Sunday, February 10, 2019

integration - Show that $intlimits_2^{+infty}frac{sin{x}}{xln{x}}, rm dx$ is conditionally convergent



The integral $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\,dx$ is conditionally convergent.



I know that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x}\, dx$
is conditionally convergent and $ {\forall}p > 1$,
$\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x^p}\, dx$ is absolute convergent,
but $\ln{x}$ is between $x$ and $x^p$, so how to prove that $\displaystyle\int_{2}^{+\infty}\frac{\sin{x}}{x\ln{x}}\, dx$ is conditionally convergent?



Answer



This is roughly integral analogue of the alternating series test. Since proving its generalization cause little harm, let me actually show




Proposition 1. Suppose that $f : [a, \infty) \to \mathbb{R}$ satisfies the following two conditions:




  1. $f$ is monotone-decreasing, i.e., $f(x) \geq f(y)$ for all $a \leq x \leq y$.

  2. $\lim_{x\to\infty} f(x) = 0$.




Then



$$ \int_{a}^{\infty} f(x)\sin(x) \, \mathrm{d}x = \lim_{b\to\infty} \int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x $$



converges. Moreover, this integral is absolutely convergent if and only if $\int_{a}^{\infty} f(x) \, \mathrm{d}x < \infty$.




The proof is quite simple. We first prove that the integral converges. Let $n$ be an integer so that $\pi n \geq a$. Then for $ b \geq \pi n$,




\begin{align*}
\int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x
&= \int_{a}^{\pi n} f(x)\sin(x) \, \mathrm{d}x + \sum_{k=n}^{\lfloor b/\pi\rfloor - 1} \int_{\pi k}^{\pi(k+1)} f(x)\sin(x) \, \mathrm{d}x \\
&\quad + \int_{\pi\lfloor b/\pi\rfloor}^{b} f(x)\sin(x) \, \mathrm{d}x.
\end{align*}



Writing $N = \lfloor b/\pi \rfloor$ and defining $a_k$ by $a_k = \int_{0}^{\pi} f(x+\pi k)\sin(x) \, \mathrm{d}x$, we find that




  1. $a_k \geq 0$, since $f(x+\pi k) \geq 0$ for all $x \in [0, \pi]$.



  2. $a_{k+1} \geq a_k$ since $f(x+\pi k) \geq f(x+\pi(k+1))$ for all $x \in [0, \pi]$.


  3. $a_k \to 0$ as $k\to\infty$, since $a_k \leq \int_{0}^{\pi} f(\pi k) \sin (x) \, \mathrm{d}x = 2f(\pi k) \to 0$ as $k \to \infty$.


  4. Bu a similar computation as in step 3, we check that $\left| \int_{\pi N}^{b} f(x) \sin (x) \, \mathrm{d}x \right| \leq 2f(\pi N)$, and so, $\int_{\pi N}^{b} f(x) \sin (x) \, \mathrm{d}x \to 0$ as $b\to\infty$.


  5. We have



    $$ \sum_{k=n}^{N - 1} \int_{\pi k}^{\pi(k+1)} f(x)\sin(x) \, \mathrm{d}x = \sum_{k=n}^{N-1} (-1)^k a_k. $$



    So, by the alternating series test, this converges as $N\to\infty$, hence as $b \to \infty$.





Combining altogether, it follows that $\int_{a}^{b} f(x)\sin(x) \, \mathrm{d}x $ converges as $b\to\infty$.



To show the second assertion, let $n$ still be an integer with $\pi n \geq a$. Then for $k \geq n$, integrating each side of the inequality $f(\pi(k+1))|\sin x| \leq f(x)|\sin x| \leq f(\pi k)|\sin x|$ for $x \in [\pi k, \pi(k+1)]$ gives



$$ 2f(\pi(k+1))
\leq \int_{\pi k}^{\pi(k+1)} f(x)|\sin(x)| \, \mathrm{d}x
\leq 2f(\pi k) $$



and similar argument shows




$$ \pi f(\pi(k+1))
\leq \int_{\pi k}^{\pi(k+1)} f(x) \, \mathrm{d}x
\leq \pi f(\pi k). $$



From this, we easily check that



$$ \frac{2}{\pi} \int_{\pi(n+1)}^{\infty} f(x) \, \mathrm{d}x
\leq \int_{\pi n}^{\infty} f(x)|\sin x| \, \mathrm{d}x
\leq 2f(\pi n) + \frac{2}{\pi} \int_{\pi(n+1)}^{\infty} f(x) \, \mathrm{d}x. $$




Therefore the second assertion follows.


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