New answers tagged

3

It's quite common for such integrals to use algebraic identities in order to solve them, see also here. We can use the first one in the link from above, namely: $$6ab^2=(a+b)^3+(a-b)^3-2a^3\Rightarrow I=\int_0^1 \frac{\ln(1-x) \ln^2(1+x)}{x} dx$$ $$=\frac16\int_0^1 \frac{\ln^3(1-x^2)}{x}dx+\frac16\int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)}{x}dx-\...


0

$$\int_{0}^{\infty }\frac{ln(1+x^2)ln^2x}{1+x^2}dx\\ \\ let\ I(a)=\int_{0}^{\infty }\frac{ln^2(x)ln(1+a^2.x^2)}{1+x^2}\\ \\ \therefore I'(a)=\int_{0}^{\infty }\frac{2ax^2ln^2(x)}{(1+a^2x^2)(1+x^2)}dx=\frac{2a}{1-a^2}\int_{0}^{\infty }\frac{ln^2(x)}{1+a^2x^2}-\frac{ln^2}{1+x^2}dx\\ \\ let\ \ G=\int_{0}^{\infty }\frac{ln^2(x))}{1+a^2x^2}dx\ \ \ \ ,\ \ but \ we ...


6

Considering the algebraic identity \begin{align*} &(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\ &\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b \end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that \begin{align*} 2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&...


2

\begin{align} \sum_{n=1}^\infty\frac{(-1)^{n-1}H_n}{(2n+1)^3}&=\sum_{n=1}^\infty(-1)^{n-1} H_n\int_0^1\frac12x^{2n}\ln^2 x\ dx\\ &=-\frac12\int_0^1\ln^2x\sum_{n=1}^\infty(-x^2)H_n\\ &=\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx\tag{1} \end{align} . Lets evaluate the integral : \begin{align} I&=\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ ...


1

Let $$I=\int_{0}^{p} \exp[-(\ln x- a)^2/(2b^2)] ~dx.$$ Let $x=e^t$, then $$I=\int_{0}^{\ln p} \exp[-(t-a)^2/(2b^2)+t]~ dt= \int_{0}^{\ln p} \exp[-(t^2-2at-2b^2 t+a^2)/(2b^2)] dt=$$ $$ I=\exp[((a+b^2)^2-a^2)/(2b^2)] \int_{0}^{p} \exp[-(t-(a+b^2))^2/(2b^2)].$$ Let $(t-a-b^2)(/b\sqrt{2})=u$, then $$\Rightarrow I= \exp[((a+b^2)^2-a^2)/(2 b^2)] \int_{-(a+b^2)/(...


0

This is not a full solution to this problem but i believe it does provide useful insight and is not a cul de sac. The following identities hold: \begin{eqnarray} \int\limits_0^1 \frac{\log(1-x)^3\cdot \log(1+x)}{x} dx + \int\limits_0^1 \frac{\log(1+x)^3\cdot \log(1-x)}{x} dx = -\frac{69}{16} \zeta(5) \quad (i) \\ \int\limits_0^1 \frac{\log(1-x)^2 \log(1+x)^...


0

\begin{align} I&=\int_0^1\frac{\operatorname{Li}_2(1-x)\ln^2(1-x)}{x}\ dx=\int_0^1\frac{\operatorname{Li}_2(x)\ln^2x}{1-x}\ dx\\ &=\sum_{n=1}^\infty H_n^{(2)}\int_0^1x^n\ln^2x\ dx=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{(n+1)^3}\\ &=2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}-2\zeta(5)\\ &=2\left(3\zeta(2)\zeta(5)-\frac92\zeta(5)\right)-2\zeta(5)\\ &...


0

Using Landen's identity $$\operatorname{Li}_2(x)+\operatorname{Li}_2\left(\frac{x}{x-1}\right)=-\frac12\ln^2(1-x)$$ set $x=\sqrt{2}-1$ we get $$\operatorname{Li}_2(\sqrt{2}-1)+\operatorname{Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}-2}\right)=-\frac12\ln^2(2-\sqrt{2})\tag{1}$$ we have $$\operatorname{Li}_2\left(\frac{\sqrt{2}-1}{\sqrt{2}-2}*\color{red}{\frac{...


