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

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.

• As modified the claim is not true. Coincidence up to 7 decimal places. Apr 23, 2021 at 14:19
• @JamesArathoon What do you mean by it's not true? It is Apr 23, 2021 at 14:20
• Sorry my numerical integration was in error. Apr 23, 2021 at 14:34
• umm wait, which series expansion are you talking about? Apr 23, 2021 at 14:37
• @YuriyS coz I have only tried integrating the whole thing over and over with respect to $x,y,z$ and then tried to simplify it but it doesn't get anywhere. Thanks. Apr 23, 2021 at 14:38

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.

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

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

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

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

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

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

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.

\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.

• Thanks, I'm quite wondering if we can do it without using polylogarithms though? Apr 24, 2021 at 1:26

\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.

• Can you please complete the prof or at least give reference to the evaluations of last 2 integrals? Thank you. Apr 10 at 5:21