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What would you recommend me for the integral below? $$ \int_{0}^{1}\ln(1 - x)\ln(x) \ln(1 + x)\,\mathrm dx $$ For instance, for the version without the last logarithm would work to use Taylor series, but in this case things are a bit more complicated and it doesn't seem to work.

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    $\begingroup$ This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$ – Carl Mummert Dec 19 '13 at 12:02
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I get (and verified by Mathematica) $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 - \log^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$


Transforming to a double sum.

As in Marvis's answer, take \begin{align*} \log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\ & = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}. \end{align*} Using the integral $$\int_0^1 x^r \log x \, dx = - \frac{1}{(1+r)^2},$$ we have $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = \sum_{k=1}^{\infty} \sum_{m=1}^{\infty} \frac{(-1)^{k+1}}{km(k+m+1)^2}.$$


Evaluating the inner sum.

Partial fractions decomposition on the summand in $m$ yields \begin{align*} \sum_{m=1}^{\infty} \frac{1}{m(k+m+1)^2} &= \sum_{m=1}^{\infty} \left(\frac{1}{(k+1)^2 m} - \frac{1}{(k+1)^2 (m+k+1)} - \frac{1}{(k+1) (m+k+1)^2}\right) \\ &= \sum_{m=1}^{k+1} \frac{1}{(k+1)^2 m} - \sum_{m=k+2}^{\infty} \frac{1}{(k+1)m^2} \\ &= \frac{H_{k+1}}{(k+1)^2} - \frac{\zeta(2)}{k+1} + \frac{H^{(2)}_{k+1}}{k+1}, \end{align*} where $H^{(r)}_n = \sum_{i=1}^n i^{-r}$, the $n$th $r$-harmonic number.

Now we're left with evaluating $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1} \zeta(2)}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)}. \tag{1}$$ We take the three sums in Eq. (1) in turn.


The first sum in Eq. (1).

For the first sum, applying partial fractions decomposition yields \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k(k+1)^2} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_{k+1}}{(k+1)^2} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} - 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} - 1 \\ &= -2 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} \\ &= -3 + 2 \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2}, \end{align*} where in the last step we used the evaluation for the second sum we're about to do.


The second sum in Eq. (1).

For the second sum in Equation (1), applying partial fractions decomposition yields \begin{align*} \zeta(2) \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)} &= \zeta(2) \left(\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} + \sum_{k=1}^{\infty}\frac{(-1)^{k+2}}{k+1}\right) \\ &= \zeta(2) \left(2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} -1\right) \\ &= 2 \zeta(2) \log 2 - \zeta(2). \end{align*}


The third sum in Eq. (1).

For the third sum in Equation (1), applying partial fractions decomposition yields \begin{align*} \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k(k+1)} &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_{k+1}}{(k+1)} \\ &= \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k(k+1)^2} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} - 1 \\ &= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k+1} - \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{(k+1)^2} \\ &= -1 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} + 2 \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} - 1 + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} - 1\\ &= -3 + 2 \log 2 + \frac{1}{2} \zeta(2) + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k} ,\\ \end{align*} where in the last step we use the identity, for $p > 1$, $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^p} = (1 - 2^{1-p}) \zeta(p)$$


Combining the results.

Putting all this together, we have that the integral evaluates to $$ -6 + 4 \log 2 + \frac{3}{2} \zeta(2) - 2 \zeta(2) \log 2 + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k} + \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H_k}{k^2} + 2\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H^{(2)}_k}{k}.$$


Evaluating the alternating Euler sums.

The three remaining sums all have similar forms. Letting $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ we have $A(1,1)$, $A(1,2)$, and $A(2,1)$ left to evaluate. Sums of the form $A(p,q)$ are known as alternating Euler sums. Euler sums and their close variants are an ongoing research area. These three, though, seem like the three simplest alternating Euler sums, and one would think that there would be relatively simple ways to evaluate them. However, despite a decent amount of work today I could not find simple proofs for any of them, either on my own or in the literature.

At any rate, we have $A(1,1) = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$, $A(1,2) = \frac{5}{8} \zeta(3)$, and $A(2,1) = \zeta(3) - \frac{1}{2}\zeta(2) \log 2$.

(Here's a link to my question where I ask for a proof for the $A(1,1)$, $A(1,2)$, and $A(2,1)$ evaluations. There are multiple nice answers there. In particular, Robjohn's three answers show how to do the evaluations by manipulating the sums, the same approach I was trying here.)


Finally.

Therefore, $$\int_0^1 \log(1-x) \log x \log(1+x) \, dx = -6 + 4 \log 2 - \log^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \log 2 + \frac{21}{8} \zeta(3).$$

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  • $\begingroup$ My answer at the end agrees with the Mathematica output given by @nbubis. $\endgroup$ – Mike Spivey Jan 11 '13 at 5:50
  • $\begingroup$ You can find some ideas for evaluating those alternating Euler sums in my recent posting, though it is not as powerful as the answer by Marvis. $\endgroup$ – Sangchul Lee Jan 11 '13 at 18:03
  • $\begingroup$ brilliant! this should be taught in textbooks. $\endgroup$ – nbubis Jan 11 '13 at 18:06
  • $\begingroup$ @sos440: That's a nice post. It's interesting that you take the derivation to the same place as in my answer here; i.e., up to the point of needing to evaluate the alternating Euler sums. I like that you give a detailed discussion of evaluating those, too. I'll have to take a closer look at the details later. In fact, I think I'm going to print out that entire document you link to; it all looks interesting. $\endgroup$ – Mike Spivey Jan 11 '13 at 18:11
  • $\begingroup$ @MikeSpivey, Thanks! I admit that my document is still far from completeness and shows many typos and errors. Anyway I'm really glad that you like it. $\endgroup$ – Sangchul Lee Jan 11 '13 at 18:20
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\begin{align} \log(1+x)\log(1-x) & = \left(\sum_{k=1}^{\infty} \dfrac{(-x)^k}k \right)\left(\sum_{k=1}^{\infty} \dfrac{x^k}k \right)\\ & = \sum_{k=1}^{\infty}\sum_{m=1}^{\infty} \dfrac{(-1)^k x^{k+m}}{km}\\ & = \sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l \\ & = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^k}{k(2l-k)} \right) x^{2l} \end{align} Now $$\int_0^1 x^{2l} \log(x) dx = -\dfrac1{(2l+1)^2}$$ Hence, the integral is $$\sum_{l=2}^{\infty} \left(\sum_{k=1}^{l-1} \dfrac{(-1)^k}{k(l-k)} \right) x^l = \sum_{l=1}^{\infty} \left(\sum_{k=1}^{2l-1} \dfrac{(-1)^{k-1}}{k(2l-k)} \right)\dfrac1{(2l+1)^2}$$

P.S: Bit too long for a comment.

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  • $\begingroup$ I think you're missing a minus sign in the first line. $\endgroup$ – Ron Gordon Jan 9 '13 at 21:42
  • $\begingroup$ @rlgordonma I think it is fine. $(-1)^{k-1} \times (-1) = (-1)^k$ $\endgroup$ – user17762 Jan 9 '13 at 21:44
  • $\begingroup$ Ah - my mistake. Thanks. $\endgroup$ – Ron Gordon Jan 9 '13 at 21:48
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Mathematica gives for the final result: $$\frac{21 \zeta (3)}{8}-6-\log ^2(2)+\log (16)-\frac{1}{12} \pi ^2 (\log(64)-5)$$ The indefinite integral is really really long..

Regarding a series what's wrong with a Maclaurin series around, say, $x=1/e$?

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  • $\begingroup$ Your best bet is to do a Maclurin series for $\log{(1-x)} \log{(1+x)}$ and perform the resulting integrals to get an infinite series which, I guess, comes out to what you wrote. I have evaluated $\int_0^1 dx \: \log{(1-x)} \log{x} = 2 - \frac{\pi^2}{6}$ in a similar manner. $\endgroup$ – Ron Gordon Jan 9 '13 at 21:35
  • $\begingroup$ @Chris'ssister - no, for $\ln(1−x)\ln(x)\ln(1+x)$. $\endgroup$ – nbubis Jan 9 '13 at 21:59
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Reduce the integral to manageable components via IBP’s \begin{align} I &= \int_0^1\ln(1-x)\ln x\ln(1+x) dx = -\int_0^1 x d[\ln(1-x)\ln x\ln(1+x)] \\ &= \int_0^1 \bigg( - \ln x\ln(1-x) - \ln x\ln(1+x) - \ln (1-x)\ln(1+x) \\ & \>\>\>\>\> \>\>\>\>\>\>\>\> +\frac{\ln x\ln(1+x)}{1-x} + \frac{\ln x\ln(1-x)}{1+x} \bigg) dx\\ &= -I_1 - I_2 -I_3 +I_4 + I_5\tag1 \end{align}

