} \\[9pt]  &= 0. \end{align*}

Since we know \int_0^{\infty} \frac{1}{x^{\frac{3}{2}}} \, dx converges the theorem establishes the convergence of

    \[ \int_0^{\infty} \frac{1}{\sqrt{x^3+1}} \, dx. \qquad \blacksquare \]