I Was Vey Amaze At The Solution 3

Using the Landen identity $\mathrm{Li}_2(z) \; = \; -\tfrac12\log^2(1-z) - \mathrm{Li}_2\big(\tfrac{z}{1-z}\big)$ we deduce that $\begin{aligned} \mathrm{Li}_2\big(\tfrac12(1+i)\big) & = -\tfrac12\log^2\big(\tfrac12(1-i)\big) - \mathrm{Li}_2(-i) \\ \mathrm{Li}_2\big(\tfrac12(1-i)\big) & = -\tfrac12\log^2\big(\tfrac12(1+i)\big) - \mathrm{Li}_2(i) \end{aligned}$ and hence $\begin{aligned} \mathrm{Li}_2\big(\tfrac12(1+i)\big) & - \mathrm{Li}_2(i) + \mathrm{Li}_2(-i) - \mathrm{Li}_2\big(\tfrac12(1-i)\big) \\ & = \tfrac12\log^2\big(\tfrac12(1+i)\big) - \tfrac12\log^2\big(\tfrac12(1-i)\big) \\ & = \tfrac12\big[\log\big(\tfrac12(1+i)\big) + \log\big(\tfrac12(1-i)\big)\big]\big[\log\big(\tfrac12(1+i)\big) - \log\big(\tfrac12(1-i)\big)\big] \\ & = \tfrac12 \times \ln\tfrac12 \times \log i \; = \; -\tfrac14\pi i\ln 2 \end{aligned}$ and hence $\tfrac12i\Big( \mathrm{Li}_2\big(\tfrac12(1+i)\big) - \mathrm{Li}_2(i) + \mathrm{Li}_2(-i) - \mathrm{Li}_2\big(\tfrac12(1-i)\big)\Big) \; =\; \tfrac18\pi \ln2$ making the answer $2+8 = \boxed{10}$ .

Using the definition of the polylogarithm function

$\operatorname{Li}_s (z) = \displaystyle \sum_{k=1}^{\infty} \dfrac{z^k}{k^s}$

Let us denote

$\mathfrak{B} = \dfrac{ i \bigg[ {\color{#D61F06} \overbrace{ \left\{ \operatorname{Li}_2 (-i) - \operatorname{Li}_2 (i) \right\} }^{S_1} } + {\color{#3D99F6} \overbrace{ \left\{ \operatorname{Li}_2 \left( \dfrac{i+1}{2} \right) - \operatorname{Li}_2 \left( - \dfrac{i-1}{2} \right) \right\} }^{S_2} } \bigg] }{2}$

$\large \text{Solving } {\color{#D61F06} S_1} :$

$\operatorname{Li}_2 (-i) = \displaystyle \sum_{k=1}^{\infty} \dfrac{(-i)^k}{k^2} = \dfrac{(-i)}{1^2} + \dfrac{(-1)}{2^2} + \dfrac{i}{3^2} + \dfrac{1}{4^2} + \cdots \\ \operatorname{Li}_2 (i) = \displaystyle \sum_{k=1}^{\infty} \dfrac{i^k}{k^2} = \dfrac{i}{1^2} + \dfrac{(-1)}{2^2} + \dfrac{(-i)}{3^2} + \dfrac{1}{4^2} + \cdots \\ \\ \implies S_1 = \operatorname{Li}_2 (-i) - \operatorname{Li}_2 (i) = -2i \left( \dfrac{1}{1^2} - \dfrac{1}{3^2} + \dfrac{1}{5^2} - \dfrac{1}{7^2} + \cdots \right) = -2i \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2}$

The sum $\displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(2k+1)^2}$ is a known constant called the Catalan's constant (denoted normally by $G$ ).

Thus, $\color{#D61F06} S_1 = -2i G$ .

$\large \text{Solving } {\color{#3D99F6} S_2} :$

We can write down

$\dfrac{i+1}{2} = \dfrac{e^{{i \pi}/4}}{\sqrt{2}} \\ \dfrac{1-i}{2} = \dfrac{e^{{-i \pi}/4}}{\sqrt{2}}$

