0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.