While dealing with a definite integral on AoPS I discovered (I have to admit by pure chance) the following relation

$$\int_0^1\log\left(\frac{(x+1)(x+2)}{x+3}\right)\frac{\mathrm dx}{1+x}~=~0\tag1$$

The proof is quite easy, but feels kind of contrived. Indeed, just apply a self-similar substitution - $x\mapsto\frac{1-x}{1+x}$ - to the auxiliary integral $\mathcal I$ given by

$$\mathcal I~=~\int_0^1\log\left(\frac{x^2+2x+3}{(x+1)(x+2)}\right)\frac{\mathrm dx}{1+x}$$

And the result follows. However, to consider precisely this integral seems highly unnatural to me (in fact, as I mentioned earlier, this integral was just a by-product while evaluating something quite different and I discovered $(1)$ when experimenting with various substitutions).

The crucial point to notice concerning $\mathcal I$ is the invariance of the polynomial $f(x)=x^2+2x+3$ regarding the self-similar substitution which allows us to deduce $(1)$. Additionally for myself I am quite surprised by the special structure of $(1)$ since we have factors of the form $(x+1)$, $(x+2)$ and $(x+3)$ combined which calls for a generalization (although I found none yet).

It there a more elementary approach, not relying on such an "accident" like examining the integral $\mathcal I$ for proving $(1)$? Additionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated!

Thanks in advance!

  • $\begingroup$ The $\displaystyle\log$ argument numerator -in the $\displaystyle\mathcal{I}$ definition- must be $\displaystyle x^{2} + 3x + 2$ instead of $\displaystyle x^{2} + 2x + 3$. $\endgroup$ – Felix Marin Aug 27 '20 at 3:27
  • $\begingroup$ @FelixMarin No. By my calculations $\left(\frac{1-x}{1+x}\right)^2+2\left(\frac{1-x}{1+x}\right)+3=2\frac{x^2+2x+3}{(1+x)^2}$ which is what the approach is all about. But $\left(\frac{1-x}{1+x}\right)^2+3\left(\frac{1-x}{1+x}\right)+2=2\frac{x+3}{(x+1)^2}$ which yields an interesting other identity but not the one the post is concerned with. $\endgroup$ – mrtaurho Aug 27 '20 at 3:42

That's quite an impressive method to show that the integral vanishes.

For the first part I'll show using a different approach that your integral vanishes. $$\mathcal J=\int_0^1 \ln\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm dx}{x+1}\overset{x+1=t}= \color{blue}{\int_1^2\ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}-\color{red}{\int_1^2 \frac{\ln t}{t}\mathrm dt}$$ Let's denote the blue integral as $\mathcal J_1$ then using the substitution $\frac{2}{t}\to t$ we get: $$\mathcal J_1=\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}=\int_1^2 \ln\left(\frac{2(t+1)}{t+2}\right)\frac{\mathrm dt}{t}$$ Adding both integrals from above gives us: $$\require{cancel} 2\mathcal J_1=\cancel{\int_1^2 \ln\left(\frac{t+2}{t+1}\right)\frac{\mathrm dt}{t}}+\int_1^2 \frac{\ln 2}{t}\mathrm dt+\cancel{\int_1^2 \ln\left(\frac{t+1}{t+2}\right)\frac{\mathrm dt}{t}}=\ln^2 2$$ $$\Rightarrow \mathcal J_1=\color{blue}{\frac{\ln^2 2}{2}}\Rightarrow \mathcal J=\color{blue}{\frac{\ln^2 2}{2}}-\color{red}{\frac{\ln^2 2}{2}}=0$$

As for the second part, a small generalization outcomes by experimenting with the blue integral.

In particular, by the same approach we have: $$\int_1^a \ln\left(\frac{x+a}{x+1}\right)\frac{\mathrm dx}{x}=\int_1^a \frac{\ln x}{x}\mathrm dx$$ Which gives us a small generalization: $${\int_0^{a-1}\ln\left(\frac{x+a+1}{(x+1)(x+2)}\right)\frac{\mathrm dx}{x+1}=0}$$ Similarly, (with the substitution $\frac{ab}{x}\to x$) we get that: $$\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)=\int_a^b \ln\left(\frac{x}{a}\right)\frac{dx}{x}$$ And the following follows: $$\int_{a-1}^{b-1} \ln\left(\frac{a(x+b+1)}{(x+1)(x+a+1)}\right)\frac{dx}{x+1}=0$$ One might be interested in the following similar generalization too: $$\int_1^{t}\ln\left(\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\right)\frac{dx}{x}=0,\quad s\in R, t>1$$

  • 1
    $\begingroup$ This is what I was looking for; I guess^^ (+1) anyway and I'm looking forward to see a generalization (if possible). I guess you know where I got this integral from :D $\endgroup$ – mrtaurho Aug 12 '19 at 15:17
  • $\begingroup$ @mrtaurho I got there a small generalization for now. However since $\int_a^b \ln\left(\frac{x+b}{x+a}\right)\frac{dx}{x}=\frac12 \ln^2 \left(\frac{b}{a}\right)$ it might be possible to obtain a better one. (I'll try later to work with it). $\endgroup$ – Zacky Aug 12 '19 at 15:38
  • $\begingroup$ It's hard to write out your name now :P But yes, this seems promising, I'm curious! As I noted above it was a rather strange by-product to discover this equality; and it was tedious to ran into it three times while hoping for something more helpful for solving the original task^^' $\endgroup$ – mrtaurho Aug 12 '19 at 15:44
  • $\begingroup$ @mrtaurho just in case you missed it in winter I'll mention that $\mathcal I$ (before the self-similar sub was applied) originates from this generalization: math.stackexchange.com/a/3049039/515527. Aka: $$\int_1^{\sqrt{t}}\ln\left(\frac{x^4+sx^2+t}{x(x^2+sx+t)}\right)\frac{dx}{x}=0$$ $\endgroup$ – Zacky Aug 12 '19 at 16:06
  • $\begingroup$ As I've upvoted both, the question you linked and your answer, I guess I've seen it at some point. But I'll take a look at it again :) $\endgroup$ – mrtaurho Aug 12 '19 at 16:23

The Answer

I have used the substitution $(x+1)(y+1)=2$ before to good effect because $$ \int_0^1f(x)\,\frac{\mathrm{d}x}{x+1}=\int_0^1f\!\left(\tfrac{1-y}{1+y}\right)\frac{\mathrm{d}y}{y+1}\tag1 $$ If $f(x)=\log\left(\frac{x+3}{(x+2)(x+1)}\right)$, then $f\!\left(\frac{1-y}{1+y}\right)=\log\left(\frac{(y+2)(y+1)}{y+3}\right)$. Therefore $$ \int_0^1\log\left(\frac{x+3}{(x+2)(x+1)}\right)\frac{\mathrm{d}x}{x+1}=\int_0^1\log\left(\frac{(y+2)(y+1)}{y+3}\right)\frac{\mathrm{d}y}{y+1}\tag2 $$ and since the two sides of $(2)$ are negatives, they are both $0$.

A Generalization

We can generalize $(1)$ by letting $(x+a)(y+a)=a(1+a)$, then we get $$ \int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3 $$

  • $\begingroup$ (+1) Interesting as well. $\endgroup$ – mrtaurho Aug 12 '19 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.