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Prove an integral formula for ∑ (sin (nx)) / n2

Prove that the series

    \[ \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2} \]

converges for all x \in \mathbb{R} and let f(x) denote the value of this sum for each x. Prove that f(x) is continuous for x \in [0, \pi] and prove that

    \[ \int_0^{\pi} f(x) \, dx = 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3}. \]


Proof. First, the series converges for all real x by the comparison test since

    \[ |\sin (nx)| \leq 1 \quad \implies \quad \left|\frac{\sin (nx)}{n^2}\right| \leq \frac{1}{n^2} \]

for all n. Therefore, the convergence of \sum \frac{1}{n^2} implies the convergence of \sum \frac{\sin (nx)}{n^2}. Furthermore, this convergence is uniform by the Weierstrass M-test with M_n given by \frac{1}{n^2}, and again \sum \frac{1}{n^2} converges. Thus, by Theorem 11.2 (page 425 of Apostol),

    \[ f(x) = \sum_{n=1}^{\infty} \frac{\sin (nx)}{n^2} \]

is continuous on the interval [0, \pi]. Therefore, we may apply Theorem 11.4 (page 426 of Apostol):

    \begin{align*}  && \sum_{k=1}^{\infty} \int_0^{\pi} \frac{\sin (kt)}{k^2} \, dt &= \int_0^{\pi} \sum_{k=1}^{\infty} \frac{ \sin (kt)}{t^2} \, dt \\[9pt]  \implies && \sum_{k=1}^{\infty} \left( \frac{-\cos (kt)}{k^3} \Bigr \rvert_0^{\pi} \right) &= \int_0^{\pi} f(x) \, dx \\[9pt]  \implies && \int_0^{\pi} f(x) \, dx &= 2 \sum_{n=1}^{\infty} \frac{1}{(2n-1)^3} \end{align*}

since \cos 0 - \cos (k \pi) = 0 if k = 2n and equals 2 if k = 2n-1. \qquad \blacksquare

Prove or disprove: ∫ f(x) converges implies lim f(x) = 0

The following function f is defined for all x \geq 1, and n is a positive integer. Prove or provide a counterexample to the following statement.

The convergence of the improper integral

    \[ \int_1^{\infty} f(x) \, dx \]

implies

    \[ \lim_{x \to \infty} f(x) = 0. \]


Counterexample. The idea of the construction is a function which has rapidly diminishing area, but has a height that is not going to 0. (So, for an idea consider triangles on the real line all with height 1, but for which the base is becoming small rapidly.) To make this concrete, define

    \[ f(x) = 1 - 2n^2 \left| x - \left( n + \frac{1}{2n^2} \right) \qquad \text{for } x \in [n,n+1), \]

for each positive integer n. Then for the improper integral we have

    \[ \int_1^{\infty} f(x) \, dx \leq \sum_{n=1}^{\infty} \frac{1}{2n^2} \]

which we know converges. On the other hand

    \[ f \left( n + \frac{1}{2n^2} \right) = 1 \]

for all positive integers n. Hence,

    \[ \lim_{x \to \infty} f(x) \neq 0 \]

(since it does not exist). Hence, the statement is false.

(Note: For more on this see this question on Math.SE.)

Prove that the improper integral ∫ f(x) dx and ∑ f(n) both converge or both diverge

  1. Assume that f is a monotonically decreasing function for all x \geq 1 and that

        \[ \lim_{x \to +\infty} f(x) = 0. \]

    Prove that the improper integral and the series

        \[ \int_1^{\infty} f(x)  \, dx \qquad \text{and} \qquad \sum_{n=1}^{\infty} f(n) \]

    both converge or both diverge.

  2. Give a counterexample to the theorem in part (a) in the case that f is not monotonic, i.e., find a non-monotonic function f such that \sum f(n) converges but \int_1^{\infty} f(x) \, dx diverges.

    Incomplete.

