Mathematics Asked by Kirito on January 5, 2022

I need to study the convergence of the following improper integral:

$$int_{0}^{infty} dfrac{sin(x)}{x+1}, mathrm dx$$

I did the following:

$$ -1 leq sin(x) leq 1 \

implies dfrac{-1}{x+1} leq dfrac{sin(x)}{x+1} leq dfrac{1}{x+1} \

implies left|dfrac{sin(x)}{x+1}right| leq dfrac{1}{x+1} \

implies int_{0}^{infty} left|dfrac{sin(x)}{x+1}right| , mathrm dx leq int_{0}^{infty}dfrac{1}{x+1}, mathrm dx = infty $$

I planned to use the comparison criterion and then the absolute convergence criterion. However, the idea did not work for me.

Let $$ a_n = int_{pi n}^{pi(n+1)}frac{|sin x|}{x+1}dx. $$ Note that $$ int_0^infty frac{sin x}{x+1}dx = sum_{n=0}^infty (-1)^n a_n. $$

So, if the series converges, the integral must also converge. For any positive integer $n$, we can see that $a_n$ is positive, and we can rewrite it the following way: $$ begin{align} a_n = int_{pi n}^{pi(n+1)}frac{|sin x|}{x+1}dx &= int_{pi n}^{pi(n+1)}frac{sin (x - pi n)}{x+1}dx\ &= int_0^pifrac{sin x}{x+1+pi n}dx. end{align} $$ This makes it clear that the denominator of $a_{n+1}$ is larger than the denominator of $a_n$ over the entire interval integrated, so $a_n$ must be decreasing. Furthermore, it is easy to see that $lim_{ntoinfty} a_n=0$. Therefore, by the alternating series test, the integral converges.

Answered by Polygon on January 5, 2022

The Cauchy criterion for improper integrals is:

An improper integral $int_0^infty f(x) , dx$ is convergent if and only if for any $epsilon > 0$ there exists $C_epsilon > 0$ such that $left|int_a^b f(x) , dx right| < epsilon$ for all $b > a> C_epsilon.$

Since $x mapsto frac{1}{1+x}$is decreasing, by the second mean value theorem for integrals, there exists $xi in (a,b)$ such that

$$left|int_a^b frac{sin x}{1+x} , dxright| = left|frac{1}{1+a}int_a^xi sin x, dxright| = frac{|cos a - cos xi|}{1+a}leqslant frac{2}{1+a}$$

For all $b > a > C_epsilon = frac{2}{epsilon}-1$ we have the RHS less than $epsilon$ and the Cauchy criterion is satisfied.

Answered by RRL on January 5, 2022

Granted, the integral does not converge in the sense of Lebesgue. As a proper Riemann integral it does.

Here is another solution based which uses elementary facts about alternating series.

The sequence $a_n=Big|int^{(n+1)pi}_{npi}frac{sin x}{x+1},dxBig|$ is non decreasing and $a_nxrightarrow{nrightarrowinfty}0$. This is because on $[pi n,pi(n+1)]$, $sin x=(-1)^n|sin x|$, and so $$ begin{align} a_{n+1}&=int^{(n+2)pi}_{(n+1)pi}frac{|sin x|}{x+1},dx=int^{(n+1)pi}_{npi}frac{|sin(x+pi)|}{x+pi+1},dx\ &leq int^{(n+1)pi}_{npi}frac{|sin x|}{x+1}=a_nleqfrac{pi}{pi n +1}xrightarrow{nrightarrowinfty}0 end{align}$$

The series $s=sum_{ngeq0}(-1)^na_n$ has partial sums $s_n=int^{npi}_0frac{sin x}{1+x},dx$. Being a nice alternating series, $s_n$ converges.

In general, for $T>0$, let $[T]$ be its integer part. Then

$$ Big|int^{Tpi}_0frac{sin x}{x+1},dx - int^{[T]pi}_0frac{sin x}{x+1},dxBig|leq int^{pi T}_{[T]pi}frac{|sin x|}{x+1}leq frac{pi}{[T]pi+1}xrightarrow{Trightarrowinfty}0$$

Therefore $lim_{Arightarrowinfty}int^{A}_0frac{sin x}{x+1},dx$ exists and equal $s$.

Answered by Oliver Diaz on January 5, 2022

Notice that $$int_0^infty frac{sin x}{x+1},dx = frac{-cos x}{x+1}Bigg|_0^infty - int_0^infty frac{cos x}{(x+1)^2},dx = 1 - int_0^infty frac{cos x}{(x+1)^2},dx$$

and the last integral converges absolutely since $$int_0^infty frac{left|cos xright|}{(x+1)^2},dx le int_0^infty frac{dx}{(x+1)^2} = int_1^infty frac{dx}{x^2} < +infty.$$

However the original integral does not converge absolutely. Namely, we have $$x in bigcup_{k in mathbb{N}_0} left[fracpi6+kpi,frac{5pi}6+kpiright] implies left|sin xright| ge frac12$$ so $$int_0^infty frac{left|sin xright|}{x+1},dx ge frac12sum_{k=0}^infty int_{fracpi6+kpi}^{frac{5pi}6+kpi} frac{dx}{x+1} = frac12sum_{k=0}^infty ln frac{frac{5pi}6+kpi+1}{fracpi6+kpi+1} = +infty.$$

Answered by mechanodroid on January 5, 2022

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