Evaluate each component integral

\begin{align} I_1 &= \int_0^1 \ln x\ln(1-x) dx =- \int_0^1 x d[\ln x\ln(1-x)]\\ &= -2\int_0^1 \ln xdx + \int_0^1 \frac{\ln x}{1-x}dx =2-\zeta(2) \\ \\ I_2 &= \int_0^1 \ln x\ln(1+x) dx =- \int_0^1 x d[\ln x\ln(1+x)]\\ &= -\int_0^1 (\ln x +\ln(1+x))dx + \int_0^1 \frac{\ln x}{1+x}dx =2-2\ln2-\frac{\zeta(2)}2 \\ \\ I_3 &= \int_0^1 \ln (1-x)\ln(1+x)dx =-\ln2 - \int_0^1 x d[\ln (1-x)\ln\frac{1+x}2]\\ &= -\ln2 - \int_0^1 \left( \ln\frac{1+x}2 -\ln(1-x) - \frac{\ln\frac{1+x}2}{1-x}+\frac{\ln(1-x)}{1+x} \right)dx \\ &=2-2\ln2+2\int_0^1\frac{\ln(1-x)}{1+x}dx=2-2\ln2+\ln^22-\zeta(2)\\ \\ I_4 &= \int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx = \zeta(3)-\frac32\ln2\zeta(2) \\ & \>\>\>\>\> J(a) = \int_0^1\frac{\ln x}{1+x}\left( \frac{ \ln(1-ax)}x+\ln(1-x) \right)dx\\ &\>\>\>\>\> J(1) = \int_0^1 \frac{\ln x\ln(1-x)}{x}dx = \frac12\int_0^1 \frac{\ln^2x}{1-x}dx= \zeta(3) \\ &\>\>\>\>\> J’(a) = \int_0^1\frac{-\ln xdx}{(1+x)(1-ax)} = -\frac1{1+a}\int_0^1 \left(\frac{\ln x}{1+x}+ \frac{a\ln x}{1-ax} \right)dx \\ &\>\>\>\>\>\>\>\>\>\>\>= \frac{\zeta(2)}{2(1+a)}- \frac1{1+a}\int_0^a\frac{\ln y}{1-y}dy - \frac{\ln a\ln(1-a)}{1+a} \\ &\>\>\>\>\>J(1) =J(0)+ \int_0^1 J’(a)da = \frac{\zeta(2)}2\ln2-\int_0^1 \frac1{1+a}\int_0^a \frac{\ln y}{1-y}dy \\ &\>\>\>\>\>\>\>\>\>\> \overset{IBP}= \frac{3}2\ln2\zeta(2)+ I_4 =\zeta(3) \\ \\ I_5 &= \int_0^1 \frac{\ln x\ln(1-x)}{1+x}dx = \int_0^1 \ln x\ln(1-x)d[\ln(1+x)] \\ &= I_4 - \int_0^1 \frac{\ln (1+x)\ln(1-x)}{x}dx \\ &= I_4 -\frac14\int_0^1 \frac{\ln^2(1-x^2)}xdx +\frac14 \int_0^1 \frac{\ln^2\frac{1-x}{1+x}}xdx\\ &=I_4 -\frac18\int_0^1 \frac{\ln^2(1-t)}t dt + \frac12\int_0^1 \frac{\ln^2t}{1-t^2}dt \\ &= I_4 - \frac18\int_0^1 \frac{\ln^2t}{1-t}dt +\frac7{16} \int_0^1 \frac{\ln^2t}{1-t}dt =I_4 +\frac58\zeta(3)\\ \end{align} Plug above individual results into (1) to obtain

$$I = -6 + 4 \ln 2 - \ln^2 2 + \frac{5}{2} \zeta(2) - 3\zeta(2) \ln 2 + \frac{21}{8} \zeta(3)$$

