$\begin{align*} \int \frac{1}{x (\log x)^s} \, dx &= \int \frac{1}{u^s} \, du \\ &= \frac{u^{1-s}}{1-s} \\ &= \frac{(\log x)^{1-s}}{1-s}. \end{align*}$

But then the limit

$\lim_{x \to +\infty} \frac{(\log x)^{1-s}}{1-s}$

is finite for $s > 1$ and diverges for $s \leq 1$ (since $\log x \to \infty$ as $x \to \infty$ and so the limit diverges when $1-s > 0$ and converges for $1-s < 0$ ). Hence, the integral

$\int_2^{\infty} \frac{dx}{x (\log x)^s}$

converges for $s > 1$ and diverges for $s < 1$ . In the case that $s = 1$ the indefinite integral is

$\int \frac{1}{x \log x} \, dx = \log (\log x)$

and $\log (\log x) \to +\infty$ as $x \to +\infty$ , so the integral divers for $s = 1$ as well. Therefore, we have the integral converges for $s > 1$ and diverges for $s \leq 1. \qquad \blacksquare$

2 log x) for convergence

by RoRi

March 31, 2016

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

by RoRi

March 31, 2016

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 ∫ e^{-x^1/2} / x^1/2 for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_{0^+}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx.$

The integral converges.

Proof. We can compute this integral directly. First, we evaluate the indefinite integral using the substitution $u = \sqrt{x}$ , $du = \frac{1}{2 \sqrt{x}} \, dx$ .

$\begin{align*} \int \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx &= 2 \int e^{-u} \, du \\[9pt] &= -2 e^{-u} \\[9pt] &= -2 e^{-\sqrt{x}} \, dx. \end{align*}$

Now, we have discontinuities at both limits of integration so we evaluate by taking two limits,

$\begin{align*} \int_{0^+}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \int_a^b \frac{e^{-\sqrt{x}}}{\sqrt{x}} \, dx \\[9pt] &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \left( -2 e^{-\sqrt{x}} \Bigr \rvert_a^b \right) \\[9pt] &= \lim_{\substack{b \to +\infty \\ a \to 0^+}} \left( -2 e^{-\sqrt{b}} + 2 e^{-\sqrt{a}} \right) \\[9pt] &= 2. \qquad \blacksquare \end{align*}$

Test the improper integral ∫ e^-x/2 for convergence

by RoRi

March 31, 2016

Test the following improper integral for convergence:

$\int_0^{\infty} \frac{1}{\sqrt{e^x}} \, dx.$

The integral converges.

Proof. We compute directly,

$\begin{align*} \int_0^{\infty} \frac{1}{\sqrt{e^x}} \, dx &= \int_0^{\infty} e^{-\frac{x}{2}} \, dx \\[9pt] &= \lim_{b \to +\infty} \left( \int_0^b e^{-\frac{x}{2}} \, dx \right) \\[9pt] &= \lim_{b \to +\infty} \left( -2 e^{-\frac{x}{2}} \Bigr \rvert_0^b \right) \\[9pt] &= \lim_{b \to +\infty} \left( -2 e^{-\frac{b}{2}} + 2 \right) \\[9pt] &= 2. \qquad \blacksquare \end{align*}$

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

by RoRi

March 31, 2016

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

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

Show that the limit of ∑ 1/k = log (p/q) where the sum is from k = qn to pn

by RoRi

March 31, 2016

For given integers $p$ and $q$ with $1 \leq q \leq p$ , prove
$\lim_{n \to \infty} \sum_{k=qn}^{pn} \frac{1}{k} = \log \frac{p}{q}.$
Consider the series
$1 + \frac{1}{3} + \frac{1}{5} - \frac{1}{2} - \frac{1}{4} + \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \frac{1}{6} - \frac{1}{8} + \ + \ + \ - \ - \ \cdots.$

This is a rearrangement of the alternating harmonic series ( $\sum \frac{(-1)^n}{n}$ ) in which there are three positive terms followed by two negative terms. Prove that the series converges and that the sum is equal to $\log 2 + \frac{1}{2} \log \frac{3}{2}$ .

Incomplete.

Find the limit of an expression involving s_n(a) = 1^a + … + n^a

by RoRi

March 31, 2016

Let $a \in \mathbb{R}$ be any real number and define

$s_n (a) = 1^a + 2^a + \cdots + n^a$

for integers $n$ . Find the limit

$\lim_{n \to \infty} \frac{s_n (a+1)}{ns_n (a)}.$

Incomplete.

Prove that the recursive sequence 1 / x_n+2 = 1 / x_n+1 + 1 / x_n converges

by RoRi

March 29, 2016

Prove that the sequence $\{ x_n \}$ be defined by the recursive relationship,

$x_0 = 1, \qquad x_1 = 1, \qquad \frac{1}{x_{n+2}} = \frac{1}{x_{n+1}} + \frac{1}{x_n}$

converges and find the limit of the sequence.

Proof. First, we show that the sequence $\{ x_n \}$ is monotonically decreasing for all $n \geq 1$ . For the base case we have $x_1 = 1$ and

$\frac{1}{x_2} = \frac{1}{1} + \frac{1}{1} = 2 \quad \implies \quad x_2 = \frac{1}{2}.$

Hence, $x_2 < x_1$ . Assume then that for all positive integers up to some $k$ we have $x_{k+1} < x_k$ . Then,

$\begin{align*} && \frac{1}{x_{k+2}} &= \frac{1}{x_{k+1}} + \frac{1}{x_k} \\[9pt] \implies && \frac{1}{x_{k+2}} &> \frac{1}{x_{k+1}} + \frac{1}{x_{k+1}} &(x_{k+1} < x_k \implies \frac{1}{x_{k+1}} > \frac{1}{x_k}) \\[9pt] \implies && \frac{1}{x_{k+2}} &> \frac{2}{x_{k+1}} \\[9pt] \implies && x_{k+2} &< \frac{x_{k+1}}{2} \\[9pt] \implies && x_{k+2} &< x_{k+1}. \end{align*}$

Thus, the sequences is monotonically decreasing. The sequence is certainly bounded below since all of the terms are greater than 0. Therefore, the sequence converges $. \qquad \blacksquare$

To compute the limit of the sequence, assume the sequence converges to a finite limit $L$ (justified since we just proved that it does indeed converge). Therefore,

$\begin{align*} && \lim_{n \to \infty} x_n &= L \\[9pt] \implies && \lim_{n \to \infty} \frac{1}{x_n} &= \frac{1}{L} \\[9pt] \implies && \lim_{n \to \infty} \left( \frac{1}{x_{n-1}} + \frac{1}{x_{n-2}} \right) &= \frac{1}{L} \\[9pt] \implies && \frac{1}{L} + \frac{1}{L} &= \frac{1}{L} \\[9pt] \implies && 2L &= L \\[9pt] \implies && L &= 0. \end{align*}$