7

In order to give some insight I will show a derivation for $a=1$, where a closed form in terms of the Inverse Tangent Integral is: $$\bbox[10pt, border:2px, lightblue]{\int_1^s \frac{\operatorname{arccosh}x}{\sqrt{(x-1)(s-x)}}dx=4\operatorname{Ti}_2\left(\sqrt{\frac{s-1}{2}}\right),\ s>1}$$ We'll start off with the substitution $\frac{s-x}{s-1}=t$ to get: ...


4

Finally I got the answer: Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4. $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=2$ we get $$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2 u}{1-xu}\ du$$ ...


1

Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4. $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=1$ and replacing $x$ with $x^2$ we get $$\operatorname{Li}_{2}(x^2)=-\int_0^1\frac{x^2\ln u}{1-x^2u}\ du$$ ...


2

\begin{align} I&=\int_0^1\operatorname{Li}_3(1-x^2)\ dx\overset{IBP}{=}2\int_0^1\frac{x^2\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx\\ &=2\int_0^1\left(\frac1{1-x^2}-1\right)\operatorname{Li}_2(1-x^2)\ dx\\ &=2\int_0^1\frac{\operatorname{Li}_2(1-x^2)}{1-x^2}\ dx-2\int_0^1\operatorname{Li}_2(1-x^2)\ dx\tag{1} \end{align} By the OP, the second integral ...


1

Let $I$ denotes $\int_0^1\frac{\ln^4(1+x)\ln x}{x}\ dx$ I proved here \begin{align} I&=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)\\ &\quad-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\...


2

In this solution, I proved $$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)\tag{1}$$ Multiply both sides of $(1)$ by $\large \frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\overset{IBP}{=}\large ...


4

In answer to your question, can the sums be evaluated separately? Yes they can. The results for each of these two Euler sums can be found in the 2016 paper Euler sums and integrals of polylogarithm functions by Ce Xu et al. The results are: $$\sum_{n = 1}^\infty \frac{H_n H^{(2)}_n}{n^6} = \frac{17}{6} \zeta (3) \zeta (6) + \frac{173}{72} \zeta (9) + \frac{...


6

The series $\sum_{n=1}^\infty\frac{H_n^2H_n^{(2)}}{n^3}$ can be written as $$\sum_{\substack{n_1\geq n_2\geq 1 \\ n_1\geq n_3\geq 1 \\ n_1\geq n_4\geq 1}}\frac{1}{n_1^3 n_2 n_3 n_4^2},$$ which can be recognized as a linear combination of multiple zeta values of weight $7$. Multiple zeta values of weight $w$ are series of the form $$\zeta(s_1, \ldots, s_k) = ...


1

Such recurrences are called full history. One way to tackle them is to consider $a_{n + 1} - a_n$, most of the sum falls away and you get something manageable. Another (more general) way is to use generating functions. Say you have: $\begin{align*} a_{n + 1} &= \sum_{0 \le k \le n} a_k \end{align*}$ define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply ...


2

Here is a path for getting $$ \sum_{\substack{0\le n\\1\le k\le(n+1)}} \frac 1k\cdot H_{k-1/2}\; x^n = \frac{2}{x(1-x)} \Big( \operatorname{arctanh}^2 \sqrt{x} + \log 2 \log (1-x) \Big)\ . $$ (We show this holds as an equality of meromorphic functions in the open unit disk in the complex plane.) Using $$H_{k-1/2}=(2H_{2k}-H_k)-2\log 2\ ,$$the ...


1

For $1\leq i \leq 2n-1$ we have $$S_i=n-|n-i|$$ Or in others words $S_i=i$ when $1\leq i\leq n$ and $S_i=2n-i$ when $n\leq i \leq 2n-1$.


9

Here is another approach: Again, we are going to estabish two relations and solve for the target sum. First Relation: From here we have $$\int_0^1x^{n-1}\ln^4(1-x)\ dx=\frac1n\left(H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}\right)$$ Divide both sides by $n^2$ then sum both sides from $n=1$ to $\infty$ to get $$R_1=\sum_{...


11

To compute the target sum, we are going to establish two relations and solve them by elimination. First Relation: From here we have $$-\int_0^1x^{n-1}\ln^3(1-x)\ dx=\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$ Multiply both sides by $\large \frac{H_n}{n^2}$ then sum both sides from $n=1$ to $\infty$ to get \begin{align} R_1&=\sum_{n=1}^\infty\frac{H_n^...


2

Using the generalized integral expression of the polylogrithmic function which can be found in the book (Almost) Impossible Integrals, Sums and series page 4. $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$ and by setting $n=2$ we get $$\operatorname{Li}_{3}(x)=\frac12\int_0^1\frac{x\ln^2(u)}{1-xu}\ du$$ Then \begin{align} I&...


0

Just some further thoughts, which neatly connect different methods of transforming the integral. Using Jack D'Aurizio's answer, we work with the integral: $$I(a)=\int_{0}^{+\infty}\frac{1}{2}\left[\psi\left(\frac{s+2}{2}\right)-\psi\left(\frac{s+1}{2}\right)\right]e^{-as}\,ds= \\ = \sum_{n=0}^\infty (-1)^n \int_{0}^{+\infty} \frac{e^{-a s} ds}{n+s+1}$$ ...


4

The question essentially boils down to finding partial fractions of $$\frac1{(2n+1)^m(2n+3)^m(2n+5)^m}=\sum_{k=1}^m\frac{a_k}{(2n+1)^k}+\frac{b_k}{(2n+3)^k}+\frac{c_k}{(2n+5)^k}$$ which can be done by multiplying both sides by $(2n+a)^m$ and computing the terms via Taylor's theorem: $$\frac1{(2n+3)^m(2n+5)^m}=\sum_{k=1}^ma_k(2n+1)^{m-k}+\frac{b_k(2m+1)^m}{...


0

You are right, the series is a little messed up, but there's no reason to try guessing the proper form. We can always use the general way to get the hypergeometric form of a series. Consider: $$f(z)=z \Gamma(3)\left[\frac{1}{\Gamma(3)}+\frac{(\gamma)_1}{\Gamma(4)}\frac{z}{1!}+\frac{(\gamma)_{2}}{\Gamma(5)}\frac{z^2}{2!}+\frac{(\gamma)_{3}}{\Gamma(6)}\frac{...


0

Playing with Pochhammer symbols, we could also compute the partial sum $$S_p=\sum_{n=0}^p \frac{a (a+1)_n}{b (b+1)_n}$$ and get $$S_p=\frac{a}{b-a-1 }-\frac{ \Gamma (b) }{(b-a-1) \Gamma (a) }\frac{(b+p+1) \Gamma (a+p+2)}{ \Gamma (b+p+2)}$$


0

As already indicated in the comment, this problem is related to the German Tank problem, from the analysis of which we obtain the more general formula $$ {{m - 1} \over m}\sum\limits_{j = 0}^n {{1 \over {\left( \matrix{ j + x \cr m \cr} \right)}}} = {1 \over {\left( \matrix{ x - 1 \cr m - 1 \cr} \right)}} - {1 \over {\left( \matrix{ n + x \cr ...


8

This identity is easy to deduce once you notice that $$\frac1{\binom nk}-\frac1{\binom{n+1}k}=\frac k{k+1}\frac1{\binom{n+1}{k+1}}$$ It thus follows that $$\sum_{n=k}^\infty\frac1{\binom nk}=\frac k{k-1}\sum_{n=k}^\infty\left(\frac1{\binom{n-1}{k-1}}-\frac1{\binom n{k-1}}\right)=\frac k{k-1}\frac1{\binom{k-1}{k-1}}=\frac k{k-1}$$ and better yet, $$\sum_{...


8

Euler is your friend. There is Gauss' Hypergeometric function (defined by Euler, that guy Euler was robbed, there isn't enough named after him): $${}_2 F_{1}(a,b;c;z) = 1 + \frac{a b z}{c} + \frac{a(a+1) b(b+1) z^2}{c(c+1) 2!} + \frac{a(a+1)(a+2) b(b+1)(b+2) z^3}{c(c+1)(c+2) 3!} + \ldots $$ and you are asking about the value of $${}_2 F_{1}(a,1;c;1) - 1.$...


5

The sum in question can actually be evaluated in quite an elementary way as follows $$\begin{align} \frac{a}{b}+\frac{a\cdot(a+1)}{b\cdot(b+1)}+\frac{a\cdot(a+1)\cdot(a+2)}{b\cdot(b+1)\cdot(b+2)}+\cdots &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac{(a+n)!}{(b+n)!}\\ &=\frac{(b-1)!}{(a-1)!}\sum_{n=0}^\infty\frac1{(n+a+1)\cdots(n+b)}\\ &=\frac{(b-1)...


