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Sunday, January 23, 2022

Integral of the day XXI

Integral found on Twitter

Integral involving binomial series


Today we show the proof of this integral proposed by @BenriBot_ccbs. 10arcsin(x)1+xdx=π2ln(8)πln(1+2) The proof relies on the use of binomial series.

Proof

I=10arcsin(x)1+xdx=210arcsin(w)w1+w2dw(wx)IBP=ln(w2+1)arcsin(w)|1010ln(w2+1)1w2dw=π2ln(2)10n=1(1)n+1nw2n1w2dw=π2ln(2)n=1(1)n+1n10w2n(1w2)12dw=π2ln(2)12n=1(1)n+1n10sn12(1s)12ds(sw2)=π2ln(2)12n=1(1)n+1nB(n+12,12)=π2ln(2)12n=1(1)n+1nΓ(n+12)πn! From the formulas Γ(n+12)=(2n1)!!π2n (2n)!!=2nn! (2nn)=22n(2n1)!!(2n)!! we have I=π2ln(2)12n=1(1)n+1n(2n1)!!π2nn! from (1)=π2ln(2)+π2n=1(1)n(2n1)!!n(2n)!! from (2)=π2ln(2)+π2n=1(1)n(2nn)22nn from (3) We previously proved that n=1(2nn)22nnxn=2ln(21x+1) If we put x=1 n=1(2nn)22nn(1)n=2ln(22+1) Hence, we can conclude 10arcsin(x)1+xdx=π2ln(8)πln(1+2)

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