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I came on the following multiple integral while renormalizing elliptic multiple zeta values: $$\int_0^1\cdots \int_0^1\int_1^\infty {{1}\over{t_n(t_{n-1}+t_n)\cdots (t_1+\cdots+t_n)}} dt_n\cdots dt_1.$$ Only the variable $t_n$ goes from $1$ to $\infty$, the others all go from $0$ to $1$. Numerically, I seem to be getting a rational multiple of $\pi^n$. I would like to prove this. Has anyone ever seen an integral like this before?

Edit: After reworking out why I thought it would be a power of $\pi$, I now have a different integral, which numerically really does seem to give a rational multiple of $\zeta(n)$ for each $n$ (though I have only been able to go up to n=4 numerically). I want to integrate $1/(z_1\cdots z_n)$ over the part of the simplex $0\le z_n\le \cdots \le z_1\le 1$ in which $|z_i -z_{i+1}|>\varepsilon$, then let $\varepsilon\rightarrow 0$. The integral is $$\int_{n\epsilon}^{1-\varepsilon}\int_{(n-1)\varepsilon}^{z_1-\epsilon}\cdots \int_{\varepsilon}^{z_{n-1}-\varepsilon} {{1}\over{z_1\cdots z_n}} dz_n\cdots dz_1.$$ It actually diverges when $\varepsilon\rightarrow 0$, but it can be regularised like ordinary multizeta values by computing the integral as a power series in $ln(\varepsilon)$ and $\varepsilon$ and then taking its constant term. It's related to the previous integral by the variable change $z_1=\varepsilon(t_1+\cdots+t_n),\ldots,z_n=\varepsilon t_n$, but the new bounds of the integral form an $(n+1)$-angled polyhedron, not a cube.

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    $\begingroup$ For $n=3$ the integral seems to be $5\zeta(3)/24$. $\endgroup$ – Peter Mueller Apr 17 '17 at 20:02
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    $\begingroup$ For $n=4$ the same method produces (after about 3 minutes) $0.081724116967726819510127555243991754831$, which doesn't seem to be a simple multiple of $\pi^4$ --- but I don't know how many of those digits are trustworthy coming from an improper triple integral . . . $\endgroup$ – Noam D. Elkies Apr 18 '17 at 4:52
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    $\begingroup$ Some say the integral will end in pi, some say it's not so nice. From what I've tasted of periods, I hold with those who favor pi. But only if n is twice An integer or else I think, The integral won't be so nice, but still pretty great, and zetas sould suffice. $\endgroup$ – Marty Apr 18 '17 at 6:24
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    $\begingroup$ @FedorPetrov No reason to ask Noam for a computation we can easily do -- the first few are 1/12, 4/49, 17/208, 55/673, 237/2900, 1240/15173, 2717/33246 . None of these are particularly accurate given their denominator. $\endgroup$ – David E Speyer Apr 18 '17 at 16:29
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    $\begingroup$ @SylvainJULIEN I'm pretty sure that this is not the case. I too computed an approximation (using quad from mpmath), which matches Noam Elkies' calculation on 18 decimal digits, while your number matches only 7 digits. $\endgroup$ – Peter Mueller Apr 18 '17 at 23:12
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Let's record what is possible, so far. The case $n=2$, from the comments: $$\int_0^1\frac{dy}{x(x+y)}=\frac1{x^2}\int_0^1\frac{dy}{1+\frac{y}x}=\int_0^1\sum_{k\geq0}\frac{(-1)^ky^k}{x^{k+2}}dy=\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac1{x^{k+2}}.$$ Hence, $$\int_1^{\infty}\sum_{k\geq0}\frac{(-1)^k}{k+1}\frac{dx}{x^{k+2}}=\sum_{k\geq0}\frac{(-1)^k}{(k+1)^2}=\frac{\zeta(2)}2.$$

The case $n=3$: $$\begin{aligned}\int_0^1\frac{dz}{x(x+y)(x+y+z)}&=\frac1{x(x+y)^2}\int_0^1\frac{dz}{1+\frac{z}{x+y}}\\&=\frac1x\sum_{k\geq0}\int_0^1\frac{(-1)^kz^kdz}{(x+y)^{k+2}}\\&=\frac1x\sum_k\frac{(-1)^k}{(k+1)(x+y)^{k+2}}.\end{aligned}$$ Hence, $$\frac1x\int_0^1\sum_k\frac{(-1)^kdy}{(k+1)(x+y)^{k+2}}=\frac1x\sum_k\frac{(-1)^k}{(k+1)^2}\left[\frac1{x^{k+1}}-\frac1{(x+1)^{k+1}}\right].$$ Now, integrate with respect to $x$ (standard): $$\int_1^{\infty}\frac{dx}{x^{k+2}}=\frac1{k+1} \qquad \text{and} \qquad \int_1^{\infty}\frac{dx}{x(x+1)^{k+1}}=\sum_{j=k+1}^{\infty}\frac1{j\cdot 2^j}.$$ Therefore, we compute the two series: $$\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^3}=\frac34\zeta(3) \qquad \text{and} \qquad \sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{j=k}^{\infty}\frac1{j\cdot 2^j}=\frac{13}{24}\zeta(3).$$ We arrived at the valued predicted by Peter Mueller, $$\int_1^{\infty}\int_0^1\int_0^1\frac{dz\,dy\,dx}{x(x+y)(x+y+z)}=\frac5{24}\zeta(3).$$

Caveat. One may anticipate higher values of $n$ to scale up the challenge.

UPDATE. Regarding Fedor's question, one contention is a follows: the sum in question is a weight 3 polylog, so it is a rational combination of $\zeta(3), \zeta(2)\log 2$ and $\log^3(2)$. Since a large numerical agreement verifies equality with only $\frac{13}{24}\zeta(3)$, it must be the exact evaluation.

UPDATE. I like to address the request from GH from MO directed to Agno. Then, a direct answer to Fedor's question.

This time, we start integrating with respect to $x$: $$\int_1^{\infty}\frac{dx}{x(x+y)(x+y+z)} =\frac{y\,\log(1+y)+z\,\log(1+y)-y\,\log(1+y+z)}{yz(y+z)}.$$ Next, integrate in the variable $y$: $$\int_0^1\frac{y\,\log(1+y)+z\,\log(1+y)-y\,\log(1+y+z)}{yz(y+z)}\,dy =\frac{\text{Li}_2(z+2)-\text{Li}_2(z+1)-\text{Li}_2(2)}z;$$ where $\text{Li}_2(z)$ is the dilogarithm function $$\text{Li}_2(z)=\int_1^z\frac{\log t}{1-t}\,dt.$$ Finally, we integrate in the last variable $z$: $$\begin{align} \int_0^1\frac{\text{Li}_2(z+2)-\text{Li}_2(z+1)-\text{Li}_2(2)}z\,dz &=\int_0^1\left(\int_1^{z+2}\frac{\log t}{1-t}\,dt-\int_1^{z+1}\frac{\log t}{1-t}\,dt-\int_1^2\frac{\log t}{1-t}\,dt \right)\frac{dz}z \\ &=\int_0^1\left(\int_{z+1}^{z+2}\frac{\log t}{1-t}\,dt-\int_1^2\frac{\log t}{1-t}\,dt \right)\frac{dz}z \\ &=\int_0^1\left(\int_2^{z+2}\frac{\log t}{1-t}\,dt-\int_1^{z+1}\frac{\log t}{1-t}\,dt\right)\frac{dz}z \\ &=\int_2^3\frac{\log t}{1-t}\left(\int_{t-2}^1\frac{dz}z\right)dt-\int_1^2\frac{\log t}{1-t}\left(\int_{t-1}^1\frac{dz}z\right)dt \\ &=\int_2^3\frac{\log t\,\log(t-2)}{t-1}\,dt-\int_1^2\frac{\log t\,\log(t-1)}{t-1}\,dt \\ &=\int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt-\int_0^1\frac{\log(t+1)\,\log t}t\,dt \\ &=\int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt+\frac34\zeta(3). \end{align}$$ For the first integral in the last equality, write $\log(t+2)=\log 2+\log(1+\frac{t}2)$ and apply Taylor series: $$\begin{align} \int_0^1\frac{\log (t+2)\,\log t}{t+1}\,dt &=\log 2\int_0^1\frac{\log t}{t+1}\,dt+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\int_0^1\frac{t^n\log t}{t+1}\,dt \\ &=-\frac12\zeta(2)\,\log2+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\int_0^1\frac{t^n\log t}{t+1}\,dt \\ &=-\frac12\zeta(2)\,\log2+\sum_{n\geq1}\frac{(-1)^{n-1}}{2^nn}\left[\frac{(-1)^{n-1}}2\zeta(2)+(-1)^{n-1}\sum_{k=1}^n\frac{(-1)^k}{k^2}\right] \\ &=-\frac12\zeta(2)\,\log2+\frac12\zeta(2)\sum_{n\geq1}\frac1{2^nn}+\sum_{n\geq1}\frac1{2^nn}\sum_{k=1}^n\frac{(-1)^k}{k^2} \\ &=-\frac12\,\zeta(2)\,\log2+\frac12\,\zeta(2)\,\log2+\sum_{n\geq1}\frac1{2^nn}\sum_{k=1}^n\frac{(-1)^k}{k^2} \\ &=\sum_{k\geq1}\frac{(-1)^k}{k^2}\sum_{n=k}^{\infty}\frac1{2^nn}. \end{align}$$ The above derivations indicate we do not need Agno's $\log(e^x\pm1)$ integrals, instead we got Robert Z's $\log$-integral which sends us to his useful link evaluating as $\frac{13}{24}\zeta(3)$.

