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I was challenged to prove this identity $$\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx=\frac{\pi}{4}\left(2+\sqrt{6}\sqrt{3-\sqrt{5}}\right).$$ I was not successful, so I want to ask for your help. Can it be somehow related to integrals listed in that question?

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  • $\begingroup$ A related question. $\endgroup$
    – Lucian
    Mar 6, 2014 at 6:05
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    $\begingroup$ I got this result from a CAS. By the way, the rhs write $\frac{\pi}{4} \left(2-\sqrt{3}+\sqrt{15}\right)$ in which appear part of the exponents. $\endgroup$ Mar 6, 2014 at 6:29
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    $\begingroup$ I am curious. Who challenged you? $\endgroup$
    – Potato
    Mar 6, 2014 at 6:29
  • $\begingroup$ See my answer below. It is pretty simple. What you need is substitution and you do not need use other things. $\endgroup$
    – xpaul
    Aug 15, 2014 at 14:50

5 Answers 5

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This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals.

Assume $a,b\in\mathbb{R}$. Note that $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\frac{1+x^b}2\right)}{\ln x}\frac{dx}{1+x^2}.\tag1$$ Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them: $$\begin{align}&\phantom=\underbrace{\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{split the region}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{change variable}\ y=1/x}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^0\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln\left(y^{-1}\right)}\frac1{1+y^{-2}}\left(-\frac1{y^2}\right)dy}_\text{flip the bounds and simplify}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\underbrace{\int_0^1\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln y}\frac{dy}{1+y^2}}_{\text{rename}\ y\ \text{to}\ x}\\&=\underbrace{\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^1\frac{\ln\left(\frac{1+x^{-a}}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{combine logarithms}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}=\underbrace{\int_0^1\frac{\ln\left(\frac{x^a\left(x^{-a}+1\right)}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{cancel}\ \ 1+x^{-a}}\\&=\int_0^1\frac{\ln\left(x^a\right)}{\ln x}\frac{dx}{1+x^2}=a\int_0^1\frac{dx}{1+x^2}=a\,\Big(\arctan1-\arctan0\Big)\\&=\vphantom{\Bigg|^0}\frac{\pi\,a}4.\end{align}\tag2$$ So, finally, $$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$

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    $\begingroup$ A very simple answer to a seemingly complicated problem. +1 $\endgroup$
    – Paramanand Singh
    Mar 7, 2014 at 3:04
  • $\begingroup$ What's up with the two in the denominator? $\endgroup$
    – chubakueno
    Mar 7, 2014 at 4:06
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    $\begingroup$ @chubakueno To avoid divergence at $x=1$. $\endgroup$ Mar 7, 2014 at 4:09
  • $\begingroup$ Thank you! I undertand it better now. $\endgroup$
    – chubakueno
    Mar 7, 2014 at 18:35
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    $\begingroup$ @Kugelblitz It is a bound variable. It can have any name provided that it does not clash with names of other variables in scope. $\endgroup$ Nov 12, 2015 at 17:39
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Let the considered integral be $I$. Just to make it easier to write, let $4+\sqrt{15}=a$ and $2+\sqrt{3}=b$. Use the substitution $x=\tan\theta$ to get: $$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)}{\ln\tan\theta}\,d\theta$$ Next, use the substitution $\theta=\pi/2-t$ to obtain: $$I=\int_0^{\pi/2} \frac{\ln\left(\dfrac{1+(\tan t)^a}{(1+(\tan t)^b)(\tan t)^{a-b}}\right)}{\ln\cot t}\,dt=\int_0^{\pi/2} \frac{-\ln\left(\dfrac{1+(\tan\theta)^a}{1+(\tan\theta)^b}\right)+(a-b)\ln\tan\theta}{\ln\tan\theta}\,d\theta$$ where I used $\ln(\tan(\pi/2-\theta))=\ln(\cot\theta)=-\ln(\tan\theta)$.

Add the two expressions for I and notice that you are left with: $$2I=\int_0^{\pi/2} \frac{(a-b)\ln \tan\theta}{\ln \tan\theta}\,d\theta=\frac{\pi}{2}(a-b)$$ $$I=\frac{\pi}{4}(a-b)$$ Therefore, $$\boxed{I=\dfrac{\pi}{4}(2+\sqrt{15}-\sqrt{3})}$$

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I have a short way to solve this problem. Let $x=\frac{1}{u}, a=4+\sqrt{15},b=2+\sqrt{3}$. Then \begin{eqnarray*} I&=&-\int_\infty^0\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{-2})\ln(u^{-1})}u^{-2}du\\ &=&-\int_0^\infty\frac{\ln\frac{1+u^{-a}}{1+u^{-b}}}{(1+u^{2})\ln u}du\\ &=&-\int_0^\infty\frac{\ln \left(u^{b-a}\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du\\ &=&-\int_0^\infty\frac{\ln \left(\frac{1+u^{a}}{1+u^{b}}\right)}{(1+u^{2})\ln u}du-\int_0^\infty\frac{(b-a)\ln u}{(1+u^{2})\ln u}du \end{eqnarray*} and hence $$ 2I = -\int_0^\infty\frac{(b-a)}{1+u^{2}}du=(a-b)\frac{\pi}{2}. $$ So $$ I = (a-b)\frac{\pi}{4}. $$