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$$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$ By using the the algebraic identity $ab=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2$ we get:

$$I=\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x^2\right)\:dx}_{t=x^2}-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\:dx$$ $$=\underbrace{\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt}_{I_1}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{I_2}+\frac{1}{2}\underbrace{\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx}_{I}-\underbrace{\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx}_{I_3}$$ $$\frac{1}{2}I=\frac{1}{16}I_1-\frac{1}{4}I_2-\frac{1}{4}I_3$$


$$I_1=\frac{1}{16}\int _0^1\frac{\ln \left(t\right)\ln ^2\left(1-t\right)}{\sqrt{t}}\:dt=\frac{1}{16}\lim_{\alpha\rightarrow 1/2\\\beta\rightarrow 1}\frac{\partial ^3}{\partial \alpha \partial \beta ^2}\text{B}\left(\alpha ,\beta \right)$$ $$=\frac{7}{4}\zeta \left(3\right)-6-\ln ^2\left(2\right)+4\ln \left(2\right)+2\zeta \left(2\right)-\frac{3}{2}\ln \left(2\right)\zeta \left(2\right)$$


$$I_2=-\frac{1}{4}\underbrace{\int _0^1\ln \left(x\right)\ln ^2\left(1-x\right)\:dx}_{t=1-x}=-\frac{1}{4}\int _0^1\ln \left(1-t\right)\ln ^2\left(t\right)\:dt=\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)^3}$$ $$=-\frac{1}{2}\zeta \left(3\right)+\frac{3}{2}-\frac{1}{2}\zeta \left(2\right)$$


For this one make use of $\ln ^2\left(1-x\right)=2\sum _{k=1}^{\infty }\left(\frac{H_k}{k}-\frac{1}{k^2}\right)x^k$. $$I_3=-\frac{1}{4}\int _0^1\ln \left(x\right)\ln ^2\left(1+x\right)\:dx$$ $$=-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k}\int _0^1x^k\ln \left(x\right)\:dx+\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2}\int _0^1x^k\ln \left(x\right)\:dx$$ $$=\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k\left(k+1\right)^2}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k^2\left(k+1\right)^2}$$ $$=\frac{1}{16}\zeta \left(3\right)-\frac{1}{4}\zeta \left(2\right)+\frac{1}{2}\ln ^2\left(2\right)+\frac{3}{2}-2\ln \left(2\right)$$


Collecting the results after multiplying by $2$ we have: $$I=\int _0^1\ln \left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)\:dx$$ $$=\frac{21}{8}\zeta \left(3\right)-3\ln \left(2\right)\zeta \left(2\right)+\frac{5}{2}\zeta \left(2\right)-\ln ^2\left(2\right)-6+4\ln \left(2\right)$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{1}\ln\pars{1 - x}\ln\pars{x} \ln\pars{1 + x}\,\dd x} \\[5mm] = &\ {1 \over 2}\,\int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 - x^{2}}\,\dd x - {1 \over 2}\int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 - x}\,\dd x \\[2mm] & \!\!\!\!\! - {1 \over 2}\int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 + x}\,\dd x \\[5mm] = &\ {1 \over 8}\ \overbrace{\int_{0}^{1}x^{-1/2}\ln\pars{x}\ln^{2}\pars{1 - x} \,\dd x}^{\ds{I_{1}}}\ -\ {1 \over 2}\ \overbrace{\int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 - x}\,\dd x} ^{\ds{I_{2}}} \\[2mm] & \!\!\!\!\!\!\! -\,{1 \over 2}\ \underbrace{\int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 + x}\,\dd x} _{\ds{I_{3}}}\ = \ {1 \over 8}\,I_{1} - {1 \over 2}\,I_{2} - {1 \over 2}\, I_{3} \label{1}\tag{1} \end{align} $\ds{I_{1}}$ and $\ds{I_{2}}$ are tipically evaluated via the use of Beta and/or Gamma Function derivatives as follows: \begin{equation} \left\{\begin{array}{rcl} \ds{I_{1}} & \ds{=} & \ds{{\partial^{3} \over \partial\mu\,\partial\nu^{2}} \bracks{\Gamma\pars{\mu + 1/2}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 3/2}}_{\ \mu\ =\ 0,\ \nu\ =\ 0}} \\ & \ds{=} &\ds{-96 + {16\pi^{2} \over 3} + 64\ln\pars{2} - 4\pi^{2}\ln\pars{2} -16\ln^{2}\pars{2} + 28\zeta\pars{3}} \\[2mm] \ds{I_{2}} & \ds{=} & \ds{{\partial^{3} \over \partial\mu\,\partial\nu^{2}} \bracks{\Gamma\pars{\mu + 1}\Gamma\pars{\nu + 1} \over \Gamma\pars{\mu + \nu + 2}}_{\ \mu\ =\ 0,\ \nu\ =\ 0} = -6 + {\pi^{2} \over 3} + 2\zeta\pars{3}} \end{array}\right.\label{2}\tag{2} \end{equation} For the the being, I have the partial value: \begin{equation} {1 \over 8}\,I_{1} - {1 \over 2}\,I_{2} = -9 + {\pi^{2} \over 2} + 8\ln\pars{2} - {\pi^{2} \over 2}\,\ln\pars{2} - 2\ln^{2}\pars{2} + {5\zeta\pars{3} \over 2} \end{equation}