Find a value of C so that the given improper integral converges

Find a value for the constant C \in \mathbb{R} so that the improper integral

    \[ \int_0^{\infty} \left( \frac{1}{\sqrt{1+x^2}} - \frac{C}{x+1} \right) \, dx \]

converges and find the value of the integral in this case.


First, we have

    \begin{align*}  \int_0^{\infty} \left( \frac{1}{\sqrt{1+2x^2}} - \frac{C}{x+1} \right) \, dx &= \int_0^{\infty} \frac{x+1-C\sqrt{1+2x^2}}{(x+1)\sqrt{1+2x^2}} \, dx \\[9pt]  &= \int_0^{\infty} \frac{x+1-xC\sqrt{2 + \frac{1}{x^2}}}{(x^2+x)\sqrt{2 + \frac{1}{x^2}}} \, dx \\[9pt]  &= \int_0^{\infty} \frac{x \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) + 1}{x^2 \sqrt{2 + \frac{1}{x^2}} + x \sqrt{2 + \frac{1}{x^2}}} \, dx. \end{align*}

For this to converge we must have the coefficient of the x in the numerator going to 0 as x \to \infty (otherwise the integral will converge by limit comparison with \int \frac{1}{x}). Hence,

    \[ \lim_{x \to \infty} \left(1 - C \sqrt{2 + \frac{1}{x^2}} \right) = 1 - \sqrt{2} C \quad \implies \quad C = \frac{\sqrt{2}}{2}. \]

Now, we need to evaluate the integral. Incomplete.

Find a value of the constant C so that the given improper integral converges

Find the value of C \in \mathbb{R} so that the improper integral

    \[ \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx \]

converges, and compute the value of the integral in this case.


First, we compute the value of the constant C,

    \begin{align*}  \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \frac{x^2+x-2C(x^2+C)}{2(x^2+C)(x+1)} \, dx \\[9pt]  &= \int_1^{\infty} \frac{(1-2C)x^2 + x - 2C^2}{2(x^2+C)(x+1)} \, dx. \end{align*}

This integral will diverge by comparison with \int_1^{\infty} \frac{1}{x} \, dx if the coefficient of the x^2 term in the numerator is not equal to zero. Hence, we must have

    \[ 1-2C = 0 \quad \implies \quad C = \frac{1}{2}. \]

Next, we evaluate the integral with C = \frac{1}{2},

    \begin{align*}  \int_1^{\infty} \left( \frac{x}{2x^2+2C} - \frac{C}{x+1} \right) \, dx &= \int_1^{\infty} \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt]  &= \lim_{a \to +\infty} \int_1^a \left( \frac{x}{2x^2+1} - \frac{1}{2(x+1)} \right) \, dx \\[9pt]  &= \lim_{a \to +\infty} \left( \frac{1}{4} \int_1^a \frac{4x}{2x^2+1} \, dx - \frac{1}{2} \int_1^a \frac{1}{x+1} \, dx \right) \\[9pt]  &= \lim_{a \to +\infty} \left( \frac{1}{4} \log(2x^2+1) - \frac{1}{2} \log |x+1| \right) \Bigr \rvert_1^a \\[9pt]  &= \frac{1}{4} \cdot \lim_{a \to +\infty} \left( \log(2a^2+1) - \log 3 - 2 \log (a+1) + 2 \log 2 \right) \\[9pt]  &= \frac{1}{4} \cdot \lim_{a\to +\infty} \left( \log \frac{2a^2+1}{(a+1)^2} + \log \frac{4}{3} \right) \\[9pt]  &= \frac{1}{4} \cdot \log \frac{8}{3}. \end{align*}

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

Test the following improper integral for convergence:

    \[ \int_2^{\infty} \frac{dx}{x (\log x)^s}. \]


The integral converges if s < 1 and diverges for s \geq 1.

Proof. We can compute this directly. We evaluate the indefinite integral (using the substitution u = \log x) for s \neq 1:

    \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