# Integral $\int_0^\infty \frac{\ln x}{(x+c)(x-1)} dx$

I've been trying to solve the following integral for days now.

$$P = \int_0^\infty \frac{\ln(x)}{(x+c)(x-1)} dx$$

with $$c > 0$$. I figured out (numerically, by accident) that if $$c = 1$$, then $$P = \pi^2/4$$. But why? And more importantly: what's the general solution of $$P$$, for given $$c$$? I tried partial fraction expansions, Taylor polynomials for $$ln(x)$$ and more, but nothing seems to work. I can't even figure out where the $$\pi^2/4$$ comes from.

(Background: for a hobby project I'm building a machine learning algorithm that predicts sports match scores. Somehow the breaking point is this integral, so solving it would get things moving again.)

• If it was me: I would do a partial fraction expansion, then integration by parts, and change of variable. YMWV! – rrogers Jul 27 at 12:51
• I would be interested in how that integral relates to machine learning. Could you give a link, or a brief comment? – thomasfermi Jul 28 at 16:54
• @thomasfermi: It's to predict sports match scores. There's a few assumptions in the model. 1. Every team has a strength s_i. 2. The probability that team i beats team j is a sigmoidal function e^(s_i)/(e^(s_i)+e^(s_j)). 3. The prior distribution of s_i is the PDF e^(s_i)/(e^(s_i)+1)^2. If you then apply Bayes' law, you can find the distribution of the full set S of all team strengths. That is, p(S|games) = p(games|S)*p(S) / p(games). If you simply take two teams, insert a 2-0 score, and try to calculate p(games), which is a double integral, then you wind up with the above integral. – Hildo Bijl Jul 29 at 22:33

$$\bbox[10pt, border:2px, lightblue]{\int_0^\infty \frac{\ln x}{(x+c)(x-1)}dx=\frac{\pi^2+\ln^2 c}{2(1+c)},\ \ c>0}$$ A nice solution can be found here due to Yaghoub Sharifi.

Perhaps it might be into your interest to see a solution for the following integral: $$I(a,b)=\int_0^\infty \frac{\ln x}{(x+a)(x+b)}dx\overset{x\to \frac{ab}{x}}=\int_0^\infty \frac{\ln\left(\frac{ab}{x}\right)}{(x+a)(x+b)}dx$$ Summing up the two integrals from above gives: $$2I(a,b)=\ln(ab)\int_0^\infty \frac{1}{(x+a)(x+b)}dx\Rightarrow \boxed{I(a,b)=\frac{\ln(ab)}{2}\frac{\ln\left(\frac{a}{b}\right)}{a-b},\ \ a,b>0}$$ One might force putting $$a=c, b=-1$$ in the above and take $$\ln(-1)=i\pi$$ (the principal value). $$\Rightarrow I(c,-1)=\frac{\ln^2 c-\ln^2 (-1)}{2(c+1)}=\frac{\pi^2 +\ln^2 c}{2(1+c)}$$

• Thanks! This is great. :-D It definitely provides me with enough inspiration for another day buried in equations. – Hildo Bijl Jul 27 at 13:22
• I don't get it how replacing $a$ with $c$ and $b=-1$ gets you from $ln(ab)ln(a/b)$ to $ln^2c-ln^2(-1)$. Because after replacing I simply get $ln(-c)ln(-c)$ which ($ln(-c)=ln(c)+ln(-1)$) is $ln^2c+2ln(c)ln(-1)+ln^2(-1)$ which gives another $2i{\pi}ln(c)$ Any explanation would be welcome. – imranfat Jul 28 at 22:11
• @imranfat I've asked a question about it here: math.stackexchange.com/questions/3306956/… hope it will be helpful. – Zacky Jul 29 at 0:03
• That's very interesting and quite coincidental. For fun I was working on $\int_0^\infty \frac{\ln x}{x^2-1}dx$, and when I subbed $x=1/t$, it lead to nothing as the integral $I$ cancels.You gave me enough food for thought for the time being. Now I can't sleep tonight. Thanks! :) – imranfat Jul 29 at 1:26
• I figured it out. Very helpful – imranfat Jul 31 at 0:20

A more general way:

Use the classic integral

$$\int_0^\infty \frac{x^p}{a+x}\;dx=-a^p\frac{\pi}{\sin(\pi p)};-1

Then

$$\int_0^\infty\frac{x^p}{(a+x)(c+x)}\;dx=\frac{\pi}{\sin(\pi p)}\frac{c^p-a^p}{c-a}$$

Now differentiate this with respect to $$p$$ and compute the limit of the result of differentiation as $$p$$ approaches $$0$$ to get

$$\int_0^\infty\frac{\ln x}{(a+x)(c+x)}\;dx=\frac{1}{2}\frac{\ln^2 c-\ln^2 a}{c-a}$$

• An interesting approach, but is there a link how to find the anti derivative of the "classic" integral? – imranfat Jul 30 at 16:19
• @imranfat There is no simple expression for the antiderivative. But if you want to compute a similar, a bit of simpler integral $\int_0^\infty \frac{x^{p-1}}{1+x}\;dx=\frac{\pi}{\sin(\pi p)}$ then the shortest way is to note that $\frac{1}{1+x}=\int_0^\infty e^{-(1+x)y}\;dy$. Then change the order of integration in double integral and use Euler's reflection formula $\Gamma(p)\Gamma(1-p)=\frac{\pi}{\sin(\pi p)}$ – Martin Gales Aug 2 at 14:50
• Ah, something with the Gamma function. I knew there had to be something to it. Thanks. – imranfat Aug 2 at 16:34