# Thread: Discussions on the convergence of integrals and series

1. 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

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

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

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

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 $[\epsilon, \infty)$.

Originally Posted by Random Variable
$\frac{e^{-x}}{x} = \frac{1}{x} \Big( 1 - x + O(x^{2}) \Big) = \frac{1}{x} - 1 + O(x)$

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

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

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 .

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

Originally Posted by ZaidAlyafey
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-$\displaystyle \int^{\infty}_0 \frac{\cos x}{x}$

6. $\int_{\frac{\pi}{2}}^{\infty} \frac{\sin x}{x} \ dx$ 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 $a > \frac{\pi}{2}$

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

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

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

Originally Posted by ZaidAlyafey
4-$\displaystyle \int_{0}^{1}\frac{\ln^{2}(x)}{x^{2}+x-2}dx$
$\displaystyle \frac{1}{x}=\frac{1}{1+x-x} = \frac{1}{1-(1-x)}$

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

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