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According to WolframAlpha the integral $$\int _0^1\frac{\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)}{t\left(\ln t\right)^2}dt$$ is shown to have a decimal expansion exactly identical to $\ln 2$.

How can we prove they are equal?

This integral came about as a result of integrating by parts

$$\int_0^x\frac{\arctan t}{\ln \left(t\right)\left(1+t\right)^2}dt$$

with

$u=\frac{1}{\ln t}$

$dv=\frac{\arctan t}{\left(1+t\right)^2}dt$

$du=-\frac{1}{t\ln^2 t}dt$

$v=\frac{1}{4}\ln \left(2\right)-\frac{1}{4}\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)-\frac{1}{2}\left(\frac{1-t}{1+t}\right)\arctan \left(t\right)$

tyobrien
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3 Answers3

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Hint: $$\int\frac{\ln(1+(\dfrac{1-t}{1+t})^2)}{t(\ln t)^2}=\int\frac{\ln2}{t(\ln t)^2}+\int\frac{\ln(1+t^2)}{t(\ln t)^2}-2\int\frac{\ln(1+t)}{t(\ln t)^2}=\int\frac{\ln2}{t(\ln t)^2}$$

Nosrati
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  • Can you elaborate on the last equality? – Chris Jul 03 '17 at 02:59
  • The second integral will be equal to the third with $t^2=u$. – Nosrati Jul 03 '17 at 03:02
  • I'm making the $u$-substitution in the second integral and while some of the terms are playing nice, in the denominator the $\ln(t)^2$ doesn't seem to go well with your substitution. – Chris Jul 03 '17 at 03:10
  • @Chris $(\log(t^2))^2=4\log^2(t)$ and $dt\to 2t,dt$ upon substitution. – Mark Viola Jul 03 '17 at 03:10
  • @MarkViola welp, hold on a sec and I'll try that! – Chris Jul 03 '17 at 03:11
  • Actually, I'm a little concerned here. Clearly $\frac{d}{dt} [\frac{-1}{\ln(t)}] = \frac{1}{t\ln(t)^2}$, but then when you go to evaluate the integral (from $0$ to $1$) I find that the value is $+\infty$. This doesn't correspond to what the OP had. – Chris Jul 03 '17 at 03:34
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    I have the same problem as Chris since $\int\frac{dt}{t(\ln t)^2}=-\frac 1 {\log(t)}$ – Claude Leibovici Jul 03 '17 at 05:03
  • @ClaudeLeibovici the "canceled" integrals also diverge (at $t=1$), one has to be really careful about the split here... – tired Jul 03 '17 at 05:47
  • @tired this is the primary reason I wanted the OP to clarify his full solution. – Chris Jul 03 '17 at 05:58
  • the upvotes are a big miracle... – tired Jul 03 '17 at 22:36
  • If the site allowed me to remove my upvote from this post, to be very frank, I would do so. I'm bothered by the lack of detail in how you deal with the cancellation of divergent integrals, and why your answer apparently has a numerical discrepancy with the OP's calculation. – Chris Jul 03 '17 at 22:41
  • @Chris you should be able to remove your upvote if you really want – tired Jul 04 '17 at 07:07
  • @tired System tells me I'm locked in unless the post is edited. I wasn't trying to be dramatic, but I don't expect the post will be edited anytime soon. – Chris Jul 04 '17 at 07:23
  • Sorry, but this doesn't work (although it's nice to see "that the second integral will be equal to the third"). Correction factors are necessary to make sense on it. As tired said: the upvotes are a big miracle ... – user90369 Jul 04 '17 at 12:19
5

The given integral equals $$\begin{eqnarray*} I &=& \int_{0}^{1}\frac{\log(2)+\log(1+t^2)-2\log(1+t)}{t\log^2t}\,dt\\&\stackrel{t\mapsto e^{-x}}{=}&\int_{0}^{+\infty}\frac{\log(2)+\log(1+e^{-2x})-2\log(1+e^{-x})}{x^2}\,dx\\&\stackrel{\text{IBP}}{=}&2\int_{0}^{+\infty}\left(\frac{1}{e^x+1}-\frac{1}{e^{2x}+1}\right)\frac{dx}{x} \tag{1}\end{eqnarray*}$$ and the claim simply follows by recalling Frullani's theorem.
As an alternative, $$\mathcal{L}\left(\frac{1}{e^x+1}\right) = \frac{1}{2}\left(H_{\frac{s}{2}}-H_{\frac{s-1}{2}}\right)\tag{2}$$ leads to

$$ I = 2\lim_{s\to +\infty}\log\left(\frac{\Gamma\left(\frac{s+2}{4}\right)\,\Gamma\left(\frac{s+2}{2}\right)}{\Gamma\left(\frac{s+4}{4}\right)\,\Gamma\left(\frac{s+1}{2}\right)}\right) = \color{red}{\log 2}\tag{3}$$ by Stirling's inequality.

Jack D'Aurizio
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1

Using $\enspace\displaystyle\zeta(0)=-\frac{1}{2}\enspace$ and $\enspace\displaystyle\lim\limits_{h\to 0}\frac{2^h-1}{h}=\ln 2$

and $\enspace\displaystyle\int _0^\infty\frac{x^{h-1}}{e^{ax}+1}dx=\frac{2^{h-1}-1}{2^{h-1}a^h}\zeta(h)\Gamma(h)\enspace$ for $\enspace h,a>0\enspace$

and $\enspace\displaystyle t:=e^{-x}\enspace$ we get:

$\displaystyle\int _0^1\frac{\ln \left(1+\left(\frac{1-t}{1+t}\right)^2\right)}{t\left(\ln t\right)^2}dt=\int _0^\infty\frac{\ln 2+\ln(1+e^{-2x})-2\ln(1+e^{-x})}{x^2}dx$

$\displaystyle=2\int _0^\infty (\frac{1}{e^{2x}+1}-\frac{1}{e^{x}+1}) \frac{dx}{x}=2\lim\limits_{h\to 0} \int _0^\infty (\frac{x^{h-1}}{e^{2x}+1}-\frac{x^{h-1}}{e^{x}+1})dx$

$\displaystyle=2\lim\limits_{h\to 0} \zeta(h)\Gamma(h+1)\frac{2^{h-1}-1}{2^{2h-1}}\frac{2^h-1}{h}=2(-\frac{1}{2}\frac{2^{-1}-1}{2^{-1}}\ln 2)=\ln 2$

user90369
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