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It has been conjectured by Kirill that $$\int_0^1\frac{\operatorname{li}^4(x)}{x^4}\,dx=\int_0^\infty\operatorname{Ei}^4(-x)\,e^{3x}\,dx\stackrel{\color{gray}?}=\frac{26\pi^4}{135}+\frac49\left(\pi^2\ln^22-\ln^42\right)-\frac{32}{3}\operatorname{Li}_4\!\left(\tfrac12\right).$$

Can we prove this conjecture?

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This is not a complete answer, but it gives an alternative representation of $I(4)$ which may be better to work with. Define $F(s)$ by $$ F(s) = \int_{0}^{1} \frac{s}{1-sx} \log(1-x)\operatorname{Li}_{2}(x) \, dx. $$

Then we have the following representation:

Claim. We have $\displaystyle I(4) = \frac{71\pi^4}{540} - \frac{8}{3}F(1/2)$.

Consequently, the conjecture is equivalent to proving

$$ \color{red}{ F(1/2) \stackrel{?}{=} 4\operatorname{Li}_4 (1/2) + \frac{1}{6}\log^4 2 - \zeta(2)\log^2 2 - \frac{11}{480} \pi^4. } $$

These coefficients are sort of familiar, so I believe that we are on a right track. On the other hand, this still seems daunting because

$$ F(1/2) = \int_{0}^{1} \frac{\log x}{1+x} \left( \zeta(2) - \log x \log(1-x) - \operatorname{Li}_2(x) \right) \, dx $$

is obviously related to the class of alternating Euler sums.


Step 1. Using @Kirill's calculation, we obtain

$$ I(n) = \int_{[0,\infty)^n} \frac{dt_1 \cdots dt_n}{(1+t_1)\cdots(1+t_n)(1+t_1+\cdots+t_n)}. $$


Step 2. We consider the following change of variables:

$$ (x, y) = \Phi(u, v) = (\tfrac{u+v}{2}, \sqrt{uv}). $$

Then $\Phi$ maps the region $\Delta = \{ (u, v) : 0 \leq v \leq u \}$ onto $\Delta$ itself, and its Jacobian is

$$ \frac{\partial(x,y)}{\partial(u,v)} = \frac{2y}{\sqrt{x^2 - y^2}}. $$

From the symmetry and this change of variables, we have

\begin{align*} I(2n) &= \int\limits_{(u_1, v_1), \cdots, (u_n, v_n) \in \Delta} \Bigg[ \prod_{i=1}^{n} \frac{2 du_i dv_i}{(1+u_i)(1+v_i)} \Bigg] \frac{1}{1+u_1+v_1+\cdots+u_n+v_n} \\ &= \int\limits_{(x_1, y_1), \cdots, (x_n, y_n) \in \Delta} \Bigg[ \prod_{i=1}^{n} \frac{4y_i \, dx_i dy_i}{(1+2x_i+y_i^2)\sqrt{x_i^2 - y_i^2}} \Bigg] \frac{1}{1+2x_1 + \cdots + 2x_n}. \end{align*}

We remark that this trick is essentially the same as the method used by @Random Variable. Now using the following formula

$$ \int_{0}^{x} \frac{2y \, dy}{(1+2x+y^2)\sqrt{x^2-y^2}} = \frac{\log(1+2x)}{1+x}, $$

we can make a further simplification and we get

$$ I(2n) = 2^n \int_{[0,\infty)^n} \Bigg[ \frac{\log(1+2x_i) \, dx_i}{1+x_i} \Bigg] \frac{1}{1+2x_1 + \cdots + 2x_n}. $$

In particular, when $n = 2$, then the substitutions $1+2x_1 = \frac{1}{1-x}$ and $1+2x_2 = \frac{1}{1-y}$ yield

$$ \color{blue}{ I(4) = \int_{0}^{1}\int_{0}^{1} \frac{\log(1-x)\log(1-y)}{(1-xy)(1-\frac{1}{2}x)(1-\frac{1}{2}y)} \, dxdy. } \tag{1} $$

This is a reminiscence of the integral

$$ \int_{0}^{1} \int_{0}^{1} \frac{\log(1-x)\log(1-y)}{1-xy} \, dxdy = \frac{17\pi^4}{360}, $$

which already required some hard works as in here. Indeed, this result will be used in our computation.


Step 3. We first remark the following identity:

Lemma 1. For any $s \in \Bbb{C}\setminus[1,\infty)$ and $m = 1, 2, \cdots$, we have $$ \int_{0}^{1} \frac{s}{1-sx} \log^{m-1}(1-x) \, dx = (-1)^{m}\Gamma(m) \operatorname{Li}_{m}\left( \frac{s}{s-1} \right). $$

(I acknowledge that this lemma is motivated by @Felix Marin's calculation.) This is easily proved by letting $w = s/(s-1)$ and noticing that

$$ \frac{s}{1-s x} = - \frac{w}{1-w(1-x)}. $$

To accompany this result, we remind some identities involving dilogarithm:

Lemma 2. For $s \in \Bbb{C} \setminus [1, \infty)$ we have $\displaystyle \operatorname{Li}_{2}\left( \frac{s}{s-1} \right) = -\operatorname{Li}_2(s) - \frac{1}{2}\log^2(1-s)$.

Then from Lemma 1 and Lemma 2, it is straightforward to check that \begin{align*} &\int_{0}^{1} \int_{0}^{1} \frac{\log(1-x)\log(1-y)}{(1-xy)(1-sx)} \, dxdy \\ &= \int_{0}^{1} \frac{\log(1-x)}{x(1-sx)}\operatorname{Li}_{2}\left( \frac{x}{x-1} \right) \, dx \\ &= - \int_{0}^{1} \left( \frac{1}{x} + \frac{s}{1-sx} \right)\left( \operatorname{Li}_{2}(x) + \frac{1}{2}\log^{2}(1-x) \right) \log(1-x) \, dx \\ &= \frac{17\pi^{4}}{360} - F(s) - 3 \operatorname{Li}_{4}\left( \frac{s}{s-1} \right). \end{align*}

Finally, in order to facilitate this calculation, we notice that

\begin{align*} &\frac{1}{(1-xy)(1-sx)(1-sy)} \\ &\qquad = \frac{1}{1-s^{2}}\left( \frac{1}{1-xy}\left( \frac{1}{1-sx} + \frac{1}{1-sy} - 1 \right) - \frac{s}{1-sx} \frac{s}{1-sy} \right). \end{align*}

Consequently it gives

\begin{align*} &\int_{0}^{1} \int_{0}^{1} \frac{\log(1-x)\log(1-y)}{(1-xy)(1-sx)(1-sy)} \, dxdy \\ &= \color{blue}{ \frac{1}{1-s^{2}}\left( \frac{17\pi^{4}}{360} - 2F(s) - 6 \operatorname{Li}_{4}\left( \frac{s}{s-1} \right) - \operatorname{Li}_{2}\left( \frac{s}{s-1} \right)^{2} \right) }. \tag{2} \end{align*}

Then plugging $s = 1/2$ proves the claim. ////

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