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    $\begingroup$ Why $\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{j=k}^{\infty}\frac1{j\cdot 2^j}=\frac{13}{24}\zeta(3)?$ $\endgroup$ – Fedor Petrov Apr 18 '17 at 11:18
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    $\begingroup$ After inserting Noam's high precision calculation from above, the inverse symbolic calculator answered "Wow, really found nothing" :( $\endgroup$ – Leila Schneps Apr 18 '17 at 13:15
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    $\begingroup$ I would also like to see the crucial step pointed out by Fedor Petrov. $\endgroup$ – GH from MO Apr 18 '17 at 16:46
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    $\begingroup$ For what it's worth. After some integration by parts and substitution of variables, I found that $\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2}\sum_{j=k}^{\infty}\frac1{j\cdot 2^j}$ could also be expressed as: $$\int_{ln(2)}^{\infty} (\ln(e^x+1)-x)\,\ln(e^x-1)\, dx$$ or even simpler, it would imply that: $$\frac{13}{24}\zeta(3)=-\int_{0}^{\ln(2)} \ln(e^x+1)\,\ln(e^x-1) \,dx$$ Note that numerically the following related integral also seems true: $$\int_{0}^{\infty} \left(\ln(e^x+1)-x\right)\ln(e^x-1) dx = \frac18\,\zeta(3)$$ $\endgroup$ – Agno Apr 25 '17 at 12:34
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    $\begingroup$ From Agno's comment, by letting $t=e^x-1$, we obtain $$-\int_{0}^{\ln(2)} \ln(e^x+1)\,\ln(e^x-1) \,dx=\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt$$ which is equal to $\frac{13}{24}\zeta(3)$ by math.stackexchange.com/questions/1344455/… $\endgroup$ – Robert Z Apr 25 '17 at 17:38
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You can easily reduce to a double integral and in a few seconds you obtain 0.0817241169677268267134627567961242656233303062217211895... (all digits correct, I checked at higher accuracy), but still cannot recognize it.

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    $\begingroup$ I take it this is for $n=4$. $\endgroup$ – Gerry Myerson Apr 24 '17 at 0:36
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The following is a somewhat simplified version of the proof by T. Amdeberhan, Agno, and Robert Z that $$\int_0^1 \int_0^1 \int_1^\infty \frac{dx\,dy\,dz}{x(x+y)(x+y+z)}=\frac{5}{24}\zeta(3).$$ We integrate with respect to $x$ first, then with respect to $y$: \begin{align*}\int_0^1\int_1^\infty \frac{dx\,dy}{x(x+y)(x+y+z)}&=\int_0^1\left(\frac{\log(1+y)}{yz}-\frac{\log(1+y+z)}{(y+z)z}\right)dy\\ &=\int_0^1\frac{\log(1+t)}{tz}\,dt-\int_z^{1+z}\frac{\log(1+t)}{tz}\,dt\\ &=\int_0^z\frac{\log(1+t)}{tz}\,dt-\int_1^{1+z}\frac{\log(1+t)}{tz}\,dt.\\ \end{align*} The integral of the right hand side with respect to $z$ equals, by Fubini, \begin{align*}&=\int_0^1\int_0^z\frac{\log(1+t)}{tz}\,dt\,dz-\int_0^1\int_1^{1+z}\frac{\log(1+t)}{tz}\,dt\,dz\\ &=\int_0^1\int_t^1\frac{\log(1+t)}{tz}\,dz\,dt-\int_1^2\int_{t-1}^1\frac{\log(1+t)}{tz}\,dz\,dt\\ &=-\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt+\int_1^2\frac{\log(t-1)\log(t+1)}{t}\,dt\\ &=-\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt+\int_0^1\frac{\log(t)\log(t+2)}{t+1}\,dt. \end{align*} On the right hand side, the second integral equals $-\frac{13}{24}\zeta(3)$ by this MSE post, while the first integral equals \begin{align*}\int_0^1\frac{\log(t)\log(t+1)}{t}\,dt &=\int_0^1\frac{\log(t)}{t}\sum_{n=1}^\infty\frac{(-1)^{n-1}t^n}{n}\,dt\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}\int_0^1 t^{n-1}\log(t)\,dt=\\ &=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}=-\frac{3}{4}\zeta(3). \end{align*} To summarize, $$\int_0^1 \int_0^1 \int_1^\infty \frac{dx\,dy\,dz}{x(x+y)(x+y+z)}=\frac{3}{4}\zeta(3)-\frac{13}{24}\zeta(3)=\frac{5}{24}\zeta(3).$$

