Interesting triple integral Some time ago I stumbled on an alleged identity
$$\int\limits_0^\infty \frac{dx}{x} \int\limits_0^x \frac{dy}{y}
\int\limits_0^y \frac{dz}{z} [\sin{x}+\sin{(x-y)}-\sin{(x-z)}-\sin{(x-y+z)}]=
-\frac{\pi^3}{12}.$$
How this identity can be proved? The context under which this integral has emerged is described in Multiple Integral (American Mathematical Monthly problem 11621 and its generalization) .
 A: I consider the function
$$f(t):=\int_0^t\frac{dx}{x}\int_0^x\frac{dy}{y}\int_0^y\frac{dz}{z}\bigl\{
\sin x+\sin(x-y)-\sin(x-z)-\sin(x-y+z)\bigr\}.$$
It has an asymptotic expansion with main terms
$$f(t)= -\frac{\cos t}{2}\frac{\log^2t}{t}+O\Bigl(\frac{\log t}{t}\Bigr).$$
Therefore, the limit $\lim_{t\to+\infty}f(t)=0\ne-\pi^3/12$. 
My proofs are too long to post them here, and I will only briefly explain the main points:
The function $f(t)$ extends to an entire function with power series
$$f(t)=\sum_{n=1}^\infty(-1)^n
\Bigl(\sum_{k=1}^{2n+1}\frac{H_{k-1}}{k}\Bigr)\frac{t^{2n+1}}{(2n+1)!(2n+1)}$$
where $H_k$ are the harmonic numbers.
The power series can be transformed into an alternative integral representation
$$f(t)=\int_0^tdu\int_0^1\int_0^1\Bigl(\frac{x\sin u}{u(1-x)(1-x y)}-\frac{\sin(ux)}{u(1-x)(1-y)}+\frac{\sin(uxy)}{u(1-y)(1-xy)}\Bigr)\,dx\,dy.$$
This can be used to write $f(t)$ as a Fourier transform
$$f(t)=\int_0^1\Bigl(\frac12\log^2(1-x)+\sum_{n=1}^\infty\frac{(1-x)^n-1}{n^2}\Bigr)
\frac{\sin(tx)}{x}\,dx.$$
Since the integrand is in $L^1(0,1)$, the
Riemann-Lebesgue Lemma implies that $\lim_{t\to\infty}f(t)=0$, and   the asymptotic expansion is obtained from this representation by standard methods.
The formulas above and the asymptotic expansion allow us to compute the 
function $f(t)$ easily.  
The details can be found in http://arxiv.org/abs/1505.00440
