# Riemann integrable over given interval?

I was looking for a brief explanation as to which of the following functions are Riemann integrable or not.

$f(x):=(1-x^2)^{-1}$if $x≠ 1$ and $x ≠1,$

$f(x):= 0,$ if $x=1$ or $x=-1.$

(Over interval: $[-1,1]$)

I can see that the function is continuous & hence Riemann integrable over $(-1,1)$ but don't know how I can explain that it is over $[-1,1]$. I am thinking I maybe can just say that it is bounded?

Also

$f(x)=\frac{1}{x^2-3x-4},$ over interval $[-3,0]$, I see in the interval when $x=-1$, $f(x)$ is undefined. Is it acceptable to say that this function is hence not bounded over this interval and hence not Riemann integrable?

## 2 Answers

Both functions are unbounded and therefore they are not Riemann-integrable.

If the function is not bounded, we cannot talk about Riemann integrability, but we can discuss about the improper Riemann integrability. In this case, even if we deal with improper Riemann integral, the following still does not exist: \begin{align*} \int_{0}^{1}\dfrac{1}{1-x^{2}}dx=\int_{0}^{1}\dfrac{1}{1+x}\dfrac{1}{1-x}dx\geq\dfrac{1}{2}\int_{0}^{1}\dfrac{1}{1-x}dx=-\dfrac{1}{2}\lim_{M\rightarrow 1^{-}}\log(1-M)=\infty. \end{align*} So the improper Riemann integral $\displaystyle\int_{-1}^{1}\dfrac{1}{1-x^{2}}dx$ does not exist.