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find-A-n-0-1-x-n-ch-x-dx-

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Question Number 37284 by abdo.msup.com last updated on 11/Jun/18
find A_n = ∫_0 ^1 (x^n /(ch(x))) dx .
findAn=∫01xnch(x)dx.
Commented by prof Abdo imad last updated on 17/Jun/18
A_n = ∫_0 ^1 ((2x^n )/(e^x +e^(−x) )) =2∫_0 ^1 ((x^n e^(−x) )/(1+e^(−2x) ))dx =2 ∫_0 ^1 (Σ_(n=0) ^∞ (−1)^n e^(−2nx) x^n e^(−x) )dx =2Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^n e^(−(2n+1)x) dx =_((2n+1)x=t) 2Σ_(n=0) ^∞ (−1)^n ∫_0 ^(2n+1) (t^n /((2n+1)^n )) e^(−t) (dt/(2n+1)) =2 Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^(n+1) )) A_n with A_n = ∫_0 ^(2n+1) t^n e^(−t) dt =[−t^n e^(−t) ]_0 ^(2n+1) +∫_0 ^(2n+1) nt^(n−1) e^(−t) dt =−(2n+1)^n e^(−(2n+1)) +n A_(n−1) be continued...
An=∫012xnex+e−x=2∫01xne−x1+e−2xdx=2∫01(∑n=0∞(−1)ne−2nxxne−x)dx=2∑n=0∞(−1)n∫01xne−(2n+1)xdx=(2n+1)x=t2∑n=0∞(−1)n∫02n+1tn(2n+1)ne−tdt2n+1=2∑n=0∞(−1)n(2n+1)n+1AnwithAn=∫02n+1tne−tdt=[−tne−t]02n+1+∫02n+1ntn−1e−tdt=−(2n+1)ne−(2n+1)+nAn−1becontinued…

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