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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-similiar 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 experimentating 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-similiar 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 $(1)$? Addionally, can this particular pattern be further generalized? Answers to both questions (also separately) are highly appreciated!

Thanks in advance!

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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: $$\int_1^{t}\ln\left(\frac{x^4+sx^2+t^2}{x^3+sx^2+t^2x}\right)\frac{dx}{x}=0,\quad s\in R, t>1$$

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    $\begingroup$ 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 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 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 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 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 at 16:23
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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 $$ \int_0^1f(x)\frac{\mathrm{d}x}{x+a}=\int_0^1f\!\left(\tfrac{a(1-y)}{a+y}\right)\frac{\mathrm{d}y}{y+a}\tag3 $$

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

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