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- Jan 17, 2013

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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

- Thread starter ZaidAlyafey
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- Jan 17, 2013

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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

- 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 $[\epsilon, \infty)$.

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

$ \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 $[\epsilon, \infty)$.

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- Thread starter
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- Jan 17, 2013

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Well, that is better . Very good .$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

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

$ \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 $[\epsilon, \infty)$.

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- Jan 17, 2013

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

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- #5

- Jan 17, 2013

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The integral has a removable singularity at the origin , so it converges there , now let us examine at infinity2-\(\displaystyle \int^{\infty}_0 \frac{\sin x}{x}\)

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

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

- Jan 31, 2012

- 253

- Jan 31, 2012

- 253

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

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- #8

- Jan 17, 2013

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4-[tex]\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx[/tex]

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- Thread starter
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- Jan 17, 2013

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\(\displaystyle \frac{1}{x}=\frac{1}{1+x-x} = \frac{1}{1-(1-x)}\)4-[tex]\int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx[/tex]

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

- Thread starter
- #10

- Jan 17, 2013

- 1,667

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

- May 12, 2013

- 84

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

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.

- Feb 13, 2012

- 1,704

According to the Diriclet test the series converges so that we have to compute its sum. Using the well known expansion...5- \(\displaystyle \sum^{\infty}_{n=1}\frac{\sin(nx)}{n}\)

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

... 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

$\chi$ $\sigma$

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- Jan 17, 2013

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

$\chi$ $\sigma$

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- Feb 7, 2012

- 2,697

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).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...

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

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

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- Jan 17, 2013

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Since the series converges for all $x$ , there may be a general solution that works for all $x$ , right ?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 $[0,2\pi].$

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- Feb 7, 2012

- 2,697

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

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- #17

- Jan 17, 2013

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

- Jan 31, 2012

- 253

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.

- 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 $

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

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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 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.

- May 12, 2013

- 84

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.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...

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.

- 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.

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.

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- 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.

- 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!

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

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!

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