Now

$\operatorname{Li}_2 \left( \dfrac{e^{{i \pi}/4}}{\sqrt{2}} \right) = \dfrac{\left( \frac{e^{{i \pi}/4}}{\sqrt{2}} \right)}{1^2} + \dfrac{ {\left( \frac{e^{{i \pi}/4}}{\sqrt{2}} \right)}^2 }{2^2} + \dfrac{ {\left( \frac{e^{{i \pi}/4}}{\sqrt{2}} \right)}^3 }{3^2} + \cdots \\ \operatorname{Li}_2 \left( \dfrac{e^{{-i \pi}/4}}{\sqrt{2}} \right) = \dfrac{\left( \frac{e^{{-i \pi}/4}}{\sqrt{2}} \right)}{1^2} + \dfrac{ {\left( \frac{e^{{-i \pi}/4}}{\sqrt{2}} \right)}^2 }{2^2} + \dfrac{ {\left( \frac{e^{{-i \pi}/4}}{\sqrt{2}} \right)}^3 }{3^2} + \cdots \\ \\ \implies S_2 = \operatorname{Li}_2 \left( \dfrac{e^{{i \pi}/4}}{\sqrt{2}} \right) - \operatorname{Li}_2 \left( \dfrac{e^{{-i \pi}/4}}{\sqrt{2}} \right) = \displaystyle \sum_{m=0}^{\infty} \dfrac{ 2i \sin \left( \frac{(2m+1) \pi}{4} \right) }{ {\sqrt{2}}^{(2m+1)} (2m+1)^2 } = i \sum_{k=0}^{\infty} {\left( - \dfrac 12 \right)}^k \dfrac{1}{(2k+1)^2}$

The sum $\displaystyle \sum_{k=0}^{\infty} {\left( - \dfrac 12 \right)}^k \dfrac{1}{(2k+1)^2}$ evaluates to $2G - \dfrac{\pi \ln(2)}{4}$ (I wasn't able to find this closed form on my own).

Thus, $\color{#3D99F6} S_2 = 2iG - \dfrac{i \pi \ln(2)}{4}$

Therefore

${\color{#D61F06} S_1} + {\color{#3D99F6} S_2} = - \dfrac{i \pi \ln(2)}{4}$

Hence

$\mathfrak{B} = \dfrac{i \left( - \dfrac{i \pi \ln(2)}{4} \right)}{2} = \dfrac{\pi \ln(2)}{8}$

which gives $q+r = 2+8 = \boxed{10}$ .

This is where this problem originally came from.

$\int_0^1 \cfrac{\ln (x + 1) }{x^2 + 1}$

Let $x = \cfrac{1-u}{1+u} \implies dx = \cfrac{-2}{(u+1)^2} \ du$ and $x + 1 = \cfrac{2}{1+u}$ and $x^2 + 1 = \cfrac{2(1 + u^2)}{(1 + u)^2 }$

$\implies \int_0^1 \cfrac{\ln (x + 1) }{x^2 + 1} \ dx = \int_1^0 \cfrac{\ln \left( \cfrac{2}{1+u}\right) }{\cfrac{2(1 + u^2)}{(1 + u)^2 }} \cdot \cfrac{-2}{(u+1)^2} \ du \\ = \int_0^1 \cfrac{\ln \left( \cfrac{2}{1+u}\right) }{1 + u^2} \ du \\ = \int_0^1 \left( \cfrac{\ln (2)}{1+u^2}\right) \ du - \int_0^1 \left( \cfrac{\ln (u+1)}{1+u^2}\right) \ du$

$\implies 2 \left(\int_0^1 \cfrac{\ln (x + 1) }{x^2 + 1} \ dx \right) = \int_0^1 \left( \cfrac{\ln (2)}{1+u^2}\right) \ du \\ \therefore \ \int_0^1 \cfrac{\ln (x + 1) }{x^2 + 1} \ dx = \cfrac{\ln (2) \cdot \arctan (u) \bigg|_0^1}{2} = \cfrac{\ln (2) \cdot \cfrac{\pi}{4}}{2} = \cfrac{\pi}{8} \cdot \ln (2)$

Now, when I put this integral to an online calculator, this is what it gives:

$\cfrac{{\pi}\ln\left(2\right)}{4}-\cfrac{\mathrm{i}\left(\operatorname{Li}_2\left(\cfrac{\mathrm{i}+1}{2}\right)-\operatorname{Li}_2\left(\mathrm{i}\right)+\operatorname{Li}_2\left(-\mathrm{i}\right)-\operatorname{Li}_2\left(-\cfrac{\mathrm{i}-1}{2}\right)\right)}{2}$

and since i knew that $\displaystyle \int_0^1 \cfrac{\ln (x + 1) }{x^2 + 1} = \cfrac{\pi}{8} \cdot \ln (2)$

$\implies \cfrac{\mathrm{i}\left(\operatorname{Li}_2\left(\cfrac{\mathrm{i}+1}{2}\right)-\operatorname{Li}_2\left(\mathrm{i}\right)+\operatorname{Li}_2\left(-\mathrm{i}\right)-\operatorname{Li}_2\left(-\cfrac{\mathrm{i}-1}{2}\right)\right)}{2} = \cfrac{{\pi}\ln\left(2\right)}{4} - \cfrac{\pi}{8} \cdot \ln (2) = \cfrac{{\pi}\ln\left(2\right)}{8}$