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Question Number 122298 by mnjuly1970 last updated on 15/Nov/20
...nice calculus... prove that : Σ_(n=1 ) ^∞ {((ζ(2n+1)−1)/(n+1))}=−γ+ln(2)✓ ..m.n.1970..
…nicecalculus…provethat:∑∞n=1{ζ(2n+1)−1n+1}=−γ+ln(2)✓..m.n.1970..
Answered by mindispower last updated on 17/Nov/20
ζ(s)−1=∫_0 ^∞ (x^(s−1) /(Γ(s)(e^x (e^x −1))))dx ⇒((ζ(2n+1)−1)/(n+1))=∫_0 ^∞ ((x^(2n) dx)/(Γ(2n+1)e^x (e^x −1))) Σ_(n≥1) ((ζ(2n+1)−1)/(n+1))=∫_0 ^∞ ((x^(2n) dx)/((n+1)Γ(2n+1)e^x (e^x −1)))..I Σ(x^(2n) /(2n!(n+1)))=(1/x^2 )Σ_(n≥1) (x^(2n+2) /(2n!(n+1)))=g(x) Σ_(n≥1) (x^(2n+2) /(2n!(n+1)))=f(x)⇒f′(x)=xΣ_(n≥1) ((2x^(2n) )/((2n!))) =2x(ch(x)−1) =2xsh(x)−2ch(x)−x^2 +2 f(x)=2xsh(x)−x^2 −2ch(x)+2 g(x)=((2sh(x))/(x ))−1−((2ch(x))/x^2 )+(2/(x^2 )) ∫_0 ^∞ ((2xsh(x)−x^2 −2ch(x)+2)/(x^2 e^x (e^x −1)))dx=S =∫_0 ^∞ ((xe^x −xe^(−x) −x^2 −e^x −e^(−x) +2)/(x^2 e^x (e^x −1)))dx =∫_0 ^∞ ((xe^(−x) −xe^(−3x) −x^2 e^(−2x) −e^(−x) −e^(−3x) +2e^(−2x) )/(x^2 (1−e^(−x) )))dx..E Ψ(z)=∫_0 ^∞ (e^(−t) /t)−(e^(−zt) /(1−e^(−t) ))dt ∫_0 ^∞ (e^(−t) /t)−(e^(−t) /(1−e^(−t) ))dt=Ψ(1) =∫_0 ^∞ ((e^(−t) −te^(−t) −e^(−2t) )/(t(1−e^(−t) )))dt..nice E=∫_0 ^∞ ((te^(−t) −t^2 e^(−t) −te^(−2t) +t^2 e^(−t) +te^(−2t) −te^(−3t) −t^2 e^(−2t) −e^(−t) −e^(−3t) +2e^(−2t) )/(t^2 (1−e^(−t) )))dt =Ψ(1)+∫_0 ^∞ ((t^2 e^(−t) (1−e^(−t) )+te^(−2t) (1−e^(−t) )−e^(−t) (1+e^(−2t) −2e^(−t) ))/(t^2 (1−e^(−t) )))dt =Ψ(1)+∫_0 ^∞ ((t^2 e^(−t) +te^(−2t) −e^(−t) (1−e^(−t) ))/t^2 )dt=Ψ(1)+T T=∫_0 ^∞ e^(−t) dt+∫_0 ^∞ ((te^(−2t) −e^(−t) +e^(−2t) )/t^2 )e^(−st) dt=1+w(s) T=w(0)+1 w′(s)=∫_0 ^∞ −e^(−(2+s)t) dt+∫_0 ^∞ ((e^(−t(1+s)) −e^(−t(2+s)) )/t) dt w′(s)=−(1/(2+s))−∫_0 ^∞ ∫_(2+s) ^(1+s) e^(−zt) dzdt w′(s)=−(1/(2+s))−∫_(2+s) ^(1+s) ∫_0 ^∞ e^(−zt) dtdz =−(1/(2+s))+ln(((2+s)/(1+s))) w(s)=−ln(2+s)+(s+2)ln(s+2)−(1+s)ln(1+s)+c w(s)=(s+1)ln(((s+2)/(s+1)))+c (s+1)ln(1+(1/(1+s)))+c lim_(s→∞) w(s)=0⇒lim_(s→∞) (s+1)ln(1+(1/(1+s)))+c=0 ⇒c=−1 ⇒w(0)=ln(1+1)−1=ln(2)−1 T=1+w(0)=ln(2) S=Ψ(1)+T=−γ+ln(2)...
ζ(s)−1=∫0∞xs−1Γ(s)(ex(ex−1))dx⇒ζ(2n+1)−1n+1=∫0∞x2ndxΓ(2n+1)ex(ex−1)∑n⩾1ζ(2n+1)−1n+1=∫0∞x2ndx(n+1)Γ(2n+1)ex(ex−1)..IΣx2n2n!(n+1)=1x2∑n⩾1x2n+22n!(n+1)=g(x)∑n⩾1x2n+22n!(n+1)=f(x)⇒f′(x)=x∑n⩾12x2n(2n!)=2x(ch(x)−1)=2xsh(x)−2ch(x)−x2+2f(x)=2xsh(x)−x2−2ch(x)+2g(x)=2sh(x)x−1−2ch(x)x2+2x2∫0∞2xsh(x)−x2−2ch(x)+2x2ex(ex−1)dx=S=∫0∞xex−xe−x−x2−ex−e−x+2x2ex(ex−1)dx=∫0∞xe−x−xe−3x−x2e−2x−e−x−e−3x+2e−2xx2(1−e−x)dx..EΨ(z)=∫0∞e−tt−e−zt1−e−tdt∫0∞e−tt−e−t1−e−tdt=Ψ(1)=∫0∞e−t−te−t−e−2tt(1−e−t)dt..niceE=∫0∞te−t−t2e−t−te−2t+t2e−t+te−2t−te−3t−t2e−2t−e−t−e−3t+2e−2tt2(1−e−t)dt=Ψ(1)+∫0∞t2e−t(1−e−t)+te−2t(1−e−t)−e−t(1+e−2t−2e−t)t2(1−e−t)dt=Ψ(1)+∫0∞t2e−t+te−2t−e−t(1−e−t)t2dt=Ψ(1)+TT=∫0∞e−tdt+∫0∞te−2t−e−t+e−2tt2e−stdt=1+w(s)T=w(0)+1w′(s)=∫0∞−e−(2+s)tdt+∫0∞e−t(1+s)−e−t(2+s)tdtw′(s)=−12+s−∫0∞∫2+s1+se−ztdzdtw′(s)=−12+s−∫2+s1+s∫0∞e−ztdtdz=−12+s+ln(2+s1+s)w(s)=−ln(2+s)+(s+2)ln(s+2)−(1+s)ln(1+s)+cw(s)=(s+1)ln(s+2s+1)+c(s+1)ln(1+11+s)+clims→∞w(s)=0⇒lims→∞(s+1)ln(1+11+s)+c=0⇒c=−1⇒w(0)=ln(1+1)−1=ln(2)−1T=1+w(0)=ln(2)S=Ψ(1)+T=−γ+ln(2)…
Commented by mnjuly1970 last updated on 18/Nov/20
bravo bravo sir mindspower . extraordinary my friend.(good)^∞
bravobravosirmindspower.extraordinarymyfriend.(good)∞

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