$\ds{\large I_{3}\ \mbox{Evaluation:}}$ \begin{align} I_{3} & \equiv \int_{0}^{1}\ln\pars{x}\ln^{2}\pars{1 + x}\,\dd x = \int_{1}^{2}\ln\pars{x - 1}\ln^{2}\pars{x}\,\dd x \\[5mm] & = \int_{1}^{1/2}\ln\pars{{1 \over x} - 1}\ln^{2}\pars{1 \over x} \pars{-\,{\dd x \over x^{2}}} \\[5mm] & = \color{red}{\int_{1/2}^{1}{\ln\pars{1 - x}\ln^{2}\pars{x} \over x^{2}} \,\dd x} - \int_{1/2}^{1}{\ln^{3}\pars{x} \over x^{2}}\,\dd x \\[5mm] & \mbox{Note that}\ \int_{1/2}^{1}{\ln^{3}\pars{x} \over x^{2}}\,\dd x = 6 - 12\ln\pars{2} + 6\ln^{2}\pars{2} - 2\ln^{3}\pars{2} \\ & \mbox{while the}\ \color{red}{\mbox{remaining integral}}\ \mbox{is evaluated with a}\ "Feynman\ Trick". \end{align} Namely, \begin{align} &\totald{}{a}\color{red}{\int_{1/2}^{1} {\ln\pars{1 - \color{black}{a}x}\ln^{2}\pars{x} \over x^{2}}\,\dd x} = -\int_{1/2}^{1}{\ln^{2}\pars{x} \over x\pars{1 - ax}}\,\dd x = -\int_{a/2}^{a}{\ln^{2}\pars{x/a} \over x\pars{1 - x}}\,\dd x \\[5mm] = & -\int_{a/2}^{a}{\ln^{2}\pars{x/a} \over x}\,\dd x -\int_{a/2}^{a}{\ln^{2}\pars{x/a} \over 1 - x}\,\dd x \\[5mm] = &\ -\ \underbrace{\int_{1/2}^{1}{\ln^{2}\pars{x} \over x}\,\dd x} _{\ds{{1 \over 3}\,\ln^{3}\pars{2}}}\ -\ \ln\pars{1 - {a \over 2}}\ln\pars{1 \over 2} + 2\int_{a/2}^{a}\mrm{Li}_{2}'\pars{x}\ln\pars{x \over a}\,\dd x \\[5mm] = &\ -\,{1 \over 3}\ln^{3}\pars{2} + \ln\pars{2}\ln\pars{1 - {a \over 2}} + 2\ln\pars{2}\,\mrm{Li}_{2}\pars{a \over 2} - 2\,\mrm{Li}_{3}\pars{a} + 2\,\mrm{Li}_{3}\pars{a \over 2} \end{align} Integrating the last expression over $\ds{a \in \pars{0,1}}$:

$\ds{\large\tt To\ be\ continued\ldots}$

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