# How can I study the convergence of the improper integral $int_{0}^{ infty} frac{sin(x)}{x+1} , mathrm dx,$?

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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