・はじめに・内容・最後に
どうも、色数です。今回はarcsinxや反復ベータ積分などを用いて級数を解いていきます。1番大変な積分は数楽さんに解いていただきました。ありがとうございます。証明などを共有する意図で記事を書いているわけではないのでおおまかな流れ以外は割愛します。
arcsin2r−1x=(2r−1)!22r−1∑0<n1<⋯<nr22nrn12⋯nr−12nr(2nrnr)x2nr−11−x2
arcsin2rx=(2r)!22r∑0<n1<⋯<nr22nrn12⋯nr2(2nrnr)x2nr
arcsin3x=6∫0xdt31−t32∫0t3dt21−t22∫0t2dt11−t1=3∫0xarcsin2t31−t32dt3=32∑n=1∞22nn2(2nn)∫0xt32n1−t32dt3=32∑n=1∞22nn2(2nn)(2nn)22n∑n<m22m2m(2mm)x2m−11−x2=34∑0<n1<n222n2n12n2(2n2n2)x2n2−11−x2
arcsin4x=4∫0xarcsin3t1−t2dt=3∑0<n1<n222n2n12n2(2n2n2)∫0xt2n2−1dt2=3∑0<n1<n222n22n12n22(2n2n2)x2n2arcsin5x=5∫0xarcsin4t1−t2dt=152∑0<n1<n222n2n12n22(2n2n2)∫0xt2n21−t2dt=154∑0<n1<n2<n322n3n12n22n3(2n3n3)x2n3−11−x2
∑0<n1<n21n12n2(2n2n2)=π31623∑0<n1<n2<n31n12n22n3(2n3n3)=π5291603⋮
一般にR({2}r,1)が求まります。
∑0<n1<⋯<nr22nrn12⋯nr−12nr(2nrnr)x2nr−11−x2=∑0≤n1<⋯<nr(2nrnr)22nr(n1+12)2⋯(nr−1+12)2(nr+12)x2nr+1
∫01t2r−1arcsin2tdt=[t2r2rarcsin2t]01−∫01t2rarcsintr1−t2dt=π28r−[(2rr)r22r∑r<m22mt2m−11−t22m(2mm)arcsint]01+(2rr)r22r∑r<m22m2m(2mm)∫01t2m−1dt=π28r−(2rr)r22r∑r<m22m4m2(2mm)∫01arcsin2tarctanhtdt=∑n=1∞12n−1∫01t2n−1arcsin2tdt=π28∑n=1∞1(2n−1)n+14∑0<n1<n2122n1−2n2n12n22(2n1n1)(2n2n2)−12∑0<n1<n2122n1−2n2(2n1−1)n22(2n1n1)(2n2n2)=π24ln2+14∑0<n1<n2122n1−2n2n12n22(2n1n1)(2n2n2)−12∑0<n1<n2122n1−2n2(2n1−1)n22(2n1n1)(2n2n2)14∑0<n1<n2122n1−2n2n12n22(2n1n1)(2n2n2)−12∑0<n1<n2122n1−2n2(2n1−1)n22(2n1n1)(2n2n2)=78ζ(3)−2ln2+14π2ln2−π24∫01arcsin2ttanh−1tdt=∑n=1∞12n−1∫01t2n−1arcsin2tdt=12∑n=1∞12n−1∑m=1∞22mm2(2mm)∫01t2m+2n−1dt=12∑0<n,m22m(2n−1)(2n+2m)m2(2mm)
78ζ(3)−2ln2+12π2ln2−π24=12∑0<n,m22m(2n−1)(2n+2m)m2(2mm)=π24ln2+14∑0<n1<n2122n1−2n2n12n22(2n1n1)(2n2n2)−12∑0<n1<n2122n1−2n2(2n1−1)n22(2n1n1)(2n2n2)
∫01t2r−1arcsin3tdt=[t2r2rarcsin3t]01−3∫01t2r2rarcsin2t1−t2=π316r−32r[∑r<m22mt2m−11−t22m(2mm)arcsin2t]01+34r∑r<m∫0122mt2m−11−t2m(2mm)arcsin2tdt=π316r+34r∑r<m22mm(2mm)([(t2m2m1−t2+12m22m+22(m+1)(2m+2m+1)∑m+1≤n22nt2n−11−t22n(2nn))arcsin2t]01−2∫01(t2m2m1−t2+12m22m+22(m+1)(2m+2m+1)∑m+1≤n22nt2n−11−t22n(2nn))arcsint1−t2dt)=π316r−34r∑r<m22mm2(2mm)(∫01t2marcsintdt+22m+22(m+1)(2m+2m+1)∑m+1≤n22n2n(2nn)∫01t2n−1arcsintdt)=π316r−34r∑r<m22mm2(2mm)(π4m+2−12m+122m+22(m+1)(2m+2m+1)+22m+22(m+1)(2m+2m+1)∑m<n22n2n(2nn)π4(1n−(2nn)n22n))
3π8∑m=1∞12m−1∑0<n1<n222n2n12n2(2n2n2)(2n2+mn2+m)22n2+2m=∑0<n1<n2<n312n1−1(π316n1−34n122n2n22(2n2n2)(π4n2+2−12n2+122n2+22(n2+1)(2n2+2n2+1)+22n2+22(n2+1)(2n2+2n2+1)22n32n3(2n3n3)π4(1n3−(2n3n3)n322n3)))
うん、反復ベータ積分凄い。
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