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Question Number 38118 by maxmathsup by imad last updated on 21/Jun/18
prove that ∫_0 ^1 (1/(1+(t^a /2)))dt =Σ_(n=0) ^∞ (((−1)^n )/(2^n (na+1))) 2) find the value of Σ_(n=0) ^∞ (((−1)^n )/(2^n (3n+1)))
$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{{a}} }{\mathrm{2}}}{dt}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} \left({na}+\mathrm{1}\right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} \left(\mathrm{3}{n}+\mathrm{1}\right)} \\ $$
Commented by abdo mathsup 649 cc last updated on 05/Jul/18
we have ∣(t^a /2)∣<1 ⇒∫_0 ^1 (1/(1+(t^a /2)))dt =∫_0 ^1 (Σ_(n=0) ^∞ (−1)^n (t^(na) /2^n ))dt =Σ_(n=0) ^∞ (((−1)^n )/2^n ) ∫_0 ^1 t^(na) dt =Σ_(n=0) ^∞ (((−1)^n )/2^n ) (1/(na +1)) 2)we have Σ_(n=0) ^∞ (((−1)^n )/(2^n (3n+1))) =∫_0 ^1 (dt/(1+(t^3 /2))) =2 ∫_0 ^1 (dt/(2+t^3 )) let α/ α^3 =2 ⇒α=^3 (√2) ∫_0 ^1 (dt/(t^3 +2)) =∫_0 ^1 (dt/(t^3 +α^3 )) let ddcompose F(t) = (1/(t^3 +α^3 )) F(t) = (1/((t+α)(t^2 −αt +α^2 ))) =(a/(t+α)) +((bt +c)/(t^2 −αt +α^2 )) ...be continued...
$${we}\:{have}\:\mid\frac{{t}^{{a}} }{\mathrm{2}}\mid<\mathrm{1}\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{{a}} }{\mathrm{2}}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{n}} \:\frac{{t}^{{na}} }{\mathrm{2}^{{n}} }\right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{{na}} {dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:\frac{\mathrm{1}}{{na}\:+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} \left(\mathrm{3}{n}+\mathrm{1}\right)}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+\frac{{t}^{\mathrm{3}} }{\mathrm{2}}} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{2}+{t}^{\mathrm{3}} }\:\:\:\:{let}\:\:\alpha/\:\alpha^{\mathrm{3}} \:=\mathrm{2}\:\Rightarrow\alpha=^{\mathrm{3}} \sqrt{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}^{\mathrm{3}} +\mathrm{2}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{t}^{\mathrm{3}} \:+\alpha^{\mathrm{3}} }\:{let}\:{ddcompose} \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{{t}^{\mathrm{3}} \:+\alpha^{\mathrm{3}} } \\ $$$${F}\left({t}\right)\:=\:\frac{\mathrm{1}}{\left({t}+\alpha\right)\left({t}^{\mathrm{2}} \:−\alpha{t}\:+\alpha^{\mathrm{2}} \right)}\:=\frac{{a}}{{t}+\alpha}\:\:+\frac{{bt}\:+{c}}{{t}^{\mathrm{2}} −\alpha{t}\:+\alpha^{\mathrm{2}} } \\ $$$$…{be}\:{continued}… \\ $$

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