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There's a nice problem that's found in textbooks and on MSE to show that $$\zeta(2) = \frac{\pi^2}{6}$$ using just a tictch of multivariable calculus and a couple trig substitutions. The main idea is to find the below integral,

$$I = \int_0^1 \int_0^1 \frac{dx\, dy}{1-xy},$$ in two different ways. The first is to Taylor expand the integrand and exchange sum and integral signs to net $$ I = \sum_{n=0}^\infty \frac{1}{(n+1)^2}.$$

The second is to transform to rotated coordinates, integrate over one of them with a trig substitution, and then integrate over the other with another such substitution. I recommend checking out the explicit steps here.

Though I'm aware of many derivations for the Basel problem (and the generating function, Poisson summation, contour integral derivations for the other $\zeta(2n)$), many from MSE, I enjoy the style of proof above.

My question is: Can we show that $$\zeta(4) = \frac{\pi^4}{90}$$ using the integral $$J = \int_0^1 \int_0^1 \int_0^1 \int_0^1 \frac{dw\, dx\, dy\, dz}{1-wxyz} = \sum_{n=0}^\infty \frac{1}{(n+1)^4}$$ through a handful of change of variables and direct integration?

I'm most interested in linear transformations followed by trig subs in the style of the above proof, but that's not a necessity.

user196574
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  • Additionally, why does it work with 2 and 4 variables but not 3 (the standard answer is that $\sum_n \frac1{(x+n)^2}$ is periodic and it is a quotient of trigonometric function whereas $\sum_{n\ge 1} \frac1{(x+n)^2}$ involves $\Gamma'/\Gamma$) – reuns Dec 22 '19 at 04:18

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