"Feel the bern" integral

Calculus Level 5

0 1 0 1 0 1 0 1 0 1 0 1 d b d e 1 d r d n d i d e 2 1 b e 1 r n i e 2 \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \dfrac{db \, de_{1} \, dr \, dn \, di \, de_{2} }{1- be_{1}rnie_{2} }

If the above thing can be expressed in the form a π b c \dfrac{a \pi^{b}}{c} , with a , c a, c as coprime positive integers, find a + b + c a+b+c .

Details and assumptions:

There's no e 2.71828... e \approx 2.71828... or i = 1 i = \sqrt{-1} . The b , e 1 , r , n , i , e 2 b, e_{1}, r, n, i, e_{2} are all variables.


The answer is 952.

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1 solution

Otto Bretscher
Apr 29, 2016

Viva Bernie! Long live the Revolution!

This integral involving n n variables is ζ ( n ) \zeta(n) ; see formula 10 here . In Bernie's case we have ζ ( 6 ) = π 6 945 \zeta(6)=\frac{\pi^6}{945} , so that the answer is 1 + 6 + 945 = 952 1+6+945=\boxed{952} .

We can derive this result by expanding the integrand into a geometric series: 1 1 x y = 1 + x y + x 2 y 2 + . . . \frac{1}{1-xy}=1+xy+x^2y^2+... in the case of n = 2 n=2 , and then integrating term by term: 1 + 1 2 2 + 1 3 2 + . . . = ζ ( 2 ) 1+\frac{1}{2^2}+\frac{1}{3^2}+...=\zeta(2) . For larger n n it's "more of the same."

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