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  • $\begingroup$ This is well done! +1. $\endgroup$ Jul 30, 2014 at 11:51
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Though this isn't an answer, this is interesting enough to me but too large for a comment. Based on Vladimir's solution, if we know $$f(a) = \int_0^\infty \frac{\ln[(1+x^a)/2]}{\ln x} \frac{1}{1+x^2}dx = a\frac{\pi}{4}$$ then we should have $$f'(a) = \int_0^\infty \frac{\ln x e^{a\ln x}}{\ln x (1+e^{a \ln x})} \frac{1}{1+x^2} dx = \pi/4$$ or $$f'(a) = \int_0^\infty \frac{x^a}{1+x^a} \frac{1}{1+x^2}dx = \int_0^\infty \left(1-\frac{1}{1+x^a} \right) \frac{1}{1+x^2}dx = \pi/4.$$ This is quite interesting because I would not expect the integral to be constant as a function of $a$. Furthermore, we should expect

$$f''(a) = \int_0^\infty \frac{\ln x \cdot x^a}{(1+x^a)^2}\frac{1}{1+x^2} dx = 0.$$ I didn't bother to check carefully for convergence issues (passing the derivative through), but I think everything cis okay. Does anyone know how to compute the above integrals without referring to Vladimir's answer?

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    $\begingroup$ Indeed, $\forall a\,\int_0^\infty\frac{dx}{\left(1+x^2\right)\,\left(1+x^a\right)}=\frac{\pi}{4}.\,$ I think this is quite well-known folklore. $\endgroup$ Mar 7, 2014 at 4:04
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    $\begingroup$ Then you might enjoy another one: $\forall{a}\,\int_{0}^{\infty}\frac{\operatorname{arccot}\left(x^{a}\right)}{1+x^{2}}dx=\frac{{\pi}^{2}}{8}.$ $\endgroup$ Mar 7, 2014 at 4:45
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    $\begingroup$ @nayrb: the integral for $f''(a)$ is easy. Split in two parts, one for $[0,1]$ and other for $[1,\infty]$ and put $x=1/y$ in second integral. $\endgroup$
    – Paramanand Singh
    Mar 7, 2014 at 4:50
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    $\begingroup$ Found the relevant question: math.stackexchange.com/questions/87735/… $\endgroup$
    – abnry
    Mar 7, 2014 at 4:59
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    $\begingroup$ @nayrb: My comment was not to find an alternative solution (because frankly speaking Vladimir's solution is best), but rather to show that the integrals $f'(a)$ and $f''(a)$ can also be evaluated directly using Vladimir technique. This was an answer to the question you asked in last line of your answer. $\endgroup$
    – Paramanand Singh
    Mar 7, 2014 at 5:14
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I do not know how to answer your question. However, in order you to challenge your challenger, I give you a few amazing results (obtained using a CAS) for $$f(n)=\int_0^\infty\frac{\log\left(\frac{1+x^a}{1+x^b}\right)}{\left(1+x^2\right)\log x}dx$$ in which $a=2n+\sqrt{4 n^2-1}$ and $b=n+\sqrt{n^2-1}$. $$f(1)=\frac{1}{4} \left(1+\sqrt{3}\right) \pi$$ $$f(2)=\frac{1}{4} \left(2-\sqrt{3}+\sqrt{15}\right) \pi$$ $$f(3)=\frac{1}{4} \left(3-2 \sqrt{2}+\sqrt{35}\right) \pi$$ $$f(4)=\frac{1}{4} \left(4+3 \sqrt{7}-\sqrt{15}\right) \pi$$ $$f(5)=\frac{1}{4} \left(5-2 \sqrt{6}+3 \sqrt{11}\right) \pi$$ $$f(6)=\frac{1}{4} \left(6-\sqrt{35}+\sqrt{143}\right) \pi$$ $$f(7)=\frac{1}{4} \left(7-4 \sqrt{3}+\sqrt{195}\right) \pi$$ $$f(8)=\frac{1}{4} \left(8-3 \sqrt{7}+\sqrt{255}\right) \pi$$ $$f(9)=\frac{1}{4} \left(9-4 \sqrt{5}+\sqrt{323}\right) \pi$$ $$f(10)=\frac{1}{4} \left(10-3 \sqrt{11}+\sqrt{399}\right) \pi$$ which are exactly what Pranav Arora answered (what I missed) $$f(n)=\frac \pi 4(a-b)$$

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  • $\begingroup$ please give me how did you got them ..please I am tired of trying. $\endgroup$ May 25, 2014 at 10:59
  • $\begingroup$ @ShivamPatel, you can see this from my answer. $\endgroup$
    – xpaul
    Apr 25, 2017 at 13:50

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