Can anyone come up with a method for integrating product of logarithms?

Question

So recently I’ve been studying integrals of products of logarithms $$\int \prod_{n=1}^N \log(x+a_n) dx$$ Can anyone come up with general method for dealing with such integral? Let's start simple, with N = 2. $$ \int \frac{\log(z+m)}{z} dz = \int \frac{\log(1-u)+\log m}{u} du = -Li_2(\frac{-z}{m}) + \log m\log z + C$$ Then we apply integration by parts to $\int \log(x+a)\log(x+b) dx$, setting $u = \log(x+b)$ and $v = (x+a)\log(x+a)-(x+b)$. Well, $x+a$ could be written as $(x+b) + (a-b)$. Yielding $$\int \log(x+a)\log(x+b) dx = (x+a)\log(x+a)\log(x+b) - (x+a)\log(x+a) - (x+b)\log(x+b) + 2x + (b-a)\int\frac{\log(x+a)}{x+b}dx$$ $$=(a-b)Li_2(\frac{x+b}{b-a}) +(x+a)\log(x+a)\log(x+b) - (x+a)\log(x+a) $$ $$-(x+(a-b)\log(a-b)+b)\log(x+b) + 2x + C$$ The case $N = 3$ is a lot more complicated. I wasn't able to come up with a solution until recently. WolframAlpha was able to give solution but cannot provide any human readable derivation to its solution. My solution to $N=3$ goes like this. We apply integration by parts to $\int\frac{log(z)log(z+m)}{z}dz$ setting $u = logz$ and $v = -Li_2(\frac{-z}{m}) + logmlogz$. Yielding $$\int\frac{log(z)log(z+m)}{z}dz = Li_3(\frac{-z}{m}) - logz Li_2(\frac{-z}{m}) + \frac{1}{2} logmlog^2z + C$$ Now, we apply integration by parts to $\int\frac{log^2z}{z+m}$ setting $u = log^2z$ and $v=log(z+m)$, hence we get $$\int\frac{log^2z}{z+m} dz = log^2zlog(z+m) - 2\int\frac{logzlog(z+m)}{z} dz = $$ $$-2Li_3(\frac{-z}{m}) + 2logz Li_2(\frac{-z}{m}) + log^2zlog(z+m) - logmlog^2z+ C$$ To keep things neat from here, $$\int log(x+a)log(x+b) dx = A(x, a, b) + C$$ $$\int\frac{log(z)log(z+m)}{z}dz = B(z, m) + C $$ $$\int\frac{log^2z}{z+m} dz = C(z, m) + C$$ We perform a substitution, $z = (y+p)/(y+q)$. $dz = (q-p)/(y+q)^2 dy$ $$\frac{dz}{z+m} = \frac{\frac{q-p}{(y+q)^2}}{\frac{(1+m)y+(mq+p)}{y+q}} dy = \frac{qdy}{(y)(y+q)} = (\frac{1}{y}-\frac{1}{y+q})dy$$ let's set $mq+p = 0 => m = \frac{-p}{q}$ $$\int(log^2(y+p)-2log(y+p)log(y+q)+log^2(y+q))(\frac{1}{y}-\frac{1}{y+q})dy = C(\frac{y+p}{y+q}, \frac{-p}{q})+C$$

$$\int\frac{log(y+p)log(y+q)}{y}dy = D(y, p, q) + C = \frac{1}{2}(-C(\frac{y+p}{y+q}, \frac{-p}{q})+C(y+p, -p) - C(y+p, q-p) + C(y+q, -q) + \frac{log^3(y+q)}{3})+B(y+q, p-q)+C$$