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As suggested by მამუკა ჯიბლაძე, for n=2,3 there is indeed an interesting connection of $$I_n=\int\limits_0^1\cdots \int\limits_0^1 dt_1\cdots dt_{n-1}\int\limits_1^\infty dt_n \frac{1}{t_n(t_n+t_{n-1})\cdots(t_n+\cdots +t_1)}$$ with the Beukers-type integrals. Using the Feynman parametrization $$\frac{1}{A_1A_2\cdots A_n}=(n-1)!\int\limits_0^1\cdots \int\limits_0^1\frac{\delta\left(1-\sum\limits_{i=1}^n x_i\right)dx_1\cdots dx_n}{(x_1A_1+x_2A_2+\cdots +x_nA_n)^n},$$ we get after the parametrization and subsequent trivial integration over $t_n$: $$I_n=\frac{(n-1)!}{n-1}\int\limits_0^1\cdots \int\limits_0^1 \frac{dx_1\cdots dx_n dt_1\cdots dt_{n-1}\;\delta(1-x_1-\cdots-x_n)}{[1+t_{n-1}(x_{n-1}+\cdots x_1)+\cdots +t_1x_1]^{n-1}}.$$ In particular, for n=2 we get immediately $$I_2=\int\limits_0^1\int\limits_0^1 \frac{dx dt}{1+xt}=\frac{1}{2}\zeta(2),$$ and for n=3 we get $$I_3=\int\limits_0^1\cdots\int\limits_0^1 \frac{dx_1 dx_2 dx_3dt_1 dt_2\;\delta(1-x_1-x_2-x_3)}{[1+t_2(x_1+x_2)+t_1x_1]^2}=$$ $$\iint\limits_0^1\frac{dx_1dx_2dx_3\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(x_1+x_2)}\left [\ln{(1+x_1+x_2)}+\ln{(1+x_1)}-\ln{(1+2x_1+x_2)}\right]=$$ $$\iiint\limits_0^1\frac{dx_1dx_2dx_3\;\delta(1-\sum\limits_{i=1}^3 x_i)}{x_1(1-x_3)}\left [\ln{(2-x_3)}+\ln{(1+x_1)}-\ln{(2+x_1-x_3)}\right].$$ Making $x=x_1,y=1-x_3,z=x_2$ change of variables in the integral, and performing integration over $z$ thanks to the $\delta$-function, we get $$I_3=\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right]\theta(y-x)=$$ $$ \frac{1}{2}\iint\limits_0^1\frac{dx dy}{xy}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right], \tag{1}$$ where in the last step we used the symmetry of the integrand and $\theta(y-x)+\theta(x-y)=1$. This result can be represented in the form $$I_3=\frac{1}{2}\iiint\limits_0^1\frac{dx dy dz}{1+x+y+xyz}. \tag{2}$$ Note the resemblance with Beukers integrals $$ \iiint\limits_0^1\frac{dx dy dz}{1-z+xyz}=2\zeta(3),\;\; \iiint\limits_0^1\frac{dx dy dz}{1+xyz}=\frac{3}{4}\iiint\limits_0^1\frac{dx dy dz}{1-xyz}=\frac{3}{4}\zeta(3).$$ Analogously, for $I_4$ we get $$I_4=\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+ \ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right]\theta(z-x)\theta(y-z).$$ $\theta(z-x)\theta(y-z)$ term is a consequence of integration with the help of the $\delta$-function and reflects $x_2=z-x>0$ and $x_3=y-z>0$ conditions. After symmetrization, we finally get $$I_4=\frac{1}{6}\iiint\limits_0^1\frac{dx dy dz}{xyz}\left[\ln{(1+x)}+\ln{(1+y)}+ \ln{(1+z)}+\ln{(1+x+y+z)}-\right . $$ $$ \left . \ln{(1+x+y)}- \ln{(1+x+z)}-\ln{(1+y+z)}\right],$$ and the analogy with Beukers-type integrals is less obvious, because it doesn't seem this can be represented in the form analogous to (2).

