oscillating integers

working with $f(x)=sin(\frac{\pi }{x})$ on interval $[0,1]$
as $x$ approaches $0$, function oscillates with ever-increasing frequency

$S={\int }_0^{1}sin(\frac{\pi }{x})dx$
$R={\int }_0^1|sin(\frac{\pi }{x})|dx$

${\int }_0^{1}sin(\frac{\pi }{x})dx\le {\int }_0^1|sin(\frac{\pi }{x})|dx$ by definition
which gives $-R<S$

zeros happen at $x=\frac{1}{n}$, for $n=1,2,3,\dots$
for each interval $(\frac{1}{n+1},\frac{1}{n})$, area has sign $(-1{)}^n$
sum is essentially $S={\sum }_{n=1}^{\infty }{\int }_{1/(n+1)}^{1/n}sin(\frac{\pi }{x})dx$
and first interval is largest, which in this case is $(\frac{1}{2},1)$
which has sign of $(-1{)}^1=-1$
all contributions are smaller in magnitude and alternate; so $S<0$

together, $-R<S<0$

basic riemann sum form is ${\sum }_{i=1}^{n}f(a+\frac{(i-1)(b-a)}{n})\cdot \frac{(b-a)}{n}$
but actually, use rectangles at middle, so ${\sum }_{i=1}^{n}f(a+\frac{(i-0.5)(b-a)}{n})\cdot \frac{(b-a)}{n}$

main problem is that whatever $\Delta x$ is chosen, it’ll be too big for the increasingly small curves when going $\to 0$
i.e., need to scale $n$ somehow

instead of having $a=0$ and $b=1$, take some $a=\frac{1}{n+1}$ and $b=\frac{1}{n}$
so now have some $g(a,b)={\sum }_{i=1}^{n}f(a+\frac{(i-0.5)(b-a)}{n})\cdot \frac{(b-a)}{n}$

and then for each section, want to have some $k$ bars, so $n=k$
so each has a $\Delta x=\frac{(b-a)}{k}$

and then have the actual function like $h(n)={\sum }_{i=1}^{n}g(\frac{1}{i+1},\frac{1}{i})$

trying to graph this out in desmos, first need to compute the position of $i$, say using function $i_x$

taking in endpoints, $i_x(a,b,x)$. trying to normalize $x$ from $[a,b]$ to $[1,k]$,
so $x-a$, $\frac{x-a}{b-a}$, $\frac{k(x-a)}{b-a}$, $i_x(a,b,x)=\frac{k(x-a)}{b-a}+1$
and since discrete, need to get nearest $i$, so $i_x(a,b,x)=\lfloor \frac{k(x-a)}{b-a}\rfloor$

the computation for a given rectangle height is the same—using some $x(a,b,i)=a+\frac{(i-0.5)(b-a)}{k}$
combining, over range $[a,b]$, use $x(a,b,i_x(a,b,x))$ to get $x$s

so now, in desmos, represent this as $r_x(a,b,x)=\{a\le x\le b:x(a,b,i_x(a,b,x))\}$
and want to put these x values into $f$, so $f(r_x(a,b,x))$

and highlight above and below axes with $\le y\le 0$ or $0\le y\le$

$S={\int }_0^{1}sin(\frac{\pi }{x})dx$
$R={\int }_0^1|sin(\frac{\pi }{x})|dx$
$T={\int }_0^{1}si{n}^2(\frac{\pi }{x})dx$

$0<|y|<1$ implies $0\le y^2\le |y|$
excluding $0$ and $1$, working with $0<|sin(\frac{\pi }{x})|<1$, so $0<si{n}^2(\frac{\pi }{x})<|sin(\frac{\pi }{x})|$
and since not all $sin(\frac{\pi }{x})=0,1$ on a set of $x$s, strictly $si{n}^2(\frac{\pi }{x})<|sin(\frac{\pi }{x})|$
and thus, $T<R$, and by extent, $0<T<R$

with $-S<T$, turns into $S+T>0$
so ${\int }_0^1[sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})]dx>0$

splitting at zeros, $I_n={\int }_{1/(n+1)}^{1/n}[sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})]dx$
pairing $\frac{1}{n}$ to $\frac{1}{n+1}$, gives $x$ → $\frac{x}{1+x}$

over a cycle, $sin(\frac{\pi }{x})$ changes sign, $si{n}^2(\frac{\pi }{x})$ stays the same, $dx$ is divided by $(1+x{)}^2$, but same otherwise
so $I_{n+1}={\int }_{1/(n+1)}^{1/n}[-sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})]\frac{dx}{(1+x{)}^2}$
as one, $I_n+I_{n+1}={\int }_{1/(n+1)}^{1/n}[sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})]dx+{\int }_{1/(n+1)}^{1/n}[-sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})]\frac{dx}{(1+x{)}^2}$
and combining, $I_n+I_{n+1}={\int }_{1/(n+1)}^{1/n}[sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})\frac{-sin(\frac{\pi }{x})+si{n}^2(\frac{\pi }{x})}{(1+x{)}^2}]dx$
and factoring, $I_n+I_{n+1}={\int }_{1/(n+1)}^{1/n}[sin(\frac{\pi }{x})(1-\frac{1}{(1+x{)}^2})+si{n}^2(\frac{\pi }{x})(1+\frac{1}{(1+x{)}^2})]dx$

when choosing $n$ s.t. $sin(\frac{\pi }{x})\ge 0$, then also $si{n}^2(\frac{\pi }{x})>0$ (as always)
and more importantly $1+\frac{1}{(1+x{)}^2},1-\frac{1}{(1+x{)}^2}>0$
so everything is positive and $I_n+I_{n+1}>0$
so $-S<T$

already know $-R<S<0$
now know $-S<T<R$
together, $0<-S<T<R$