Finally, we apply integration by parts to $\int log(x+a)log(x+b)log(x+c)$ setting $u=log(x+a)log(x+b)log(x+c)$ and $v=x+c$ $$\int log(x+a)log(x+b)log(x+c)dx = (a-c)D(x+a, b-a, c-a) + (b-c)D(x+b, a-b, c-b) - A(x, a, b) - A(x,a, c) - A(x, b, c) + (x+c)log(x+a)log(x+b)log(x+c)+ C$$ This can be generalized to a method that reduces $\int\frac{\prod_{n=1}^N log(x+a_n)}{x}dx$ to a bunch of integrals of the form $\int\frac{log^2(y+b)\prod_{n=1}^{N-2}log(y+b_n)}{y}dy$ $$\frac{log(y+p)log(y+q)}{y} = \frac{1}{2}(log^2(y+p)+log^2(y+q) - log^2(\frac{y+p}{y+q}))(\frac{1}{y} - \frac{1}{y+q}) + \frac{log(y+p)log(y+q)}{y+q} = $$ $$\frac{q}{2}\frac{log(\frac{y+p}{y+q})}{y(y+q)}+\frac{1}{2}(log^2(y+p)+log^2(y+q))(\frac{1}{y}-\frac{1}{y+q})+\frac{log(y+p)log(y+q)}{y+q}$$ We split $\int\frac{f(y)log(y+p)log(y+q)}{y}dy$ using this. $$\int\frac{f(y)log^2(\frac{y+p}{y+q})}{y(y+q)}dy = \int\frac{f(y)log^2(\frac{y+p}{y+q})}{\frac{y}{y+q}}*\frac{dy}{(y+q)^2} = \frac{1}{q}\int_{z=\frac{y+p}{y+q}}\frac{f(\frac{-qz+p}{z-1})log^2z}{z-\frac{p}{q}}dz$$ Substitute $z = \frac{y+p}{y+q} = 1+\frac{p-q}{y+q} => y = \frac{-qz+p}{z-1}$. $dz = \frac{q-p}{(y+q)^2}dy$ $$\frac{1}{\frac{y}{y+q}}*\frac{dy}{(y+q)^2}=\frac{\frac{1}{q}}{\frac{(q-p)y}{q(y+q)}}*\frac{(q-p)dy}{(y+q)^2} = \frac{\frac{1}{q}}{\frac{y+p}{y+q}-\frac{p}{q}} dz = \frac{1}{q} \frac{1}{z-\frac{p}{q}} dz$$

We apply integration by parts to $\int\frac{f(y)log(y+p)log(y+q)}{y+q}dy$ setting $u = f(y)log(y+p)$ and $v = \frac{1}{2} log^2(y+q)$. We have $$\int\frac{f(y)log(y+p)log(y+q)}{y+q} dy = \frac{1}{2}(f(y)log(y+p)log^2(y+q) - \int f'(y) log(y+p) log^2(y+q)dy-\int\frac{f(y)log^2(y+q)}{y+p})$$

$$\int\frac{f(y)log(y+p)log(y+q)}{y}dy = \frac{1}{2} (-\int_{z=\frac{y+p}{y+q}} \frac{f(\frac{-qz+p}{z-1})log^2z}{z-\frac{p}{q}}dz + \int(log^2(y+p)-log^2(y+q))(\frac{1}{y}-\frac{1}{y+q})dy+$$ $$f(y)log(y+p)log^2(y+q) - \int f'(y) log(y+p) log^2(y+q)dy-\int\frac{f(y)log^2(y+q)}{y+p})$$ Setting $f(y) = log(y+r)$ allows us decompose $\int\frac{log(y+p)log(y+q)log(y+r)}{y}dy$ into bunch of integrals of the form $\int\frac{log(w+P)log^2(w+Q)}{w}dw$. Solving this integral will let us solve the case $N = 4$. For the case $N=4$, WolframAlpha fails to give any answer. I am also stuck when it comes to the case $N=4$

Answer 1

Using $\log(x+a)=\lim_{\alpha\rightarrow 0} {(x+a)^\alpha -1 \over \alpha}$ we can write the product as $$\lim_{\alpha\rightarrow 0}{1 \over \alpha^N}\int dx\prod((x+a_i)^\alpha -1) $$ Expanding the product we see that we need to evaluate terms of the form $$\int dx (x+a)^\alpha (x+b)^\alpha\cdots$$ and Wolfram Alpha does not supply a closed form beyond the third power, so I guess $N=4$ is still open, unless someone can find a closed solution to $$\int dx (x+a)^\alpha (x+b)^\alpha (x+c)^\alpha (x+d)^\alpha $$

Answer 2

$$I(u) = \int \prod_{n=u}^N \ln(x+a_n) \,\,dx $$

By integration by parts:

$$ I (u) = \ln(x+a_1)I (u+1)-\int \frac{I(u+1)}{x+a_1}dx = \ln(x+a_1)\left(\ln(x+a_2)I (u+2)-\int \frac{I(u+2)}{x+a_2}dx\right)-\int \frac{\ln(x+a_2)I (u+2)-\int \frac{I(u+2)}{x+a_1}dx }{x+a_1}dx $$ And we could expand this all the way to $N$, but we would have $2^N$ $I$ functions in our equation, which can be a lot

I'd say this is the idea.