Discussions on the convergence of integrals and series

[alyafey22]

This thread will be dedicated to discuss the convergence of various definite integrals and infinite series , if you have any question to post , please don't hesitate , I hope someone make the thread sticky.

1- \(\displaystyle \int^{\infty}_0 \left(\frac{e^{-x}}{x} \,-\,\frac{1}{x(x+1)^2}\right)\,dx\,=1-\gamma\)

Let us have some ideas

 

[polygamma]

$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

$\lim_{x \to 0} \Big( \frac{e^{-x}}{x} - \frac{1}{x(x+1)^{2}} \Big) = \lim_{x \to 0} \Big( \frac{1}{x} - 1 + O(x) - \frac{1}{x} + \frac{1}{x+1} + \frac{1}{(x+1)^{2}} \Big)$

$ \lim_{x \to 0} \Big(- 1 + O(x) + \frac{1}{x+1} + \frac{1}{(x+1)^{2}} \Big) =1$

So the singularity at $x=0$ is removable.

EDIT: And there is no issue at $\infty$ since the integral can be separated into two integrals that both converge on $[\epsilon, \infty)$.

 

[alyafey22]

$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

$\lim_{x \to 0} \Big( \frac{e^{-x}}{x} - \frac{1}{x(x+1)^{2}} \Big) = \lim_{x \to 0} \Big( \frac{1}{x} - 1 + O(x) - \frac{1}{x} + \frac{1}{x+1} + \frac{1}{(x+1)^{2}} \Big)$

$ \lim_{x \to 0} \Big(- 1 + O(x) + \frac{1}{x+1} + \frac{1}{(x+1)^{2}} \Big) =1$

So the singularity at $x=0$ is removable.

EDIT: And there is no issue at $\infty$ since the integral can be separated into two integrals that both converge on $[\epsilon, \infty)$.

Well, that is better . Very good .

 

[alyafey22]

2-\(\displaystyle \int^{\infty}_0 \frac{\sin x}{x}\)

 

[alyafey22]

2-\(\displaystyle \int^{\infty}_0 \frac{\sin x}{x}\)

The integral has a removable singularity at the origin , so it converges there , now let us examine at infinity

\(\displaystyle \int^{\infty}_{\frac{\pi}{2}} \frac{\sin x}{x}\)

Integrating by parts we get

\(\displaystyle \int^{\infty}_{\frac{\pi}{2}} \frac{\sin x}{x}= \frac{-\cos x}{x} \biggr]^ {\infty }_{\frac{\pi}{2}} -\int^{\infty}_{\frac{\pi}{2}} \frac{\cos x}{x^2}\)

The first term vanishes , for the second one

Because the integral is absolutley convergent it converges

\(\displaystyle \int^{\infty}_{\frac{\pi}{2}} \frac{1}{x^2}< \infty\)

Now Let us look at another form

3-\(\displaystyle \int^{\infty}_0 \frac{\cos x}{x}\)

 

[polygamma]

$\int_{\frac{\pi}{2}}^{\infty} \frac{\sin x}{x} \ dx$ is also convergent by Dirichlet's convergence test since $\frac{1}{x}$ is bounded, monotonic, and tends to zero, while $\int_{\frac{\pi}{2}}^{a} \sin x \ dx $ is bounded for any $a > \frac{\pi}{2}$

 

[polygamma]

$\frac{\cos x}{x} = \frac{1}{x} \big( 1-\frac{x^{2}}{2!} + O(x^{4}) \Big)= \frac{1}{x} - \frac{x}{2!}+ O(x^{3})$

Since $\frac{1}{x}$ is not integrable at zero, $\frac{\cos x}{x}$ is not integrable at zero.

 

[alyafey22]

4-$$\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx$$

 

[alyafey22]

4-$$\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx$$

\(\displaystyle \frac{1}{x}=\frac{1}{1+x-x} = \frac{1}{1-(1-x)}\)

\(\displaystyle \frac{1}{x}=\sum^{\infty}_{n=0}(1-x)^n\) converges \(\displaystyle \forall \, x \, : \, \,\, |1-x|<1\)

\(\displaystyle \ln(x) =-\sum^{\infty}_{n=0} \frac{(1-x)^{n+1}}{n+1}\)

At $1$ we have a removable singularity .

