Convergence Problem of Improper Integrals In this post (c.f. Existence of improper integral.), in Parcly Taxel's Answer, he concluded that the improper integral $\int_1^\infty\frac{\sin x}{x(1+x)}\,dx$ converges by the estimate
$$\left|\int_1^\infty\frac{\sin x}{x(1+x)}\,dx\right|\le\int_1^\infty\frac1{x(1+x)}\,dx\le\int_1^\infty\frac1{x^2}\,dx.$$
I don't really understand why the boundedness of $\int_1^\infty\frac{\sin x}{x(1+x)}\,dx$ is sufficient to ensure the convergence of it, because I think it may oscillate.
Thanks for help.
 A: What is really happening is an analogue of absolutely convergent series. If a series is absolutely convergent, then the original series is convergent. One proves this by showing that if the series is absolutely convergent, then the originial series satifies a Cauchy condition.
In a similar fashion, if an improper integral is absolutely convergent, then the original improper integral is convergent. So, what do we mean by an absolutely convergent improper integral? We say that the improper integral $\int_a^{\to b}f(x)\,dx$ converges absolutely if
$$
  \int_a^{\to b}|f(x)|\,dx
$$
converges. (What "$\to b$" means is simply "take the limit $\lim_{t\to b}\int_a^tf(x)\,dx$".)
Now, to prove that if an improper integral is absolutely convergent then it is convergent we prove it as if we were dealing with series: we prove that the original improper integral, which is really a limit, satisfies a Cauchy condition. The difference is that this Cauchy condition is for limits and not just limits of sequences. (Maybe this question
can help clarifying what I mean by Cauchy condition for a limit.)
Going back to your problem, what Taxel really did was proving that the improper integral is absolutely convergent:
$$
  \int_1^\infty \left|\frac{\sin x}{x(1+x)}\right|\,dx\leq
  \int_1^\infty \frac{1}{x(1+x)}\,dx \leq\int_1^\infty\frac{1}{x^2}\,dx
  ,
$$
and so, the original improper integral converges!
A: If it oscillates it helps the convergence. When one limit becomes $\infty$ the integral is improper, for it to be convergent  the integrand near $x \sim infty$ must fall faster than $\frac{1}{|x|}$ when $x\pm \infty.$  This is so because
$$\int_{1}^\infty \frac{dx}{x^\beta} < \infty, ~ if~\beta>1.$$
You may check that $$I_1=\int_{1}^{\infty} \frac{dx}{x^{1.01}} =\infty$$ but
$$I_2=\int_{1}^{\infty} \frac{dx}{x^{0.99}}$$ is finite.
Here in your case $\beta=2$, so your integral is improper but convergent.
