MassIntegration

Calculus Level 5

∫ 0 1 ∫ 0 1 ∫ 0 1 ⋯ ∫ 0 1 d x 1 d x 2 d x 3 ⋯ d x n 1 + x 1 x 2 x 3 ⋯ x n = f ( n ) ζ ( n ) \large \displaystyle \int_0^1\int_0^1\int_0^1 \cdots \int_0^1\dfrac{dx_1\; dx_2\; dx_3\; \cdots\; dx_n}{1+x_1\, x_2\, x_3\, \cdots\, x_n}=f(n)\, \zeta(n) f ( 2 ) + f ( 3 ) + f ( 4 ) + ⋯ + f ( 2020 ) = A + 2 B f(2)+f(3)+f(4)+\cdots+f(2020)=\text{A}+2^{\text{B}}

If A \text{A} and B \text{B} are integers, Find |A| + |B| . \text{|A|}+\text{|B|}.

Inspiration


The answer is 4037.

Digvijay Singh by Digvijay Singh , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

First Last
Feb 8, 2018

∫ 0 1 ∫ 0 1 . . . ∫ 0 1 d x 1 d x 2 . . . d x n 1 + x 1 x 2 . . . x n = ∫ 0 1 ∫ 0 1 . . . ∫ 0 1 ∑ k = 0 ∞ ( − x 1 x 2 . . . x n ) k d x 1 d x 2 . . . d x n = \displaystyle\int_0^1\int_0^1...\int_0^1\frac{dx_1dx_2...dx_n}{1+x_1x_2...x_n}=\int_0^1\int_0^1...\int_0^1\sum_{k=0}^\infty(-x_1x_2...x_n)^kdx_1dx_2...dx_n=

∑ k = 0 ∞ ∫ 0 1 ∫ 0 1 . . . ∫ 0 1 ( − x 1 x 2 . . . x n ) k d x 1 d x 2 . . . d x n = ∑ k = 0 ∞ ( − 1 ) k ( k + 1 ) n Now note this is the Dirichlet Eta Function \displaystyle\sum_{k=0}^\infty\int_0^1\int_0^1...\int_0^1(-x_1x_2...x_n)^kdx_1dx_2...dx_n=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^n}\quad\text{Now note this is the Dirichlet Eta Function}

∑ k = 0 ∞ ( − 1 ) k ( k + 1 ) n = η ( n ) = ( 1 − 2 n − 1 ) ζ ( n ) And so f ( n ) = ( 1 − 2 n − 1 ) \displaystyle\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)^n} = \eta(n)=(1-2^{n-1})\zeta(n)\quad\text{And so }f(n)=(1-2^{n-1})

∑ n = 2 2020 1 − 2 n − 1 = 2019 − 1 2 − 1 4 − . . . − 1 2 2019 = 2018 + 1 2 2019 \displaystyle\sum_{n=2}^{2020}1-2^{n-1}=2019-\frac1{2}-\frac1{4}-...-\frac1{2^{2019}} = 2018 + \frac{1}{ 2^{2019}}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Why is η ( n ) = ( 1 − 2 n − 1 ) ζ ( n ) \eta(n)=(1-2^{n-1})\zeta(n) ?

Digvijay Singh - 3 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...