1. ## [SOLVED] Improper integrals

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent."

I don't get what to do next, but turns out the integral is

arctan(x) I don't know where to continue from there.

2. Because $\displaystyle \frac{1}{1+x^{2}}$ is an 'even function' we can swap x with -x and obtain...

$\displaystyle \int_{-\infty}^{1}\frac{dx}{1+x^{2}} = \int_{-1}^{\infty}\frac{dx}{1+x^{2}}= \int_{-1}^{0}\frac{dx}{1+x^{2}} + \int_{0}^{\infty}\frac{dx}{1+x^{2}}$ (1)

Because for $\displaystyle n-1<x<n$, $\displaystyle n\ge 1$ integer is...

$\displaystyle \frac{1}{1+x^{2}} < \frac{1}{n^{2}}$ (2)

... is...

$\displaystyle \int_{0}^{\infty}\frac{dx}{1+x^{2}} < \sum_{n=1}^{\infty}\frac{1}{n^{2}} = \frac{\pi^{2}}{6}$ (3)

... so that the integral (1) converges. Its value can be computed in 'standard form'...

$\displaystyle \int_{-1}^{\infty} \frac{1}{1+x^{2}} = \lim_{x \rightarrow \infty} (tan^{-1} x + \frac{\pi}{4}) = \frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$ (4)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$