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Request If \$L_n=displaystylefrac1piint_0^pi|D_n(t)|dt\$, my spouse and I have \$L_n=dfrac4pi^2log n+O(1) \$.

Proof Let’s start by dividing the interval \$[0,pi]\$ into subintervals, where \$sinleft(n+dfrac 1 2right)\$ ends with a sign. Since \$sin t/2\$ will always be positive in this region, our team can write }intlimits_0^pi \$\$frac1{pi D_nleft( t right) right means frac1 pi intlimits_0^frac2pi 2n + 1 fracsin t/2dt + frac1pi sumlimits_k is 1^n – 1 intlimits_frac2kpi 2n + 1^ frac2left ( k + 1 right)pi 2n + 1 frac sin left( n + frac12 right)t rightsin t/2dt + frac1pi intlimits_frac2n pi 2n + 1 ^ pi frac sin left( n + frac12 right)t rightsin t/2dt \$\$ The last integral helps you \$0\$ as \$n\$ goes to infinity for you. We can pay attention to the middle of the noise. The correct results \$g(t)=dfrac1sin t/2 – dfrac1t/2\$ are continuous with respect to \$[0,pi]\$, so the integrals mean \$\$c_n frac 1 pi int limits_0 ^frac2n pi 2n + 1 gleft( t \$\$ right)dt must exist and be bounded above by any constant for any \$n\$. Thenwe could assume that \$\$d_n=frac2pi sumlimits_k is 1^n – 1 intlimits_frac2kpi 2n + 1^frac2left( k + 1 right) pi 2n + 1 fracleftt dt\$\$ But \$t^-1\$ is positive and decreasing since we have limits \$\$frac2pi sumlimits_k equals 1^n – 1 frac2n + 12 left ( k + 1 right)pi intlimits_frac2kpi 2n + 1^frac2left( k + 1 right)pi 2n + left sin left( n + frac12 right ) t right dt leqslant d_n leqslant frac2pi sumlimits_k = 1^n – 1 frac2n + 12kpi intlimits_frac2kpi 2n + 1^frac2left( k + 1 right) pi 2n + 1 sin left( n + frac12 right)t right dt\$\$ But directly \$\$intlimits_frac2kpi 2n + 1^ dt= frac42n +1 \$\$ where \$\$ frac4pi ^2sumlimits_k = 1^n – 1 frac1k + 1 leqslant d_n leqslant frac4pi ^2sumlimits_k implies 1^n – 1 frac1k \$\$

You probably know that \$displaystylesumlimits_k = 1^n – 1 frac1k = log deborah + Oleft( 1 right)\$ (just implement \$=log +n+gamma + o ( 1)\$), so we can say that \$d_n = dfrac4pi ^2log s + Oleft( 1 right)\$, i.e. \$d_n-dfrac4pi^ 2 log n\$ is bounded by a new constant of the form \$ntoinfty\$. Finally, we check the \$\$intlimits_0^frac2pi 2n + 1 left( fracsin n + frac12 right)tt dt\$\$ integral we got from our sum. The integrand is confident, stable and decreasing over the period under consideration, reaching a maximum \$t=0\$ at \$n+dfrac definitely 2\$, from which the following rough estimate can be given: \$\$int limit_0^ frac2pi 2n + 1 fracsin left( n + frac12 right)tt dt leqslant left( n + frac12 right)intlimits_0^frac2pi 2n + 1 dt implies pi \$\$

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