I don't know whether there are some further deeper connections to Beukers-type integrals, or to its generalizations (for example Vasilenko integrals - see section 8 in https://eudml.org/doc/249095).

P.S. From (1) we have $$I_3=\frac{1}{2}\int\limits_0^1f(x)\;d\ln{x}=-\frac{1}{2}\int\limits_0^1 \ln{x}\,\frac{df(x)}{dx}\;dx,$$ where $$f(x)=\int\limits_0^1\frac{dy}{y}\left[\ln{(1+x)}+\ln{(1+y)}-\ln{(1+x+y)}\right],$$ and $$\frac{df(x)}{dx}=\int\limits_0^1 \frac{dy}{y}\left[\frac{1}{1+x}-\frac{1}{1+x+y}\right]=\frac{\ln{(2+x)}-\ln{(1+x)}}{1+x}.$$ Therefore $$I_3=-\frac{1}{2}\left[\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}-\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}\right]dx=-\frac{1}{2}\left[-\frac{13}{24}+\frac{1}{8}\right]\zeta(3)=\frac{5}{24}\zeta(3),$$ because $$\int\limits_0^1 \frac{\ln{x}\ln{(2+x)}}{1+x}dx=-\frac{13}{24}\zeta(3),$$ and $$\int\limits_0^1 \frac{\ln{x}\ln{(1+x)}}{1+x}dx=-\frac{1}{8}\zeta(3).$$ In the case of $I_4$, we analogously get $$I_4=\frac{1}{6}\int\limits_0^1\int\limits_0^1\frac{\ln{x}\ln{y}}{(1+x+y)^2}\left[\ln{(2+x+y)}-\ln{(1+x+y)}+\frac{1}{2+x+y}\right],$$ however further progress doesn't seem feasible.

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    $\begingroup$ Just like to share a (maybe trivial) observation on the beautiful integral for $\zeta(3)$ in your arxiv paper. When we slightly alter it into: $$f(n):=\frac{1}{7}\,\int_0^{\pi} x^n\,\frac{\pi-x}{\sin(x)}dx$$ we get closed forms expressed as finite series of $\zeta(2k+1)$ that are all weighted by $\pi^k$ and a rational. E.g. $f(2) = \frac12\pi\zeta(3)$, $f(3) = \frac32\pi^2\zeta(3)-\frac{93}{7}\zeta(5)$, $f(4) = 2\,\pi^3\zeta(3)-\frac{279}{14}\zeta(5)$, etc. $\endgroup$ – Agno May 9 '17 at 13:23
  • $\begingroup$ Interesting observation. How do you got them? By the way while checking f(2), due to an accidental error, I got (numerically) $$\int\limits_0^1\frac{x^2(\pi-x)}{\pi \sin{x}}dx=\sin\left(\frac{13\pi}{46}\right)-\sin\left(\frac{6\pi}{53}\right).$$ How this identity can be proved? $\endgroup$ – Zurab Silagadze May 10 '17 at 5:25
  • $\begingroup$ It seems the last identity is only correct up to the precision $10^{-10}$ but not exactly: mathoverflow.net/questions/269405/interesting-identity $\endgroup$ – Zurab Silagadze May 10 '17 at 7:20
  • $\begingroup$ I have used Maple to evaluate $f(n)$ and it came back with the closed forms. Not exactly sure how it does the evaluation, however I have the strong impression that it first expresses the integrals as a finite series of weighted polylogarithms (+other some components) that only for this specific integrals can be simplified and reduced to the form with only weighted $\zeta(2k+1)$s. $\endgroup$ – Agno May 10 '17 at 9:13
  • $\begingroup$ I have changed my comment into a follow-up question here: mathoverflow.net/questions/270286/… $\endgroup$ – Agno May 21 '17 at 12:17

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