2

\begin{align}\lambda=\frac{1+\frac{\ln a-\mu}{\sigma^2}}{a}&\implies\lambda=\frac{\sigma^2+\ln a-\mu}{\sigma^2a}\\&\implies\lambda\sigma^2a=(\sigma^2-\mu)+\ln a\\&\implies\frac1{e^{\sigma^2-\mu}}=ae^{-\lambda\sigma^2a}\\&\implies-\frac{\lambda\sigma^2}{e^{\sigma^2-\mu}}=-\lambda\sigma^2ae^{-\lambda\sigma^2a}\\&\implies W\left(-\frac{\...


1

As $a$ appears outside and inside a logarithm, there is no closed-form expression using the standard functions. This can probably be solved with $W$. https://en.wikipedia.org/wiki/Lambert_W_function


5

By the Cauchy product we have, $$\operatorname{Li}_3^2(x)=\sum_{n=1}^\infty\left(\frac{12H_n}{n^5}+\frac{H_n^{(2)}}{n^4}+\frac{2H_n^{(3)}}{n^3}-\frac{20}{n^6}\right)x^n\tag{1}$$ Divide both sides of $(1)$ by $x$ then integrate from $x=0$ to $1$ to get \begin{align} S&=\sum_{n=1}^\infty\left(\frac{12H_n}{n^6}+\frac{6H_n^{(2)}}{n^5}+\frac{2H_n^{(3)}}{n^...


1

\begin{align} S_1&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=\sum_{n=1}^\infty\frac1{n^7}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)\\ &=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{1}{n^7(n+k)^2}\right)\\ &\small{=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{7}{k^8}\left(\frac1n-\frac1{n+k}\right)-\...


4

We can solve each of the sums seperately by considering the sum as the real or imaginary part of $$\sum_{n=1}^\infty \frac{e^{inx}}n=-\ln{(1-e^{ix})},\qquad x\in\mathbb{R}\setminus \{0\}$$ Then we can use the fact that $$\ln{(z)}=\ln{(|z|)}+i\arg{(z)}$$ to seperate the above function into its real and imaginary parts; $$\begin{align} -\ln{(1-e^{ix})} &=-\...


1

$$ f(x) = - \frac{\ln(2-2\cos(x))}{2} + \arctan\left(\frac{\sin(x)}{1-\cos(x)}\right)$$


2

Partial solution to the second double sum: Lets work on the inner sum first \begin{align} S_n=\sum_{k=1}^{n+1}\frac{H_{k-\frac12}}{k}=\sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}+\frac{H_{n+\frac12}}{n+1} \end{align} by substituting $\ H_{k-\frac12}=2H_{2k}-H_k-2\ln2$, to have \begin{align} \sum_{k=1}^{n}\frac{H_{k-\frac12}}{k}&=2\sum_{k=1}^{n}\frac{H_{2k}}{...


1

Generating function of finite sum I would like to take the opportunity of this self answer to generalize the brilliant idea of Yurij S to calculate the generating function (g.f.) of the sum of interest. I suggest that a closed form of the g.f. of a finite (Euler) sum might be considered as a kind of substitute for the original closed form of the sum which ...


3

Nothing about the closed form, but a generalization that might be helpful: Consider the following function: $$f(a,s)=\int_0^1\text{arcsinh}^2\left(\sqrt{ax}\right) x^{s-1} dx$$ It is known that: $$\text{arcsinh} \sqrt{ax}=\text{arctanh} \frac{\sqrt{ax}}{\sqrt{1+ax}}$$ Using a generating function from this answer we can write: $$\text{arctanh}^2 \frac{\...


2

Yet another way: $$\sum_{1\leq k<m<n}\frac{1}{kmn(k+1)(m+1)(n+1)} = [x^3]\prod_{h\geq 1}\left(1+\frac{x}{h(h+1)}\right)=\frac{1}{\pi}[x^4]\cos\left(\frac{\pi}{2}\sqrt{1-4x}\right)$$ where the first equality follows from Viète's theorem and the second one from the Weierstrass product for the cosine function. The Maclaurin series of the RHS immediately ...


