Question Number 37284 by abdo.msup.com last updated on 11/Jun/18
$${find}\:\:{A}_{{n}} \:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{x}^{{n}} }{{ch}\left({x}\right)}\:{dx}\:. \\ $$
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...](https://www.tinkutara.com/question/Q37763.png)
$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{x}^{{n}} }{{e}^{{x}} \:+{e}^{−{x}} }\:=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{{n}} \:{e}^{−{x}} }{\mathrm{1}+{e}^{−\mathrm{2}{x}} }{dx} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \:{e}^{−\mathrm{2}{nx}} \:{x}^{{n}} \:{e}^{−{x}} \right){dx} \\ $$$$=\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{n}} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right){x}} {dx} \\ $$$$=_{\left(\mathrm{2}{n}+\mathrm{1}\right){x}={t}} \:\:\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}{n}+\mathrm{1}} \:\frac{{t}^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} }\:{e}^{−{t}} \:\frac{{dt}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\mathrm{2}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}+\mathrm{1}} }\:{A}_{{n}} \:\:{with} \\ $$$${A}_{{n}} =\:\int_{\mathrm{0}} ^{\mathrm{2}{n}+\mathrm{1}} \:{t}^{{n}} \:{e}^{−{t}} {dt} \\ $$$$=\left[−{t}^{{n}} \:{e}^{−{t}} \right]_{\mathrm{0}} ^{\mathrm{2}{n}+\mathrm{1}} \:+\int_{\mathrm{0}} ^{\mathrm{2}{n}+\mathrm{1}} \:{nt}^{{n}−\mathrm{1}} \:{e}^{−{t}} {dt} \\ $$$$=−\left(\mathrm{2}{n}+\mathrm{1}\right)^{{n}} \:{e}^{−\left(\mathrm{2}{n}+\mathrm{1}\right)} \:\:+{n}\:{A}_{{n}−\mathrm{1}} \:\:\:{be}\:{continued}… \\ $$