# Thread: Discussions on the convergence of integrals and series

1. Originally Posted by Prove It
Why do you consider the Alternating Series test not elegant? I personally find the most simple solution to be the most elegant, because it is the most likely to be understood by others...
I believe I said unsatisfying, not inelegant. At any rate, in my 1:30 AM internet-browsing state, I was annoyed at not being able to immediately see what the series should converge to. As evidenced by the solutions that followed, it seems that there was a concise, complete, and more satisfying answer all along.

Also, I'm not sure that this series technically falls under the purview of the alternating series test, but as $\chi\sigma$ pointed out, the Dirichlet test works here.

It should be pointed out though that simply recognizing that the series conforms to a Fourier series takes for granted that at some point, somebody had to show that Fourier series fulfill a whole bunch of nice properties, including convergence providing that the emulated function is bounded. That process itself resulted in the restructuring of some areas of mathematics, analysis in particular.

2. No one has attempted the integrals I posted a few days ago.

Here's my attempt.

$\displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx + \sum_{n=1}^{\infty} \int_{(n-\frac{1}{2}) \pi}^{(n+\frac{1}{2}) \pi } \frac{\sin (\tan x)}{x} \ dx$

Since $\displaystyle \frac{\sin (\tan x)}{x}$ has a removable singularity at $x=0$ and is bounded near $x= \frac{\pi}{2}$, $\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin (\tan x)}{x} \ dx$ converges.

And since $\displaystyle \frac{\sin (\tan x)}{x}$ is bounded near $(n-\frac{1}{2}) \pi$ and $(n+\frac{1}{2}) \pi$, $\displaystyle \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2}) \pi} \frac{\sin (\tan x)}{x} \ dx$ converges.

So we need to show that $\displaystyle \sum_{n=1}^{\infty} \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi} \frac{\sin (\tan x)}{x} \ dx$ converges.

$\displaystyle \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi} \frac{\sin (\tan x)}{x} \ dx = \int_{(n-\frac{1}{2})\pi }^{n \pi} \frac{\sin (\tan x)}{x} \ dx + \int^{(n+\frac{1}{2})\pi}_{n \pi } \frac{\sin (\tan x)}{x} \ dx = \int_{\frac{\pi}{2}}^{0} \frac{\sin (\tan u)}{n \pi -u} \ du + \int^{\frac{\pi}{2}}_{0} \frac{\sin (\tan v)}{n \pi + v} \ dv$

$\displaystyle = \int_{0}^{\frac{\pi}{2}} \Big( \frac{1}{n \pi + u} - \frac{1}{n \pi -u} \Big) \sin (\tan u) \ du = 2 \int_{0}^{\frac{\pi}{2}} \frac{u \sin (\tan u)}{u^{2} - n^{2} \pi^{2}} \ du$

And $\displaystyle \sum_{n=1}^{\infty} \Big| \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi} \frac{\sin (\tan x)}{x} \ dx \Big| = \sum_{n=1}^{\infty} \Big| 2 \int_{0}^{\frac{\pi}{2}} \frac{x \sin (\tan x)}{x^{2} - n^{2} \pi^{2}} \ dx \Big| \le 2 \sum_{n=1}^{\infty} \int_{0}^{\frac{\pi}{2}} \Big| \frac{x \sin (\tan x)}{x^{2} - n^{2} \pi^{2}}\Big| \ dx$

$\displaystyle \le 2 \sum_{n=1}^{\infty} \frac{\pi}{2} \max \Big| \frac{x \sin(\tan x)}{x^{2}-n^{2} \pi^{2}} \Big|\le \pi \sum_{n=1}^{\infty} \frac{\frac{\pi}{2}}{n^{2} \pi^{2} -\frac{\pi^{2}}{4}} < \infty$

Therefore $\displaystyle \sum_{n=1}^{\infty} \int_{(n-\frac{1}{2}) \pi}^{(n+\frac{1}{2})\pi} \frac{\sin (\tan x)}{x} \ dx$ converges by the absolute convergence test.

3. Originally Posted by Opalg
To take that a bit further, the expression in (2) can be simplified as $$\tan^{-1} \Bigl(\frac{\sin x}{1-\cos x}\Bigr) = \tan^{-1} \biggl(\frac{2\sin\frac x2\cos\frac x2}{2\sin^2\frac x2}\biggr) = \tan^{-1}\bigl(\cot\tfrac x2\bigr) = \tfrac\pi2 - \tfrac x2.$$ Hence $$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = \tfrac12(\pi-x).$$ But that only works provided that $0<x< 2\pi$. At the endpoints of the interval, when $x=0$ or $2\pi$, the sum $\sum _{n=1}^{\infty} \frac{\sin n x}{n}$ is obviously $0$ (since each term vanishes).

As ZaidAlyafey points out, this sum is a Fourier series, namely for the function $\tfrac12(\pi-x)$ over the interval $[0,2\pi].$

HINT: consider the Bernoulli Polynomial or order 1:

$\displaystyle B_1(x) = x-\frac{1}{2}$

The connection is there.

4. Given $\displaystyle n\ge 1$ and $\displaystyle 0\le x\le1$, or, alternatively, $\displaystyle n=1$ and $\displaystyle 0<x<1$, then

$\displaystyle B_n(x) = -\frac{2\, (n!)}{(2\pi)^n}\, \sum_{k=1}^{\infty} \frac{1 }{k^n} \cos \left(2\pi kx-\frac{\pi n}{2}\right)$

See eqn. 9.622 on (approx) page 1628 of the Maths Bible that is Gradshteyn & Ryzhik; the squirrel's guide to life, the universe, and everything:

Nom nom nom!

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