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Let $a$ and $b$ be real numbers. The question is how do we compute the following anti-derivative:

\begin{equation} {\mathfrak I}^{(a,b)}(x):=\int\frac{\log(x+a) \log(x+b)^2}{x} dx=? \end{equation}

Inspired by David H's answer to The quadratic and cubic versions of a tough intregral we have found the result in case $a+b=0$. We have:

\begin{eqnarray} {\mathfrak I}^{(a,b)}(x) &=& \frac{1}{6} \underbrace{\int\frac{[\log(x+a)(x+b)]^3}{x}dx}_{I_1} + \frac{1}{6} \underbrace{\int\frac{\log(\frac{x+a}{x+b})^3}{x}dx}_{I_2}-\frac{1}{3} \underbrace{\int\frac{\log(x+a)^3}{x}dx}_{I_3} \end{eqnarray}

where we used the identity

\begin{equation} 6 a b^2 = (a+b)^3+(a-b)^3-2 a^3 \end{equation} for $a\leftarrow (x+a)$ and $b \leftarrow (x+b)$. Now we have: \begin{eqnarray} I_1&\underbrace{=}_{u=(x+a)(x+b)}& 2\int \frac{\log(u)^3}{-(a+b)\pm \sqrt{(a-b)^2+4 u}}\cdot\frac{\pm du}{\sqrt{(a-b)^2+4 u}} \\ &\underbrace{=}_{a+b=0}& 2\int\frac{\log(u)^3}{(a-b)^2+4 u}du\\ &=&-\frac{1}{2} \sum\limits_{j=1}^4 \log(u)^{4-j} (-1)^{j-1} 3_{(j-1)} \cdot Li_j(-\frac{4}{(a-b)^2} u) \end{eqnarray} The remaining integrals are simpler and in their evaluation we do not have to assume that $a+b=0$. We have: \begin{eqnarray} I_2&\underbrace{=}_{u=(x+a)/(x+b)}&\sum\limits_{j=1}^4 \log(u)^{4-j} (-1)^{j-1} 3_{(j-1)} \cdot \left( Li_j(u)-Li_j(\frac{b}{a} u)\right)\\ I_3&\underbrace{=}_{u=x+a}&-\sum\limits_{j=1}^4 \log(u)^{4-j} (-1)^{j-1} 3_{(j-1)} \cdot Li_j(\frac{1}{a} u) \end{eqnarray} Bringing everything together we have: \begin{eqnarray} &&6{\mathfrak J}^{(a,b)}(x)= \sum\limits_{j=1}^4 (-1)^{j-1} 3_{(j-1)} \cdot \\ &&\left( -\frac{1}{2} \text{Li}_j\left(\frac{a^2-x^2}{a^2}\right) \log ^{4-j}\left(x^2-a^2\right)+\right.\\ &&\left. \left(\text{Li}_j\left(\frac{a+x}{x-a}\right)-\text{Li}_j\left(-\frac{a+x}{x-a}\right)\right) \log ^{4-j}\left(\frac{a+x}{x-a}\right)+\right.\\ &&\left.2 \text{Li}_j\left(\frac{a+x}{a}\right) \log ^{4-j}(a+x) \right) \end{eqnarray} Now let us assume that $a\ge 0$ .Then we get a closed form for the following definite integral: \begin{eqnarray} &&(-6)\int\limits_0^1 \frac{\log(\frac{1}{t+a}) \log(t+a+2)^2}{t+a+1} dt=\\ && \sum\limits_{j=1}^4 (-1)^{j-1} 3_{(j-1)}\left(\right.\\ &&\left. \frac{1}{2} \left(\text{Li}_j(-a (a+2)) \log ^{4-j}(a (a+2))-\text{Li}_j((-a-1) (a+3)) \log ^{4-j}((a+1) (a+3))\right)+\right.\\ &&\left. \left(\text{Li}_j\left(\frac{a+1}{a+3}\right)-\text{Li}_j\left(-\frac{a+1}{a+3}\right)\right) \log ^{4-j}\left(\frac{a+1}{a+3}\right)-\left(\text{Li}_j\left(\frac{a}{a+2}\right)-\text{Li}_j\left(-\frac{a}{a+2}\right)\right) \log ^{4-j}\left(\frac{a}{a+2}\right)+\right.\\ &&\left. 2 \left(\text{Li}_j(-a-1) \log ^{4-j}(a+1)-\text{Li}_j(-a) \log ^{4-j}(a)\right)\right.\\ &&\left. \right) \end{eqnarray} Now by taking the limit $a\rightarrow 0$ we get: \begin{eqnarray} &&\int\limits_0^1 \frac{\log(1/t) \log(t+2)^2}{t+1}dt=\\ &&-\frac{\text{Li}_4\left(-\frac{1}{3}\right)}{2}+\text{Li}_4\left(\frac{1}{3}\right)-\frac{1}{4} \left(\text{Li}_2\left(-\frac{1}{3}\right)-2 \text{Li}_2\left(\frac{1}{3}\right)\right) \log ^2(3)-\frac{1}{2} \left(\text{Li}_3\left(-\frac{1}{3}\right)-2 \text{Li}_3\left(\frac{1}{3}\right)\right) \log (3)-\frac{7 \pi ^4}{720}+\frac{\log ^4(3)}{16} \end{eqnarray}

Now, having said all this the ultimate question is of course how do we proceed in the case $a+b\ne 0$. Can in that case the integral be reduced to poly-logarithms as well or some new functions need to be introduced?

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