Welcome to our community

Be a part of something great, join today!

Discussions on the convergence of integrals and series

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

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

$ \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 $[\epsilon, \infty)$.
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

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

$ \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 .
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
2-\(\displaystyle \int^{\infty}_0 \frac{\sin x}{x}\)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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-\(\displaystyle \int^{\infty}_0 \frac{\cos x}{x}\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$\int_{\frac{\pi}{2}}^{\infty} \frac{\sin x}{x} \ dx$ 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 $a > \frac{\pi}{2}$
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$\frac{\cos x}{x} = \frac{1}{x} \big( 1-\frac{x^{2}}{2!} + O(x^{4}) \Big)= \frac{1}{x} - \frac{x}{2!}+ O(x^{3})$

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
4-[tex]\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx[/tex]
 
Last edited:

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
4-[tex]\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx[/tex]
\(\displaystyle \frac{1}{x}=\frac{1}{1+x-x} = \frac{1}{1-(1-x)}\)

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
5- \(\displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}\)
 

TheBigBadBen

Active member
May 12, 2013
84
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

Well-known member
Feb 13, 2012
1,704
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

$\chi$ $\sigma$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
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 $[0,2\pi].$
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

MHB Oldtimer
Staff member
Feb 7, 2012
2,681
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.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
6- \(\displaystyle \int^{\infty}_0 \frac{x}{\sqrt{e^x-1}}\, dx\)
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$\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.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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 $
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

Active member
May 12, 2013
84
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.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
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.
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
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. :rolleyes:
 

DreamWeaver

Well-known member
Sep 16, 2013
337
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).pdf


Nom nom nom! (Heidy)(Heidy)(Heidy)
 
Last edited: