# Two unequal trilog integral twins $\int_0^1 \log^2(x)\frac{ \log(1\pm \sigma x)}{1\mp x}\,dx$

Trying to generalize this problem [1] I came up with these two integrals

$$i_{\pm}(\sigma) = \int_0^1 \log^2(x)\frac{ \log(1\pm \sigma x)}{1\pm x}\,dx\tag{1}$$

Here $$-1 \le \sigma \le 1$$ is a parameter. For $$\sigma = \pm 1$$ we are led back to the problem [1].

Although these two integrals look like twins they are quite different.

Notice that the essential difference is in the denominator; under the $$\log$$ the sign change can be accomodated in $$\sigma$$.

While I was able to solve $$i_{-}(\sigma)$$ with the methods generalizing the solution to [1] I find $$i_{+}(\sigma)$$ tough.

Can you find to the closed forms of both integrals?

What I did so far:

The solution is based of the concept of the unified harmonic number $$U_n$$ I have introduced earlier [2] as a generalization of the harmonic number and the alternating harmonic number.

It is defined as

$$U_n(\sigma) = \sum_{k=1}^n \frac{\sigma^k}{k} = \sigma \int_0^1 \frac{1-(\sigma z)^n}{1-\sigma z}\,dz\tag{2}$$

The generating function is

$$g_{-}(x,\sigma) = \sum_{k=1}^\infty x^n U_n (\sigma) = -\frac{\log(1-\sigma x)}{1-x}\tag{3}$$

Here we recognize a part of the integrand of $$i_{-}(\sigma)$$.

Hence multiplying $$(3)$$ with $$\log(x)^2$$ and integrating from $$x=0$$ to $$x=1$$ we have

$$-i_{-}(\sigma) = \sum_{n=1}^\infty U_n(\sigma) \int_0^1 x^n \log(x)^2\,dx \\= \sum_{k=1}^\infty U_n(\sigma) \frac{2}{(1+n)^3} \\=\sigma \int_0^1 \sum_{n=1}^\infty \frac{1-(\sigma z)^n}{1-\sigma z}\frac{2}{(1+n)^3}\,dz\tag{4}$$

$$=2\sigma \int_0^1\frac{ \sigma z \zeta (3)-\text{Li}_3(z \sigma )}{z (1-\sigma z)}\,dz\tag{5}$$

$$-i_{-}(\sigma)= \operatorname{Li}_2(\sigma ){}^2+2 \operatorname{Li}_3(\sigma ) \log (1-\sigma )\\ -2 (\operatorname {Li}_4(\sigma )+\zeta (3) \log (1-\sigma )) \tag{6}$$

Repeating the same for $$i_{+}$$ we have now

$$g_{+}(x,\sigma) = \sum_{k=1}^\infty (-1)^n x^n U_n (\sigma) = -\frac{\log(1+\sigma x)}{1+x}\tag{3a}$$

Multiplying $$(3a)$$ with $$\log(x)^2$$ and integrating from $$x=0$$ to $$x=1$$ we have

$$-i_{+}(\sigma) = \sum_{n=1}^\infty (-1)^n U_n(\sigma) \int_0^1 x^n \log(x)^2\,dx \\= \sum_{k=1}^\infty (-1)^n U_n(\sigma) \frac{2}{(1+n)^3} \\=\sigma \int_0^1 \sum_{n=1}^\infty (-1)^n \frac{1-(\sigma z)^n}{1-\sigma z}\frac{2}{(1+n)^3}\,dz\tag{4a}$$

$$=-2\sigma \int_0^1\frac{4 \operatorname{Li}_3(-z \sigma )+3 \sigma z \zeta (3)}{2 z (1-\sigma z)}\,dz\tag{5a}$$

and here I am stuck.

While $$(5)$$ could be easily solved $$(5a)$$ is resistant.

My blocking point is this integral

$$\int_0^s \frac{\operatorname{Li}_3(t)}{1+t} \, dt\tag{7}$$

Ali Shather kindly pointed out to me that the integral for $$s=1$$ is well known, and he provided the useful hint that expanding the denominator gives

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x} \, dt=\sum_{n=1}^{\infty} (-1)^{n-1}\int_0^1 x^{n-1}\operatorname{Li}_3(x)\,dx \\= \sum_{n=1}^{\infty} (-1)^{n-1}\left(\frac{ \zeta(3)}{ n} -\frac{\zeta(2)}{ n^2}+\frac{H_{n}}{n^3}\right)\tag{8}$$

so that, finally borrowing the result for $$\sum_{k=1}^{\infty}(-1)^{k+1} \frac{H_{k}}{k^3}$$ from Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$ I find

$$\int_0^1 \frac{\operatorname{Li}_3(x)}{1+x} \, dx= -2 \text{Li}_4\left(\frac{1}{2}\right)-\frac{3}{4} \zeta (3) \log (2)+\frac{\pi ^4}{60}-\frac{1}{12} \log ^4(2)+\frac{1}{12} \pi ^2 \log ^2(2)\simeq 0.339545\tag{9}$$

But, unfortunately, this does not solve the problem in question here which requires the integral $$(7)$$ as a function of $$s$$.

References

• nice problem Wolf (+1) . Regarding $i_{-}(\sigma)$, just use taylor series for $\ln(1-\sigma x)$ and use $\int_0^1\frac{x^n \ln^2x}{1-x}\ dx=2\zeta(3)-2H_n^{(3)}$. – Ali Shather Mar 20 at 18:30
• Thanks, Ali. What do you think about the integral $\int \frac{\operatorname{Li}_3(t)}{1+t} \, dt$? – Dr. Wolfgang Hintze Mar 21 at 16:57
• You welcome. I know this integral if its definite from zero to one. – Ali Shather Mar 21 at 17:47
• I am moving to and fro between the integral and the sum $\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}$ Could you give me please a hint to $\int_0^1 \frac{\operatorname{Li}_3(t)}{1+t} \, dt$? – Dr. Wolfgang Hintze Mar 21 at 18:24
• Sure Wolf just a min – Ali Shather Mar 21 at 18:26

$$\text{Li}_3(x)=\frac14\text{Li}_3(x^2)-\text{Li}_3(-x)$$

$$\int\frac{\text{Li}_3(x)}{1+x}\ dx=\frac14\int\frac{\text{Li}_3(x^2)}{1+x}\ dx-\int\frac{\text{Li}_3(-x)}{1+x}\ dx=\frac14I_1-I_2$$

For $$I_2$$, apply integration by parts twice we get

$$I_2=\ln(1+x)\text{Li}_3(-x)+\frac12\text{Li}_2^2(-x)$$

For $$I_1$$

$$I_1=\int\frac{\text{Li}_3(x^2)}{1+x}\ dx=\int\frac{(1-x)\text{Li}_3(x^2)}{1-x^2}\ dx$$

$$=\int\frac{\text{Li}_3(x^2)}{1-x^2}\ dx-\int\frac{x\text{Li}_3(x^2)}{1-x^2}\ dx=I_1'-I_1''$$

where

$$I_1''=\int\frac{x\text{Li}_3(x^2)}{1-x^2}\ dx\overset{x^2=y}{=}\frac12\int\frac{\text{Li}_3(y)}{1-y}\ dy$$

$$\overset{IBP}{=}-\frac12\ln(1-y)\text{Li}_3(y)-\frac14\text{Li}_2^2(y)$$

$$=-\frac12\ln(1-x^2)\text{Li}_3(x^2)-\frac14\text{Li}_2^2(x^2)$$

For $$I_1'$$ we can simply use the genrating function

$$\sum_{n=1}^\infty H_n^{(a)}x^n=\frac{\text{Li}_a(x)}{1-x}$$

Set $$a=3$$ and replace $$x$$ with $$x^2$$ we get that

$$I_1'=\int\frac{\text{Li}_3(x^2)}{1-x^2}\ dx=\sum_{n=1}^\infty H_n^{(3)}\int x^{2n}\ dx=\sum_{n=1}^\infty \frac{H_n^{(3)}}{2n+1}x^{2n+1}$$

Unfortunately I was not able to find the closed form of the latter generating sum. I hope you find this approach helpful.

I also found that

$$2\int_0^s\frac{\text{Li}_3(t)}{1+t}\ dt=\ln(1+s)\text{Li}_3(s)+\ln(1-s)\text{Li}_3(-s)+\text{Li}_2(-s)\text{Li}_2(s)+\int_{-s}^s\frac{\text{Li}_3(t)}{1+t}\ dt$$

• @ Ali Shather That' nice. I have also found your last line. I have tried this: $\frac{\text{Li}_3(t)}{t+1} =\frac{1-t}{1+t} \sum _{n=1}^{\infty } H_n^{(3)} t^n$ Integrating leads to incomplete Beta functions: $\int_0^z \frac{1-t}{1+t} t^n \, dt = (-1)^{n+1} (B_{-z}(n+1,0)+B_{-z}(n+2,0))$. But finally I get into a circle. – Dr. Wolfgang Hintze Mar 23 at 15:23
• @Dr. Wolfgang Hintze Thank you. yes it seems tough. maybe there is no closed form? – Ali Shather Mar 24 at 4:13