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=2∫10arcsin(w)w1+w2dw(w↦√x)IBP=ln(w2+1)arcsin(w)|10−∫10ln(w2+1)√1−w2dw=π2ln(2)−∫10∞∑n=1(−1)n+1nw2n√1−w2dw=π2ln(2)−∞∑n=1(−1)n+1n∫10w2n(1−w2)−12dw=π2ln(2)−12∞∑n=1(−1)n+1n∫10sn−12(1−s)−12ds(s↦w2)=π2ln(2)−12∞∑n=1(−1)n+1nB(n+12,12)=π2ln(2)−12∞∑n=1(−1)n+1nΓ(n+12)√πn! From the formulas Γ(n+12)=(2n−1)!!√π2n (2n)!!=2nn! (2nn)=22n(2n−1)!!(2n)!! we have I=π2ln(2)−12∞∑n=1(−1)n+1n(2n−1)!!π2nn! from (1)=π2ln(2)+π2∞∑n=1(−1)n(2n−1)!!n(2n)!! from (2)=π2ln(2)+π2∞∑n=1(−1)n(2nn)22nn from (3) We previously proved that ∞∑n=1(2nn)22nnxn=2ln(2√1−x+1) If we put x=−1 ∞∑n=1(2nn)22nn(−1)n=2ln(2√2+1) Hence, we can conclude ∫10arcsin(√x)1+xdx=π2ln(8)−πln(1+√2)
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