Proving $\int_{1}^{\infty}\int_{0}^{1}\int_{0}^{1}\frac{dz\,dy\,dx}{x(x+y)(x+y+z)} = \frac{5\,\zeta(3)}{24}$

Question

I'm studying advanced analytic number theory and the following identity surprised me the most:

$$\int_{1}^{\infty}\int_{0}^{1}\int_{0}^{1}\dfrac{dz\,dy\,dx}{x(x+y)(x+y+z)} = \dfrac{5\,\zeta(3)}{24}$$

My idea is to use triple integral representation of $\zeta(3)$, that is

$$\zeta(3) = \int_{0}^{1}\int_{x}^{1}\int_{y}^{1}\dfrac{dx\,dy\,dz}{(1-x)\,yz}$$

however it appears to be a rather tricky problem to relate the triple integral of $\zeta(3)$.

Integrating with respect to $x$ we get

$$\int_{1}^{\infty}\int_{0}^{1}\int_{0}^{1}\dfrac{dz\,dy\,dx}{x(x+y)(x+y+z)} =\int_{0}^{1}\int_{0}^{1}\left(\dfrac{\log(y+1)}{yz} - \dfrac{\log(y+z + 1)}{z(y+z)}\right)dy\,dz$$

I'm unable to make any further progress and I think integrating this way doesn't get us anywhere so I'd not recommend anyone to work on this problem as I did (by integrating as above).

Out of curiosity: Can we prove the required using contour integration or residue theorem?

I'm in search of a neat detailed answer with proper references for tools used directly.

Thanks in advance.

Answer 1

Not a complete answer, but I show how to reduce the form, derived by @\Zacky in his comment, to a single integral, which I'm pretty sure is solved somewhere on this site.

We start with the following:

$$I=\int_0^1\int_1^{x+1}\int_1^{x+1} \frac{dudvdx}{xuv(u+v-1)}=$$

$$=\int_0^1\int_0^x\int_0^x \frac{dudvdx}{x(u+1)(v+1)(u+v+1)}=$$

$$=2\int_0^1\int_0^x\int_0^v \frac{dudvdx}{x(u+1)(v+1)(u+v+1)}$$

$$=2\int_0^1 \frac{dx}{x}\int_0^x \frac{dv}{v(v+1)}\int_0^v \left( \frac{du}{u+1}-\frac{du}{u+v+1} \right)=$$

$$=2\int_0^1 \frac{dx}{x}\int_0^x \frac{dv}{v(v+1)}\int_0^v \left( \frac{du}{u+1}-\frac{du}{u+v+1} \right)=$$

$$=2\int_0^1 \frac{dx}{x}\int_0^x \frac{dv}{v(v+1)} \left[ 2\log(v+1)-\log(2v+1) \right]=$$

---

Let's use integration by parts. Then we have:

$$=2 \log (x) \int_0^x \frac{dv}{v(v+1)} \left[ 2\log(v+1)-\log(2v+1) \right] \bigg|_0^1-$$

$$-2 \int_0^1 \log(x) \left[ 2\log(1+x)-\log(1+2x) \right] \frac{dx}{x(1+x)}$$

It's pretty easy to prove that the first term is zero, so we have:

$$I=-2 \int_0^1 \log(x) \left[ 2\log(1+x)-\log(1+2x) \right] \frac{dx}{x(1+x)}$$

Now we need to prove that:

$$\color{blue}{\int_0^1 \log(x) \left[ \log(1+2x)-2\log(1+x) \right] \frac{dx}{x(1+x)}=\frac{\zeta(3)}{12}}$$

Wolfram Alpha confirms this numerically, and again, I'm pretty sure every part of this integral is already solved on this site. We can try series expansion either on the logarithms or on the $1/(1+x)$ part.

Answer 2

\begin{align}J&=\int_{0}^{1}\int_{0}^{1}\left(\dfrac{\log(y+1)}{yz} - \dfrac{\log(y+z + 1)}{z(y+z)}\right)dy\,dz\\ &=\int_0^1 \frac{1}{z}\Big(\text{Li}_2(-1-z)-\text{Li}_2(-1)-\text{Li}_2(-z)\Big)dz\\ &=\int_0^1 \frac{\text{Li}_2(-z)}{z}dz+\int_0^1 \frac{\text{Li}_2(-1-z)-\text{Li}_2(-1)}{z}dz\\ &=\Big[\text{Li}_3(-z)\Big]_0^1+\int_0^1 \frac{\text{Li}_2(-1-z)-\text{Li}_2(-1)}{z}dz\\ &=\frac{3\zeta(3)}{4}+\int_0^1 \frac{\text{Li}_2(-1-z)-\text{Li}_2(-1)}{z}dz\\ &\overset{IBP}=\frac{3\zeta(3)}{4}+\Big[\ln z\left(\text{Li}_2(-1-z)-\text{Li}_2(-1)\right)\Big]_0^1+\int_0^1 \frac{\ln x\ln(2+x)}{1+x}dx\\ &=\frac{3\zeta(3)}{4}+\int_0^1 \frac{\ln x\ln(2+x)}{1+x}dx\\ \end{align}

See Prove: $\int_0^1 \int_0^1 \frac{\ln{\left(2+yx\right)}}{1+yx} \; \mathrm{d}y\; \mathrm{d}x = \frac{13}{24} \zeta(3)$ for the computation of the last integral.

Actually, $\displaystyle \int_0^1 \frac{\text{Li}_2(-1)-\text{Li}_2(-1-z)}{z}dz$ is computed.

Answer 3

\begin{align} I&=\int_1^\infty\int_0^1\int_0^1\frac{dzdydx}{x(x+y)(x+y+z)},\quad x\to 1/x\\ &=\int_0^1\int_0^1\int_0^1\frac{xdzdydx}{(1+xy)(1+xy+xz)}\\ &=\int_0^1\int_0^1\left[\int_0^1\frac{xdz}{1+xy+xz}\right]\frac{dydx}{1+xy}\\ &=\int_0^1\int_0^1\frac{\ln\left(\frac{1+x+xy}{1+xy}\right)}{1+xy}dydx,\quad y=t/x\\ &=\int_0^1\int_0^x\frac{\ln\left(\frac{1+x+t}{1+t}\right)}{x(1+t)}dtdx\\ &=\int_0^1\left[\int_t^1\frac{\ln\left(\frac{1+x+t}{1+t}\right)}{x}dx\right]\frac{dt}{1+t}\\ &=\int_0^1\left[\text{Li}_2\left(\frac{-t}{1+t}\right)-\text{Li}_2\left(\frac{-1}{1+t}\right)\right]\frac{dt}{1+t},\quad IBP\\ &=\int_0^1\ln(1+t)\left[\frac{t\ln\left(\frac{2+t}{1+t}\right)+\ln\left(\frac{1+2t}{1+t}\right)}{t(1+t)}\right],\quad x=1/(1+t)\\ &=-\int_{1/2}^1\ln(x)\left[\frac{\ln(1+x)}{x}+\frac{\ln(2-x)}{1-x}\right]dx. \end{align}

These two integrals are manageable.