I ran into a few problems where i had to check absolute\conditional convergence of a few integrals.
I'm sure theres a method to check this, i just can't find the trick.
I wan't help with one of the problems so that i can get the rest on my own.
Check absolute or conditional convergence of the integral:
$$\int_{0}^{\infty}\frac{\sin x}{1+x^{2}}dx$$
thanks in advance,
yaron
On
This is absolutely convergent. The intuition here is that the absolute value of our function is "small" when $x$ is large.
Since $|\sin x| \le 1$ for all $x$, we have $$\left|\frac{\sin x}{1+x^2}\right| \le \frac{1}{1+x^2}.$$ The integral $$\int_0^\infty \frac{dx}{1+x^2}$$ converges. If we want to be formal about the proof, note that $$\int_0^M \frac{dx}{1+x^2}=\arctan M,$$ and $\arctan M \to \frac{\pi}{2}$ as $M\to\infty$.
So by Comparison $\int_0^\infty \frac{|\sin x|}{1+x^2}\,dx$ converges, and therefore $\int_0^\infty \frac{\sin x}{1+x^2}\,dx$ converges absolutely.
We give a somewhat more messy alternate approach. Break up our original integral as $$\int_0^1 \frac{\sin x}{1+x^2}\,dx +\int_1^\infty \frac{\sin x}{1+x^2}\,dx.$$
The first integral is no problem, a respectable function over a finite interval. For the second, note that $$\left|\frac{\sin x}{1+x^2}\right| \le \frac{1}{x^2}$$ if $x\ge 1$. But $\int_1^\infty \frac{dx}{x^2}$ converges.
In order to make comparisons, we need to have a collection of integrals that we know converge or diverge. One of the most important classes of integrals to compare with are the integrals $$\int_1^\infty \frac{dx}{x^p}.$$ This converges if $p\gt 1$, and diverges if $p\le 1$.
Compare with $$\int_0^\infty {dx\over 1 + x^2}$$.