s-s-0-x-s-1-e-x-1-dx-Prove-that-And-prove-1-2-3-4-5-6-7-1-12-

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

Γ (s) ζ (s) = \int_{0}^{\infty} \frac{x^{s - 1}}{e^{x} + 1} d x (P r o v e t h a t)

A n d p r o v e 1 + 2 + 3 + 4 + 5 + 6 + 7 + \dots . \infty = - \frac{1}{12}

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

d o y o u l e a r n t h e s e t h i n g s i n y o u r

h i g h s c h o o l r e a l l y ?

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

The second question

1 + 2 + 3 + 4 + \dots = - \frac{1}{12} is proved using

analytical comtinuity . Search forum

for analytical continuity to get

same question answered earlier .

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

N o s i r . C u r i o s i t y

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

the question (1) is done see the platform

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

\int_{0}^{\infty} \frac{x^{s - 1}}{e^{x} + 1} dx = \int_{0}^{\infty} \frac{x^{s - 1} e^{- x}}{1 + e^{- x}} dx = \int_{0}^{\infty} x^{s - 1} e^{- x} (\sum_{n = 0}^{\infty} {(- 1)}^{n} e^{- nx}) dx

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} x^{s - 1} e^{- (n + 1) x} dx =_{(n + 1) x = t} \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\infty} {(\frac{t}{n + 1})}^{s - 1} e^{- t} \frac{dt}{n + 1}

= \sum_{n = 0}^{\infty} {(- 1)}^{n} \times \frac{1}{{(n + 1)}^{s}} \int_{0}^{\infty} t^{s - 1} e^{- t} dt = Γ (s) δ (s) with

δ (s) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(n + 1)}^{s}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{s}}

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

1 + 2 + 3 + \dots . = - \frac{1}{12} is a no sense because 1 + 2 + 3 + \dots . > 0 and - \frac{1}{12} < 0

from where come be equality \dots