Question Number 65352 by mathmax by abdo last updated on 28/Jul/19
$${give}\:{the}\:{integralA}_{{n}} =\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dt}}{\mathrm{1}+{x}^{{n}} }\:\:{with}\:{n}\:{integr}\:{and}\:{n}\geqslant\mathrm{2} \\ $$$${at}\:{form}\:{of}\:{serie}. \\ $$
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)) −.....](Q65385.png)
$${A}_{{n}} =\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{dx}}{\mathrm{1}+{x}^{{n}} }\:\:{changement}\:{x}\:=\frac{\mathrm{1}}{{t}}\:{give}\: \\ $$$${A}_{{n}} =−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{t}^{{n}} }}\:\left(−\frac{{dt}}{{t}^{\mathrm{2}} }\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{{n}} }{\left(\mathrm{1}+{t}^{{n}} \right){t}^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{{n}−\mathrm{2}} }{\mathrm{1}+{t}^{{n}} }{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}−\mathrm{2}} \left(\sum_{{k}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:{t}^{{kn}} \right){dt}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{n}−\mathrm{2}+{kn}} \:{dt} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{k}} \:\:\left[\frac{\mathrm{1}}{{n}−\mathrm{2}+{kn}\:+\mathrm{1}}\:{t}^{{n}−\mathrm{2}+{kn}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} }{{n}\left({k}+\mathrm{1}\right)−\mathrm{1}}\:=\sum_{{k}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}−\mathrm{1}} }{{nk}−\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}}{{n}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:+\frac{\mathrm{1}}{\mathrm{3}{n}−\mathrm{1}}\:−..... \\ $$