I have no idea how an iterated integral is related to an infinite series. I guess I should change $\frac 1 {1+x^4y^4z^4}$ into an infinite geometric series $1+r+r^2+\cdots$ where $r=-x^4y^4z^4$, and then interchange the integral sign and the summation sign. But it seems to be useless to make up the term $\frac {(-1)^n} {(4n+1)^3}$. Hope someone could help me with this one. Thanks!
2026-03-25 14:25:12.1774448712
Prove that $\int_0^1\int_0^1\int_0^1\frac 1{1+x^4y^4z^4} \, dx \, dy \, dz=\sum_{n=1}^{\infty}\frac {(-1)^n}{(4n+1)^3}$
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Using this method one gets the integral as $$\sum_{n=0}^\infty(-1)^n\int_0^1\int_0^1\int_0^1x^{4n}y^{4n}z^{4n} \,dx\,dy\,dz.$$ This triple integral is the cube of $\int_0^1 x^{4n}\,dx$ which is really elementary.