I have been trying to evaluate the following family of integrals:

$$ f:(0,\infty)^2 \rightarrow \mathbb{R} \, , \, f(\alpha,\beta) = \int \limits_0^\infty \frac{\ln (1+x^\alpha) \ln (1+x^{-\beta})}{x} \, \mathrm{d} x \, . $$

The changes of variables $\frac{1}{x} \rightarrow x$, $x^\alpha \rightarrow x$ and $x^\beta \rightarrow x$ yield the symmetry properties $$ \tag{1} f(\alpha,\beta) = f(\beta,\alpha) = \frac{1}{\alpha} f\left(1,\frac{\beta}{\alpha}\right) = \frac{1}{\alpha} f\left(\frac{\beta}{\alpha},1\right) = \frac{1}{\beta} f\left(\frac{\alpha}{\beta},1\right) = \frac{1}{\beta} f\left(1,\frac{\alpha}{\beta}\right) $$ for $\alpha,\beta > 0$ .

Using this result one readily computes $f(1,1) = 2 \zeta (3)$ . Then $(1)$ implies that $$ f(\alpha,\alpha) = \frac{2}{\alpha} \zeta (3) $$ holds for $\alpha > 0$ . Every other case can be reduced to finding $f(1,\gamma)$ for $\gamma > 1$ using $(1)$.

An approach based on xpaul's answer to this question employs Tonelli's theorem to write $$ \tag{2} f(1, \gamma) = \int \limits_0^\infty \int \limits_0^1 \int \limits_0^1 \frac{\mathrm{d}u \, \mathrm{d}v \, \mathrm{d}x}{(1+ux)(v+x^\gamma)} = \int \limits_0^1 \int \limits_0^1 \int \limits_0^\infty \frac{\mathrm{d}x \, \mathrm{d}u \, \mathrm{d}v}{(1+ux)(v+x^\gamma)} \, .$$ The special case $f(1,2) = \pi \mathrm{C} - \frac{3}{8} \zeta (3)$ is then derived via partial fraction decomposition ($\mathrm{C}$ is Catalan's constant). This technique should work at least for $\gamma \in \mathbb{N}$ (it also provides an alternative way to find $f(1,1)$), but I would imagine that the calculations become increasingly complicated for larger $\gamma$ .

Mathematica manages to evaluate $f(1,\gamma)$ in terms of $\mathrm{C}$, $\zeta(3)$ and an acceptably nice finite sum of values of the trigamma function $\psi_1$ for some small, rational values of $\gamma > 1$ (before resorting to expressions involving the Meijer G-function for larger arguments). This gives me some hope for a general formula, though I have not yet been able to recognise a pattern.

Therefore my question is:

How can we compute $f(1,\gamma)$ for general (or at least integer/rational) values of $\gamma > 1$ ?

Update 1:

Symbolic and numerical evaluations with Mathematica strongly suggest that $$ f(1, n) = \frac{1}{n (2 \pi)^{n-1}} \mathrm{G}_{n+3, n+3}^{n+3,n+1} \left(\begin{matrix} 0, 0, \frac{1}{n}, \dots, \frac{n-1}{n}, 1 , 1 \\ 0,0,0,0,\frac{1}{n}, \dots, \frac{n-1}{n} \end{matrix} \middle| \, 1 \right) $$ holds for $n \in \mathbb{N}$ . These values of the Meijer G-function admit an evaluation in terms of $\zeta(3)$ and $\psi_1 \left(\frac{1}{n}\right), \dots, \psi_1 \left(\frac{n-1}{n}\right) $ at least for small (but likely all) $n \in \mathbb{N}$ .

Interesting side note: The limit $$ \lim_{\gamma \rightarrow \infty} f(1,\gamma+1) - f(1,\gamma) = \frac{3}{4} \zeta(3) $$ follows from the definition.

Update 2:

Assume that $m, n \in \mathbb{N} $ are relatively prime (i.e. $\gcd(m,n) = 1$). Then the expression for $f(m,n)$ given in Sangchul Lee's answer can be reduced to \begin{align} f(m,n) &= \frac{2}{m^2 n^2} \operatorname{Li}_3 ((-1)^{m+n}) \\ &\phantom{=} - \frac{\pi}{4 m^2 n} \sum \limits_{j=1}^{m-1} (-1)^j \csc\left(j \frac{n}{m} \pi \right) \left[\psi_1 \left(\frac{j}{2m}\right) + (-1)^{m+n} \psi_1 \left(\frac{m + j}{2m}\right) \right] \\ &\phantom{=} - \frac{\pi}{4 n^2 m} \sum \limits_{k=1}^{n-1} (-1)^k \csc\left(k \frac{m}{n} \pi \right) \left[\psi_1 \left(\frac{k}{2n}\right) + (-1)^{n+m} \psi_1 \left(\frac{n + k}{2n}\right) \right] \\ &\equiv F(m,n) \, . \end{align} Further simplifications depend on the parity of $m$ and $n$.

This result can be used to obtain a solution for arbitrary rational arguments: For $\frac{n_1}{d_1} , \frac{n_2}{d_2} \in \mathbb{Q}^+$ equation $(1)$ yields \begin{align} f\left(\frac{n_1}{d_1},\frac{n_2}{d_2}\right) &= \frac{d_1}{n_1} f \left(1,\frac{n_2 d_1}{n_1 d_2}\right) = \frac{d_1}{n_1} f \left(1,\frac{n_2 d_1 / \gcd(n_1 d_2,n_2 d_1)}{n_1 d_2 / \gcd(n_1 d_2,n_2 d_1)}\right) \\ &= \frac{d_1 d_2}{\gcd(n_1 d_2,n_2 d_1)} f\left(\frac{n_1 d_2}{\gcd(n_1 d_2,n_2 d_1)},\frac{n_2 d_1}{\gcd(n_1 d_2,n_2 d_1)}\right) \\ &= \frac{d_1 d_2}{\gcd(n_1 d_2,n_2 d_1)} F\left(\frac{n_1 d_2}{\gcd(n_1 d_2,n_2 d_1)},\frac{n_2 d_1}{\gcd(n_1 d_2,n_2 d_1)}\right) \, . \end{align}

Therefore I consider the problem solved in the case of rational arguments. Irrational arguments can be approximated by fractions, but if anyone can come up with a general solution: you are most welcome to share it. ;)

  • 1
    $\begingroup$ A note: $\enspace$ With the formula I've given with my comment below it's trivial to calculate $\,\displaystyle \lim_{\gamma \rightarrow \infty} f(1,\gamma+a) - f(1,\gamma) = \frac{3a}{4} \zeta(3) \,$ . :) $\endgroup$
    – user90369
    Commented Jul 4, 2018 at 17:26
  • $\begingroup$ @user90369 You're right, that is immediately apparent from your answer. Nice! :) $\endgroup$ Commented Jul 4, 2018 at 20:16

5 Answers 5


Only a comment. We have

$$ \int_{0}^{\infty} \frac{\log(1+\alpha x)\log(1+\beta/x)}{x} \, dx = 2\operatorname{Li}_3(\alpha\beta) - \operatorname{Li}_2(\alpha\beta)\log(\alpha\beta) $$

which is valid initially for $\alpha, \beta > 0$ and extends to a larger domain by the principle of analytic continuation. Then for integers $m, n \geq 1$ we obtain

\begin{align*} f(m, n) &=\int_{0}^{\infty} \frac{\log(1+x^m)\log(1+x^{-n})}{x}\,dx \\ &\hspace{6em} = \sum_{j=0}^{m-1}\sum_{k=0}^{n-1} \left[ 2\operatorname{Li}_3\left(e^{i(\alpha_j+\beta_k)}\right) - i(\alpha_j+\beta_k)\operatorname{Li}_2\left(e^{i(\alpha_j+\beta_k)}\right) \right], \end{align*}

where $\alpha_j = \frac{2j-m+1}{n}\pi$ and $\beta_k = \frac{2k-n+1}{n}\pi$. (Although we cannot always split complex logarithms, this happens to work in the above situation.) By the multiplication formula, this simplifies to

\begin{align*} f(m, n) &= \frac{2\gcd(m,n)^3}{m^2n^2}\operatorname{Li}_3\left((-1)^{(m+n)/\gcd(m,n)}\right) \\ &\hspace{2em} - \frac{i}{n} \sum_{j=0}^{m-1} \alpha_j \operatorname{Li}_2\left((-1)^{n-1}e^{in\alpha_j}\right) \\ &\hspace{2em} - \frac{i}{m} \sum_{k=0}^{n-1} \beta_k \operatorname{Li}_2\left((-1)^{m-1}e^{im\beta_k}\right). \end{align*}

Here, $\gcd(m,n)$ is the greatest common divisor of $m$ and $n$.

The following code tests the above formula.

Numerical computation

  • $\begingroup$ Thank you very much! This is clearly more than just a comment. There is a small typo in the definition of $\alpha_j$. The simplification using the $\gcd$ is just brilliant (and took me quite a while to understand)! $\operatorname{Li}_s (1) = \zeta (s)$ and $\operatorname{Li}_s (-1) = -\eta (s)$ then yield the $\zeta(3)$-terms with the correct prefactors. In the special case $m = 1$ and $n$ odd I have been able to reduce the remaining sum of dilogarithms to the expression from my own answer. I will try to find similar simplifications for the general case. $\endgroup$ Commented Jul 5, 2018 at 13:07

Only a comment.

For $\,z> 0\,$ I’ve got:

$$f(1,z) = \frac{3}{4}\zeta(3)\left(z+\frac{1}{z^2}\right) + 2 g(z)$$

with $\enspace\displaystyle g(z):=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} \sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k(zn+k)}\enspace$ and $\enspace\displaystyle g(\frac{1}{z})=zg(z)$

Maybe someone can take it from here to create formulas.

Note: $\enspace\displaystyle f(1,\frac{1}{z})=zf(1,z)\enspace$ which is equivalent to $\enspace\displaystyle \frac{1}{\alpha}f(1,\frac{\beta}{\alpha})=\frac{1}{\beta}f(1,\frac{\alpha}{\beta})$


$\displaystyle zg(z)=\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\left(H(zn)- H(\frac{zn}{2})\right)=\int\limits_0^1\frac{Li_2(-t^z) - Li_2(-t^{z/2})}{1-t}dt $

$\hspace{1.1cm}\displaystyle = \frac{\pi^2\ln 2}{12} + \int\limits_0^1\frac{Li_2(-x^z)}{1+x}dx = z \int\limits_0^1\frac{\ln(1+x)\ln(1+x^z)}{x}$

with $\enspace\displaystyle H(x):=\sum\limits_{k=1}^\infty \frac{x}{k(x+k)}=\gamma + \psi(1+x) =\int\limits_0^1\frac{1-t^x}{1-t}dt$

It's trivial that $\,f(1,z)\,$ can be split:

$\hspace{1.1cm}\displaystyle \int\limits_0^1 \frac{\ln(1+x)\ln(1+x^{-z})}{x}dx = \frac{3z}{4}\zeta(3) + g(z) = g(-z)$

$\hspace{1.1cm}\displaystyle \int\limits_1^\infty \frac{\ln(1+x)\ln(1+x^{-z})}{x}dx = \frac{3}{4z^2}\zeta(3) + g(z)$

  • $\begingroup$ Thanks a lot! I have only been able to evaluate $g(1) = \frac{1}{4} \zeta(3)$ so far, but this looks promising. I think that there is an extra $2$ in the last line; it should be $H (z n) - H\left(\frac{z n}{2}\right)$ . Then the Euler-Mascheroni constant cancels (as is to be expected, since it does not appear in the final results). $\endgroup$ Commented Jul 4, 2018 at 20:35
  • $\begingroup$ @ComplexYetTrivial : Thanks, of course. Corrected. :) - Hope you'll find a formula for $\displaystyle \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}H(zn)$ . $\endgroup$
    – user90369
    Commented Jul 4, 2018 at 20:39
  • $\begingroup$ Thanks again for your work! I will try to figure out which of the various representations is best suited for an explicit evaluation of $g$ . $\endgroup$ Commented Jul 5, 2018 at 17:17
  • $\begingroup$ @ComplexYetTrivial : I think your own calculations for $\,f(1,n)|_{n\in\mathbb{N}}\,$ are the best till now. :) $\endgroup$
    – user90369
    Commented Jul 5, 2018 at 17:23

The integral of $x^p$ times a product of two Meijer G-functions gives a Fox H-function: $$\int_0^\infty \frac {\ln(1+x^\alpha) \ln(1+x^{-\beta})} x dx = \frac 1 \alpha \int_0^\infty G_{2,2}^{1,2} \left( x \middle| {1, 1 \atop 1, 0} \right) G_{2,2}^{2,1} \left( x^{\beta/\alpha} \middle| {0, 1 \atop 0, 0} \right) \frac {dx} x = \\ \frac 1 \alpha H_{4,4}^{4,2} \left( 1 \middle| {(0, 1), (0, \frac \beta \alpha), (1, \frac \beta \alpha), (1, 1) \atop (0, 1), (0, 1), (0, \frac \beta \alpha), (0, \frac \beta \alpha)} \right) = \frac 1 {2 \pi i } \int_\mathcal L \frac {\pi^2 \csc \pi s \csc \frac {\pi \beta s} \alpha} {\beta s^2} ds.$$ When $\beta/\alpha$ is rational, the H-function reduces to a G-function.


This is not an answer since just based on a CAS.

For integer values of $\gamma$, we have $$f(1,1)=2 \zeta (3)$$ $$f(1,2)=\pi C-\frac{3 \zeta (3)}{8}$$ $$f(1,3)=\frac{2}{27} \left(3 \zeta (3)+\sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)\right)\right)$$ $$f(1,4)=\frac{1}{64} \left(-16 \pi C-6 \zeta (3)+\sqrt{2} \pi \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi ^{(1)}\left(\frac{3}{8}\right)-\psi ^{(1)}\left(\frac{5}{8}\right)-\psi ^{(1)}\left(\frac{7}{8}\right)\right)\right)$$

I have been able to produce the results up to $f(1,6)$ without noticing any pattern; for $\gamma > 6$ start to appear Meijer G functions (which are not the most pleasant - at least to me).

For some rational values of $\gamma$, I (say the CAS) obtained $$f\left(1,\frac{1}{2}\right)=2 \pi C-\frac{3 \zeta (3)}{4}$$ For $f\left(1,\frac{3}{2}\right)$ (no room to put it on a line) $$\frac{1}{108} \left(-72 \pi C-9 \zeta (3)+2 \sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{6}\right)+\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)-\psi ^{(1)}\left(\frac{5}{6}\right)\right)\right)$$

For $f\left(1,\frac{5}{2}\right)$, I also got a result (quite long !).

$$f\left(1,\frac{1}{3}\right)=\frac{2}{9} \left(3 \zeta (3)+\sqrt{3} \pi \left(\psi ^{(1)}\left(\frac{1}{3}\right)-\psi ^{(1)}\left(\frac{2}{3}\right)\right)\right)$$

Now, I really need a break (my computer too).

I hope and wish that this could be of some interest for you.

  • $\begingroup$ Thank you very much, this is really helpful! These results suggest quite a few things: The prefactor of $\zeta(3)$ in $f(1,n)$ is apparently given by $\frac{(-1)^{n-1}}{n^3} a_n$ for $n \in \mathbb{N}$. The sequence $(a_n)_{n \in \mathbb{N}}$ starts with $2,3,6,6,10,9,14$, which may be described by $a_{2k-1} = 4k - 2$ and $a_{2k} = 3k$ for $k \in \mathbb{N}$. Catalan's constant seems to appear for even $n$ only, which is probably due to the fact that $\psi_1 (1/4) = \pi^2 + 8 \mathrm{C}$ and $\psi_1 (3/4) = \pi^2 - 8 \mathrm{C}$ . $\endgroup$ Commented Jul 4, 2018 at 11:34

I have found a way to evaluate $f(1,n)$ for $n \in \mathbb{N}$ . Here's a sketch:

Start by letting $x = v^{1/n} t$ in equation $(2)$. Then $$ f(1,n) = \int \limits_0^1 \int \limits_0^1 \int \limits_0^\infty \frac{\mathrm{d} t}{(t + (u v^{1/n})^{-1}) (1 + t^n)} \, \frac{\mathrm{d} u}{u} \, \frac{\mathrm{d} v}{v} \equiv \int \limits_0^1 \int \limits_0^1 X((u v^{1/n})^{-1}) \, \frac{\mathrm{d} u}{u} \, \frac{\mathrm{d} v}{v} \, .$$ A partial fraction decomposition yields \begin{align} X(w) &\equiv \int \limits_0^\infty \frac{\mathrm{d} t}{(t + w) (1 + t^n)} = \frac{1}{1+(-w)^n} \int \limits_0^\infty \left[ \frac{1}{t+w} - \frac{1}{1+t^n} \sum \limits_{k=0}^{n-1} (-w)^{n-1-k} t^k\right] \, \mathrm{d} t \\ &\equiv Y(w) + Z(w) \, . \end{align} Here $$ Y(w) = \frac{1}{1+(-w)^n} \int \limits_0^\infty \left[ \frac{1}{t+w} - \frac{t^{n-1}}{1+t^n} \right] \, \mathrm{d} t = \frac{-\ln(w)}{1+(-w)^n} $$ and $$ Z(w) = - \frac{1}{1+(-w)^n} \sum \limits_{k=0}^{n-2} (-w)^{n-1-k} \int \limits_0^\infty \frac{t^k}{1 + t^n} \, \mathrm{d} t = - \frac{\pi}{n} \sum \limits_{m=1}^{n-1} \csc\left(\frac{m}{n} \pi\right)\frac{(-w)^{n-m}}{1+(-w)^n} $$ have been introduced (the last step follows from this question). Now check that \begin{align} \int \limits_0^1 \int \limits_0^1 Y((u v^{1/n})^{-1}) \, \frac{\mathrm{d} u}{u} \, \frac{\mathrm{d} v}{v} &= \frac{1}{n} \int \limits_0^1 \int \limits_0^1 \frac{\ln(u^n v) u^n v}{u^n v +(-1)^n} \, \frac{\mathrm{d} v}{v} \, \frac{\mathrm{d} u}{u} \\ &\stackrel{(\mathrm{a})}{=} \frac{(-1)^{n-1}}{n} \int \limits_0^1 \int \limits_{-n \ln(u)}^\infty \frac{s}{\mathrm{e}^s +(-1)^n} \, \mathrm{d} s \, \frac{\mathrm{d} u}{u} \\ &\stackrel{(\mathrm{b})}{=} \frac{(-1)^{n-1}}{n} \int \limits_0^\infty \frac{s}{\mathrm{e}^s +(-1)^n} \int \limits_{\mathrm{e}^{-s/n}}^1 \frac{\mathrm{d} u}{u} \, \mathrm{d} s \\ &= \frac{(-1)^{n-1}}{n^2} \int \limits_0^\infty \frac{s^2}{\mathrm{e}^s +(-1)^n} \, \mathrm{d} s \\ &\stackrel{(\mathrm{c})}{=} \begin{cases} \frac{2}{n^2} \zeta (3) \, , & n ~ \mathrm{odd} \\ - \frac{3}{2 n^2} \zeta (3) \, , & n ~ \mathrm{even} \end{cases} \end{align} holds (explanation: $(\mathrm{a})$ $v = u^{-n} \mathrm{e}^{-s}$ , $(\mathrm{b})$ Tonelli, $(\mathrm{c})$ integral representations of $\zeta$). Similarly, \begin{align} \int \limits_0^1 \int \limits_0^1 Z((u v^{1/n})^{-1}) \, \frac{\mathrm{d} u}{u} \, \frac{\mathrm{d} v}{v} &= - \frac{\pi}{n} \sum \limits_{m=1}^{n-1} (-1)^{n-m} \csc\left(\frac{m}{n} \pi\right) \int \limits_0^1 \int \limits_0^1 \frac{(u^n v)^{m/n}}{(-1)^n + u^n v} \, \frac{\mathrm{d} v}{v} \, \frac{\mathrm{d} u}{u} \\ &\stackrel{(\mathrm{d})}{=} - \frac{\pi}{n} \sum \limits_{m=1}^{n-1} (-1)^{n-m} \csc\left(\frac{m}{n} \pi\right) \int \limits_0^1 \int \limits_0^{u^n} \frac{s^{\frac{m}{n} -1}}{(-1)^n + s} \, \mathrm{d} s \, \frac{\mathrm{d} u}{u} \\ &\stackrel{(\mathrm{e})}{=} - \frac{\pi}{n^2} \sum \limits_{m=1}^{n-1} (-1)^{n-m} \csc\left(\frac{m}{n} \pi\right) \int \limits_0^1 \frac{- \ln(s) s^{\frac{m}{n} -1}}{(-1)^n + s} \, \mathrm{d} s \\ &\stackrel{(\mathrm{f})}{=} - \frac{\pi}{n^2} \sum \limits_{m=1}^{n-1} (-1)^{n-m} \csc\left(\frac{m}{n} \pi\right) \begin{cases} - \operatorname{\psi_1} \left(\frac{m}{n}\right) , & \!\!\! n ~ \mathrm{odd} \\ \frac{1}{4} \left[\operatorname{\psi_1} \left(\frac{m}{2 n}\right) - \operatorname{\psi_1} \left(\frac{n+m}{2 n}\right) \right] , & \!\!\! n ~ \mathrm{even} \end{cases} \\ \end{align} can be computed (explanation: $(\mathrm{d})$ $v = u^{-n} s$ , $(\mathrm{e})$ Tonelli, $(\mathrm{f})$ integral representation for $\psi_1$).

Combine these results to find $$ f(1,n) = \frac{2}{n^2} \zeta (3) - \frac{\pi}{n^2} \sum \limits_{m=1}^{n-1} (-1)^{m} \csc\left(\frac{m}{n} \pi\right) \operatorname{\psi_1} \left(\frac{m}{n}\right) $$ for odd and $$ f(1,n) = - \frac{3}{2 n^2} \zeta (3) - \frac{\pi}{4 n^2} \sum \limits_{m=1}^{n-1} (-1)^{m} \csc\left(\frac{m}{n} \pi\right) \left[\operatorname{\psi_1} \left(\frac{m}{2 n}\right) - \operatorname{\psi_1} \left(\frac{n+m}{2 n}\right) \right] $$ for even $n \in \mathbb{N}$ .

Judging from the result for $f(1,\frac{3}{2})$ in Claude Leibovici's answer, a similar approach should work for rational values of $\gamma$, but I am not that optimistic about irrational arguments.

  • $\begingroup$ I haven't checked the result now, but you have made interesting calculations. :) $\endgroup$
    – user90369
    Commented Jul 5, 2018 at 8:43

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