2

A third answer attempt for the lack of space in the two previous ones. I hope the community forgives me this one time. Let's try dealing with the following integral, since it all comes down to it anyway: $$R(z)=z \int_0^1 u \psi (z u) du$$ I want to use the exhaustion formula referenced in this question: $$\int_0^1f(x)\,dx=-\sum_{n=1}^\infty\sum_{m=1}^{2^...


0

\begin{eqnarray*} I=\int_{0}^{\infty} \text{erf} \left( \frac{1}{x} \right) \text{erfc} \left( \frac{1}{x} \right) \frac{dx}{x} = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{1/x} e^{-t^2} dt \int_{1/x}^{\infty} e^{-u^2} du \frac{dx}{x} \end{eqnarray*} substitute $X=1/x$ \begin{eqnarray*} I = \frac{4}{\pi} \int_{0}^{\infty} \int_{0}^{X} e^{-t^2} dt \int_{X}^{...


3

I'm not sure what definition of the error function you have, but I will use: $$\operatorname{erf}(x)=\frac{2x}{\sqrt \pi}\int_0^1 e^{-x^2z^2}dz,\quad \operatorname{erfc}(x)=\frac{2x}{\sqrt \pi}\int_1^\infty e^{-x^2y^2}dy$$ Now notice that via the substitution $x\to \frac{1}{x}$ your integral is: $$I=\int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{...


0

Here is my solution that is slightly different from that of metamorphy: As metamorphy explained in his answer, we are looking to find the value of $$V(a)=\int_{0}^{\infty}\frac{x}{2}\ln\left|\frac{1+x}{1-x}\right|e^{-ax^2}\,dx-\frac{\pi i}{2}\int_{1}^{\infty}xe^{-ax^2}\,dx =: J(a)-\frac{\pi i}{2}\int_{1}^{\infty}xe^{-ax^2}\,dx=J(a) - \frac{e^{-a}}{4a}\pi i.$...


2

Adding another answer with a different attempt, this time using the series. From one of the linked questions we find out the Taylor series representation: $$\log \Gamma(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} (1-z)^{k} +\gamma (1-z)$$ $$I(z)=\sum_{k=2}^{\infty} \frac{\zeta(k)}{k} \frac{1- (1-z)^{k+1}}{k+1} +\frac{\gamma}{2} z (2-z)$$ Comparing with the ...


2

This answer is meant to connect the ones given by @Harry Peter and @Start wearing purple, clarifying a few questions emerged in the comments. The integral of interest can be evaluated in the way pointed out by @Harry Peter, without forgetting to set some conditions on the parameters. First of all, for $\left|z\right|<1$ we can use the power series ...


7

Similarly, one may obtain the following equality, used by Apery to prove the irrationality of $\zeta(3)$: $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3 \binom{2n}{n}}=\frac25\sum_{n=1}^\infty \frac{1}{n^3}$$ Using $\arcsin^2 \sqrt{-z}=-\operatorname{arcsinh}^2z $ we get:$$S=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^32^k {2k\choose k}}=-2\int_0^{-1}\frac{\arcsin^2\...


7

One possible way is to use $$S=\sum_{k=1}^\infty \frac{x^k}{k^3 \binom{2 k}{k}}=\frac{x}{2} \, _4F_3\left(1,1,1,1;\frac{3}{2},2,2;\frac{x}{4}\right)$$ which is $$S=2 \text{Li}_3\left(-\frac{x}{2}-\frac{1}{2} i \sqrt{(4-x) x}+1\right)+4 i \text{Li}_2\left(-\frac{x}{2}-\frac{1}{2} i \sqrt{(4-x) x}+1\right) \csc ^{-1}\left(\frac{2}{\sqrt{x}}\right)+\...


8

Let's use integration by parts: $$I(z)=\int_0^z\ln\Gamma(t)~dt=z \ln\Gamma(z)-\int_0^z t \psi(t) dt$$ $$\psi(t)=\log t-\frac{1}{2t}-2 \int_0^\infty \frac{udu}{(u^2+t^2)(e^{2 \pi u}-1)}$$ $$\int_0^z t \log t dt=\frac{z^2}{4} (2 \log z-1)$$ $$\frac{1}{2}\int_0^z dt=\frac{z}{2}$$ $$2 \int_0^z \frac{t dt}{u^2+t^2}=\log \left(1+ \frac{z^2}{u^2} \right)$$ ...


2

As an addendum to the previous answer, it can be shown (through the FL-expansion of $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$) that $$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$


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