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give-the-integralA-n-1-dt-1-x-n-with-n-integr-and-n-2-at-form-of-serie-




Question Number 65352 by mathmax by abdo last updated on 28/Jul/19
give the integralA_n = ∫_1 ^(+∞)  (dt/(1+x^n ))  with n integr and n≥2  at form of serie.
$${give}\:{the}\:{integralA}_{{n}} =\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\mathrm{1}+{x}^{{n}} }\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$${at}\:{form}\:{of}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 29/Jul/19
A_n =∫_1 ^(+∞)   (dx/(1+x^n ))  changement x =(1/t) give   A_n =−∫_0 ^1   (1/(1+(1/t^n ))) (−(dt/t^2 )) =∫_0 ^1   (t^n /((1+t^n )t^2 ))dt =∫_0 ^1    (t^(n−2) /(1+t^n ))dt  =∫_0 ^1  t^(n−2) (Σ_(k=0) ^∞ (−1)^k  t^(kn) )dt =Σ_(k=0) ^∞  (−1)^k  ∫_0 ^1   t^(n−2+kn)  dt  =Σ_(k=0) ^∞  (−1)^k   [(1/(n−2+kn +1)) t^(n−2+kn+1) ]_0 ^1   =Σ_(k=0) ^∞    (((−1)^k )/(n(k+1)−1)) =Σ_(k=1) ^∞   (((−1)^(k−1) )/(nk−1)) ⇒  A_n =(1/(n−1))−(1/(2n−1)) +(1/(3n−1)) −.....
$${A}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{{n}} }\:\:{changement}\:{x}\:=\frac{\mathrm{1}}{{t}}\:{give}\: \\ $$$${A}_{{n}} =−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{{n}} }}\:\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{{n}} }{\left(\mathrm{1}+{t}^{{n}} \right){t}^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{{n}−\mathrm{2}} }{\mathrm{1}+{t}^{{n}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}−\mathrm{2}} \left(\sum_{{k}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:{t}^{{kn}} \right){dt}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}−\mathrm{2}+{kn}} \:{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:\:\left[\frac{\mathrm{1}}{{n}−\mathrm{2}+{kn}\:+\mathrm{1}}\:{t}^{{n}−\mathrm{2}+{kn}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{n}\left({k}+\mathrm{1}\right)−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{nk}−\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\:−….. \\ $$

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