A calculus problem by John M.

When we deal with multi-variable integration, it may seem scary at first. However, it sometimes can be easily solved by evaluating the inner most integral and then move on to evaluate the next layer of integral. It is just like when you throw a stone into the water, the wavefronts move radially outwards from the source. Here is the evaluation of the integral above. $\begin{aligned} \int^1_0\int^1_0\int^1_0\frac{1}{1+xyz}\ dx\ dy\ dz&=&\int^1_0\int^1_0\left[\frac{\ln\left|1+xyz\right|}{yz}\right]^1_0\ dy\ dz\\ &=&\int^1_0\int^1_0\frac{\ln(1+yz)}{yz}\ dy\ dz\\ &=&\int^1_0\int^1_0\frac{yz-\frac{1}{2}(yz)^2+\frac{1}{3}(yz)^3-\frac{1}{4}(yz)^4+\cdots}{yz}\ dy\ dz\\ &&\color{#3D99F6}\textrm{(By Maclaurin series expansion of }\ln(1+x))\\ &=&\int^1_0\int^1_01-\frac{1}{2}yz+\frac{1}{3}(yz)^2-\frac{1}{4}(yz)^3+\cdots\ dy\ dz\\ &=&\int^1_0\left[y-\frac{y^2z}{2^2}+\frac{y^3z^2}{3^2}-\frac{y^4z^3}{4^2}+\cdots\right]^1_0\ dz\\ &=&\int^1_01-\frac{z}{2^2}+\frac{z^2}{3^2}-\frac{z^3}{4^2}+\cdots\ dz\\ &=&\left[z-\frac{z^2}{2^3}+\frac{z^3}{3^3}-\frac{z^4}{4^3}+\cdots\right]^1_0\\ &=&1-\frac{1}{2^3}+\frac{1}{3^3}-\frac{1}{4^3}+\cdots\\ &=&\left(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\right)-2\left(\frac{1}{2^3}+\frac{1}{4^3}+\frac{1}{6^3}+\cdots\right)\\ &=&\left(1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots\right)-\frac{1}{4}\left(1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots\right)\\ &=&\zeta(3)-\frac{1}{4}\zeta(3)\ \ \ \ \color{#3D99F6}\left(\textrm{Take note that } 1+\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+\cdots=\sum^{\infty}_{k=1}\frac{1}{k^3}=\zeta(3)\right)\\ &=&\frac{3}{4}\zeta(3)\ \ \blacksquare \end{aligned}$

First of all, obviously

$\frac{1}{1+xyz} = \sum_{n=0}^{\infty} (-1)^{n} x^{n} y^{n} z^{n}$

And we can do that because all of the integrals are bounded in the the interval $\left [ 0, 1 \right ] \supset \left [ -1, 1 \right ]$ . The integral is now easy and it is equal to

$S = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(n+1)^{3}}$

Which looks pretty much like $\zeta(3)$ . Now define (for fun)

$f(a) = S + a \zeta(3) = (a+1)\left [ \sum_{n=1}^{\infty} \frac{1}{(2n+1)^{3}} \right ] + (a-1)\left [ \sum_{n=1}^{\infty} \frac{1}{(2n)^{3}} \right ]$

Notice

$\sum_{n=1}^{\infty} \frac{1}{(2n)^{3}} = \frac{1}{8} \zeta(3)$

$\sum_{n=1}^{\infty} \frac{1}{(2n+1)^{3}} = \zeta(3) - \frac{1}{8} \zeta(3) = \frac{7}{8} \zeta(3)$

Therefore

$f(a) = \frac{\zeta(3)}{8} (8a + 6)$

And

$f(0) = S = \frac{3}{4} \zeta(3)$