Determine if each of the following integral converges or diverges (1) $\int_{10}^\infty \frac{x^2+x+2}{x^4+x^2-1}dx$
(2) $\int_{0}^\infty \frac{sinx}{x}dx$
I'm learning the convergence of series in my analysis course but I can't figure out how to test the convergence of these integrals. How can I apply the methods for series convergence into these integrals?
Thanks.
 A: "How can I apply the methods for series convergence into these integrals?" - the short answer is that you can't, integrals and series are different things.  For example, there is nothing like the ratio test for integrals as far as I know.
However, one respect in which they are similar is that both can be done by a comparison test or a limit form of the comparison test.  I'll give brief solutions and leave you to fill in the details.
For your first integral,
$$\frac{x^2+x+2}{x^4+x^2-1}\bigg/\frac{1}{x^2}\to1$$
as $x\to\infty$, and it is a standard result that
$$\int_{10}^\infty \frac{1}{x^2}\,dx$$
converges, so your integral converges by the limit form of the comparison test.
The second is harder but you could try this.  For $n=1,2,\ldots$, consider the integral
$$\eqalign{I=\int_{2n\pi}^{(2n+2)\pi}\frac{\sin x}{x}\,dx
  &=\int_{2n\pi}^{(2n+1)\pi}\frac{\sin x}{x}\,dx
    +\int_{(2n+1)\pi}^{(2n+2)\pi}\frac{\sin x}{x}\,dx\cr
  &=A_1+A_2\ .\cr}$$
From a diagram you can see that $A_1>0$ and $A_2<0$, so the integral is
$$I=A_1-|A_2|\ .$$
Also,
$$A_1=\int_{2n\pi}^{(2n+1)\pi}\frac{\sin x}{x}\,dx
  \le\int_{2n\pi}^{(2n+1)\pi}\frac{\sin x}{2n\pi}\,dx=\frac{1}{n\pi}$$
and
$$|A_2|=\int_{(2n+1)\pi}^{(2n+2)\pi}\frac{|\sin x|}{x}\,dx
  \ge\int_{(2n+1)\pi}^{(2n+2)\pi}\frac{|\sin x|}{(2n+2)\pi}\,dx
  =\frac{1}{(n+1)\pi}\ .$$
Therefore
$$I\le\frac{1}{\pi}\Bigl(\frac{1}{n}-\frac{1}{n+1}\Bigr)\ ,$$
and adding up a number of integrals gives
$$\int_{2\pi}^{2n\pi}\frac{\sin x}{x}\,dx
  \le\frac{1}{\pi}\Bigl(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n-1}-\frac{1}{n}\Bigr)<\frac{1}{\pi}\ .$$
From a diagram, the sequence of integrals
$$\int_{2\pi}^{2n\pi}\frac{\sin x}{x}\,dx$$
is monotonic increasing, and this shows it is bounded above, so it approaches a limit.
Now for any $t$ between $2n\pi$ and $(2n+2)\pi$, you can see from a diagram, using the above calculations, that
$$\int_{2\pi}^{2n\pi}\frac{\sin x}{x}\,dx\le\int_{2\pi}^t\frac{\sin x}{x}\,dx
  \le\int_{2\pi}^{2n\pi}\frac{\sin x}{x}\,dx+\frac{1}{n\pi}\ .$$
Letting $n\to\infty$ shows that
$$\int_{2\pi}^t\frac{\sin x}{x}\,dx$$
converges.
You still have to think about the bit from $0$ to $2\pi$; the point here is that $(\sin x)/x$ approaches a (finite) limit as $x\to0$.  Good luck!
A: (1): Converges. $\dfrac{x^2 + x + 2}{x^4 + x^2 - 1} \leq \dfrac{2}{x^2}$ when $x$ is large enough and $\displaystyle \int_{x = 10}^\infty \dfrac{2}{x^2}dx = \dfrac{1}{5}$ which is converges. So by comparison test, the integral converges.
(2): Converges by Jordan's lemma or can be proven by splitting up to $2$ integrals: one from $0$ to $\pi$ and the other from $\pi$ to $\infty$.
