# Improper Integral - Multiple Choice Problem - $I$

Let $f$ be a function defined $\forall~ x\geq 1$.Let $n$ denote a positive integer and let $I_n$ denote the integral $\int_1^nf(x)dx$ which is always assumed to exist. Which of the following statements are true?

$(i)$ If $f$ is a monotonic decreasing function and if $\lim_{n \rightarrow \infty} I_n$ exists, then the integral $\int_1^\infty f(x) dx$ converges

$(ii)$ If $\lim_{x \rightarrow \infty} f(x)=0$ and $\lim_{n \rightarrow \infty}I_n=A$, then $\int_1^\infty f(x)dx$ converges and has the value $A$

$(iii)$ If the sequence $\{I_n\}$ converges, then the integral $\int_1^\infty f(x)dx$ converges $(iv)$ If $f$ is positive and if $\lim_{n \rightarrow \infty} I_n=A$, then $\int_1^\infty f(x)dx$ converges and has the value $A$

$(v)$ Assume $f'(x)$ exists $~\forall~ x \geq 1$ and $\exists~M>0$ such that $|f~'(x)| \leq M~\forall~x\geq 1$. If $\lim_{n \rightarrow \infty} I_n=A$, then $\int_1^\infty f(x)dx$ converges and has the value $A$

$(vi)$ If $\int_1^\infty f(x)dx$ converges, then $\lim_{x \rightarrow \infty} f(x)=0$

Attempt: $(i)$ True. By definition, This should be true whether $f$ is monotonic or not. Is $f$ given as monotonic just extra information?

$(ii)$ True. By definition. What could be the use of the information $\lim_{n \rightarrow \infty} f(x)=0$ ?

$(iii)$ True. Sequence $\{I_n\}$ converges to lets say, $l$. Hence, $\int_1^\infty f(x)dx$ must be equal to $l$?

$(iv)$ True. By definition. Is $f$ given as positive just extra information?

$(v)$ Again, True. If $A$ is finite, then all information about $f~'$ should be superfluous.

$(vi)$ Not Sure. I have been trying to use the same logic as done for series. Let $h \rightarrow 0$ such that $h>0$. Since, $\int_1^\infty f(x)dx$ converges :

$\lim_{x \rightarrow \infty} \int_1^{x-h} f(x) dx = \lim_{x \rightarrow \infty} \int_1^{x } f(x) dx$

$\implies \lim_{x \rightarrow \infty} \int_{x-h}^{x } f(x) dx =0$. From the gien information, I am unable to move further.

However if $f$ is given to be continuous, then, by the mean value theorem : $\exists~c\in (x-h,x)$ such that $f(c)h=0$. Since, $h \rightarrow 0$, essentially this condition is the same as $\lim_{x \rightarrow \infty} f(x) ~h = 0 .$

Are my attempts correct ?

Could someone please take some time to give me a direction for the questions posed for options $(i),(ii)$ and $(iii),(iv),(v),(vi)$.

Thank you very much for your help in this regard .

• As far as I see everything is true by definition and the information you specified is superfluous. However if $A$ is allowed to be $+\infty$, then the second one is false. You mean $\lim \limits_{\color{red}x \rightarrow \infty} f(x)=0$, right? – Git Gud Jan 27 '15 at 23:24
• Yeah, I meant $\lim_{x \rightarrow \infty} f(x)=0$. If $A$ is a finite quantity, then the information $\lim_{x \rightarrow \infty} f(x)=0$ is superfluous right? – MathMan Jan 27 '15 at 23:27
• I guess "the integral converges" means that $\int |f| < \infty$, otherwise as you said some of these are void. Then the answers are not always true. – Sary Jan 27 '15 at 23:27

Ah I had missed the point. Saying that $\int_1^\infty f(x){\rm d}x$ converges is the same as saying that "$\lim_{y\to\infty}\int_1^y f(x){\rm d}x$ converges to a finite limit". Then you might want to compare that with "$\lim_{n\to\infty}I_n$ converges".