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Question Number 102115 by Dwaipayan Shikari last updated on 06/Jul/20

Γ(s)ζ(s)=∫_0 ^∞ (x^(s−1) /(e^x +1))dx (Prove that) And prove 1+2+3+4+5+6+7+....∞=−(1/(12))

$$\Gamma\left({s}\right)\zeta\left({s}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}−\mathrm{1}} }{{e}^{{x}} +\mathrm{1}}{dx}\:\:\left({Prove}\:{that}\right) \\ $$$${And}\:{prove}\:\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+\mathrm{5}+\mathrm{6}+\mathrm{7}+....\infty=−\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$ \\ $$

Commented by mr W last updated on 06/Jul/20

do you learn these things in your high school really?

$${do}\:{you}\:{learn}\:{these}\:{things}\:{in}\:{your} \\ $$$${high}\:{school}\:{really}? \\ $$

Commented by prakash jain last updated on 07/Jul/20

The second question 1+2+3+4+...=−(1/(12)) is proved using analytical comtinuity. Search forum for analytical continuity to get same question answered earlier.

$$\mathrm{The}\:\mathrm{second}\:\mathrm{question}\: \\ $$$$\mathrm{1}+\mathrm{2}+\mathrm{3}+\mathrm{4}+...=−\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{is}\:\mathrm{proved}\:\mathrm{using} \\ $$$$\mathrm{analytical}\:\mathrm{comtinuity}.\:\mathrm{Search}\:\mathrm{forum} \\ $$$$\mathrm{for}\:\mathrm{analytical}\:\mathrm{continuity}\:\mathrm{to}\:\mathrm{get} \\ $$$$\mathrm{same}\:\mathrm{question}\:\mathrm{answered}\:\mathrm{earlier}. \\ $$

Commented by Dwaipayan Shikari last updated on 07/Jul/20

No sir. Curiosity

$${No}\:{sir}.\:{Curiosity} \\ $$

Answered by mathmax by abdo last updated on 06/Jul/20

the question (1)is done see the platform

$$\mathrm{the}\:\:\mathrm{question}\:\left(\mathrm{1}\right)\mathrm{is}\:\mathrm{done}\:\mathrm{see}\:\mathrm{the}\:\mathrm{platform} \\ $$

Answered by mathmax by abdo last updated on 08/Jul/20

∫_0 ^∞ (x^(s−1) /(e^x +1))dx =∫_0 ^∞ ((x^(s−1) e^(−x) )/(1+e^(−x) ))dx =∫_0 ^∞ x^(s−1) e^(−x) (Σ_(n=0) ^∞ (−1)^n e^(−nx) )dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^∞ x^(s−1) e^(−(n+1)x) dx =_((n+1)x =t) Σ_(n=0) ^(∞ ) (−1)^n ∫_0 ^∞ ((t/(n+1)))^(s−1) e^(−t ) (dt/(n+1)) =Σ_(n=0) ^∞ (−1)^n ×(1/((n+1)^s )) ∫_0 ^∞ t^(s−1) e^(−t) dt =Γ(s) δ(s) with δ(s) =Σ_(n=0) ^∞ (((−1)^n )/((n+1)^s )) =Σ_(n=1) ^∞ (((−1)^(n−1) )/n^s )

$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{s}−\mathrm{1}} }{\mathrm{e}^{\mathrm{x}} +\mathrm{1}}\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{s}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{x}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{x}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \mathrm{x}^{\mathrm{s}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{x}} \left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{nx}} \right)\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{s}−\mathrm{1}} \mathrm{e}^{−\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}} \:\mathrm{dx}\:=_{\left(\mathrm{n}+\mathrm{1}\right)\mathrm{x}\:=\mathrm{t}} \:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{t}}{\mathrm{n}+\mathrm{1}}\right)^{\mathrm{s}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}\:} \frac{\mathrm{dt}}{\mathrm{n}+\mathrm{1}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:×\frac{\mathrm{1}}{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{s}} }\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}^{\mathrm{s}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\Gamma\left(\mathrm{s}\right)\:\delta\left(\mathrm{s}\right)\:\mathrm{with} \\ $$$$\delta\left(\mathrm{s}\right)\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{s}} }\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}−\mathrm{1}} }{\mathrm{n}^{\mathrm{s}} } \\ $$

Answered by mathmax by abdo last updated on 08/Jul/20

1+2+3+....=−(1/(12)) is a no sense because 1+2+3 +....>0 and −(1/(12))<0 from where come be equality...

$$\mathrm{1}+\mathrm{2}+\mathrm{3}+....=−\frac{\mathrm{1}}{\mathrm{12}}\:\mathrm{is}\:\mathrm{a}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{because}\:\mathrm{1}+\mathrm{2}+\mathrm{3}\:+....>\mathrm{0}\:\mathrm{and}\:−\frac{\mathrm{1}}{\mathrm{12}}<\mathrm{0} \\ $$$$\mathrm{from}\:\mathrm{where}\:\mathrm{come}\:\mathrm{be}\:\mathrm{equality}... \\ $$

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