\(\displaystyle \lim_{x \to 1}\frac{ \left( (1-x) + \frac{(1-x)^2}{2}+ \cdots \right)^2 }{ (x+2) (x-1) } < \infty\)

To examine the integral near zero , let us make the substitution

\(\displaystyle \ln(x) =-t \)

\(\displaystyle -\int_{\epsilon}^{\infty} \frac{t^2 \, e^{t}}{2e^{2t}-e^{t}-1}< - \frac{1}{2}\, \int^{\infty}_{\epsilon} t^2 e^{-t}< \infty\)

The integral converges ...

 

[alyafey22]

5- \(\displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}\)

 

[TheBigBadBen]

5- \(\displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}\)

I have seen a series like this before. The simple (but unsatisfying) explanation is that it must converge by the alternating series test, which may be extended to such unconventionally oscillating terms.

However, I'm sure there's a more elegant underlying structure if you use some decomposition of Euler's formula. Having suggested it, I will look into it if I have the inclination later.

 

[chisigma]

5- \(\displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}\)

According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

$$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

... we arrive to write...

$$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$

Kind regards

$\chi$ $\sigma$

 

[alyafey22]

According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

$$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

... we arrive to write...

$$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$

Kind regards

$\chi$ $\sigma$

Yes, I think also it is solvable by Fourier series .

 

[Opalg]

According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...

$$ \sum_{n=1}^{\infty} \frac{z^{n}}{n} = - \ln (1-z)\ (1)$$

... we arrive to write...

$$\sum _{n=1}^{\infty} \frac{\sin n x}{n} = - \mathcal {Im} \{\ln (1-e^{i x})\} = \tan^{-1} \frac{\sin x}{1-\cos x}\ (2)$$

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 $[0,2\pi].$

 

[alyafey22]

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 $[0,2\pi].$

Since the series converges for all $x$ , there may be a general solution that works for all $x$ , right ?

 

[Opalg]

Since the series converges for all $x$ , there may be a general solution that works for all $x$ , right ?

Yes, it is the $2\pi$-periodic function defined on the interval $[0,2\pi)$ by $f(x) = \begin{cases}0&(x=0),\\ \frac12(\pi-x)&(0<x<2\pi). \end{cases}$ It is an example of what is often called a sawtooth function.

 

[alyafey22]

6- \(\displaystyle \int^{\infty}_0 \frac{x}{\sqrt{e^x-1}}\, dx\)

 

[polygamma]

$\displaystyle \int_{0}^{\infty} \frac{x}{\sqrt{e^{x}-1}} \ dx = \int_{0}^{\infty} \frac{x e^{- \frac{x}{2}}} {\sqrt{1-e^{-x}}} \ dx $

Let $ \displaystyle u = e^{-\frac{x}{2}}$

$ \displaystyle = - 4 \int_{0}^{1} \frac{\ln{u}}{\sqrt{1-u^{2}}} \ du $

Let $v = \arcsin u$

$ \displaystyle = -4 \int^{\frac{\pi}{2}}_{0} \ln (\sin v) \ dv $I'm sure we've all seen evaluations of that last integral. So I'm just going to argue that it converges.The only potential issue is at $x=0$.

But $\displaystyle \ln (\sin x) = \ln \Big( \frac{x \sin x}{x} \Big) = \ln(x) + \ln \Big(\frac{\sin x}{x} \Big) $.

So near $x=0$, $ \ln(\sin x)$ behaves like $\ln x$, and thus $\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin x) \ dx $ converges.

 

[polygamma]

7- $ \displaystyle \int_{0}^{\infty} \frac{\ln (\tan^{2} x)}{1+x^{2}} \ dx$

8- $ \displaystyle \int_{0}^{\infty} \frac{\sin (\tan x)}{x} \ dx $

 

[Prove It]

I have seen a series like this before. The simple (but unsatisfying) explanation is that it must converge by the alternating series test, which may be extended to such unconventionally oscillating terms.

However, I'm sure there's a more elegant underlying structure if you use some decomposition of Euler's formula. Having suggested it, I will look into it if I have the inclination later.

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...

 

[TheBigBadBen]

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.

 

[polygamma]

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.

 

[DreamWeaver]

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.

 

[DreamWeaver]

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:http://f3.tiera.ru/ShiZ/math/MRef_R...ies, and products (5ed., AP, 1996)(1762s).pdfNom nom nom! (Heidy)(Heidy)(Heidy)