I need help with this integral:

$$I=\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx.$$

The integrand graph looks like this:

$\hspace{1in}$The integrand graph

The approximate numeric value of the integral: $$I\approx8.372211626601275661625747121...$$

Neither Mathematica nor Maple could find a closed form for this integral, and lookups of the approximate numeric value in WolframAlpha and ISC+ did not return plausible closed form candidates either. But I still hope there might be a closed form for it.

I am also interested in cases when only numerator or only denominator is present under the logarithm.

  • 204,278
  • 154
  • 264
  • 488
Laila Podlesny
  • 12,227
  • 4
  • 58
  • 64
  • 57
    Do you have any reason to believe there is a closed form for that horrid-looking thing? – dfeuer Nov 11 '13 at 17:12
  • 53
    In the meantime, I have been able to manipulate the integral into the following form: $$8 \int_0^{\infty} du \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log{u}$$ from which I may deduce that there is in fact a closed form (because the roots of the denominator are expressible in closed form, a little messy but not bad). But because there are eight roots, a residue-based approach will be very very messy and almost hopeless without some computer algebra. – Ron Gordon Nov 12 '13 at 00:21
  • 6
    A [related problem](http://math.stackexchange.com/questions/541751/how-prove-this-i-int-0-infty-frac1x-ln-left-frac1x1-x-right2/541861#541861). – Mhenni Benghorbal Nov 19 '13 at 18:46
  • 22
    @MhenniBenghorbal: Related how? – Ron Gordon Nov 19 '13 at 21:02
  • 1
    @user88377 Why is there a bounty on this? It seems Ron Gordon has clearly provided a complete solution already... Are you looking for different solutions?? – Jeff Faraci Apr 06 '14 at 22:10
  • 40
    Out of curiosity, is there an application for this integral? – linuxfreebird Nov 16 '14 at 00:35
  • 2
    @Integrals People put bounties on the question **just** to give it to him, other than the OP –  Nov 05 '15 at 01:05
  • 2
    I chanced on Ron Gordon's afterthoughts on Reddit and wouldn't have known of it otherwise: https://old.reddit.com/r/math/comments/1qpus4/master_of_integration/cdfysit/. I don't think it was linked here? It'd be such a shame not to be aware of it! – NNOX Apps Apr 22 '19 at 08:25
  • Where did you get this problem?.. I am sure who made this problem somehow knows the solution or some background because it doesn't look like a random formula. – SGKw Apr 22 '22 at 05:13

10 Answers10


I will transform the integral via a substitution, break it up into two pieces and recombine, perform an integration by parts, and perform another substitution to get an integral to which I know a closed form exists. From there, I use a method I know to attack the integral, but in an unusual way because of the 8th degree polynomial in the denominator of the integrand.

First sub $t=(1-x)/(1+x)$, $dt=-2/(1+x)^2 dx$ to get

$$2 \int_0^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} $$

Now use the symmetry from the map $t \mapsto 1/t$. Break the integral up into two as follows:

\begin{align} & 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} + 2 \int_0^{1} dt \frac{t^{1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \\ &= 2 \int_0^{1} dt \frac{t^{-1/2}}{1-t} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )} \end{align}

Sub $t=u^2$ to get

$$4 \int_0^{1} \frac{du}{1-u^2} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}$$

Integrate by parts:

$$\left [2 \log{\left (\frac{1+u}{1-u} \right )} \log{\left (\frac{5-2 u^2+u^4}{1-2 u^2 +5 u^4} \right )}\right ]_0^1 \\- 32 \int_0^1 du \frac{\left(u^5-6 u^3+u\right)}{\left(u^4-2 u^2+5\right) \left(5 u^4-2 u^2+1\right)} \log{\left (\frac{1+u}{1-u} \right )}$$

One last sub: $u=(v-1)/(v+1)$ $du=2/(v+1)^2 dv$, and finally get

$$8 \int_0^{\infty} dv \frac{(v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v}$$

With this form, we may finally conclude that a closed form exists and apply the residue theorem to obtain it. To wit, consider the following contour integral:

$$\oint_C dz \frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} \log^2{z}$$

where $C$ is a keyhole contour about the positive real axis. This contour integral is equal to (I omit the steps where I show the integral vanishes about the circular arcs)

$$-i 4 \pi \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} + 4 \pi^2 \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1}$$

It should be noted that the second integral vanishes; this may be easily seen by exploiting the symmetry about $v \mapsto 1/v$.

On the other hand, the contour integral is $i 2 \pi$ times the sum of the residues about the poles of the integrand. In general, this requires us to find the zeroes of the eight degree polynomial, which may not be possible analytically. Here, on the other hand, we have many symmetries to exploit, e.g., if $a$ is a root, then $1/a$ is a root, $-a$ is a root, and $\bar{a}$ is a root. For example, we may deduce that

$$z^8+4 z^6+70z^4+4 z^2+1 = (z^4+4 z^3+10 z^2+4 z+1) (z^4-4 z^3+10 z^2-4 z+1)$$

which exploits the $a \mapsto -a$ symmetry. Now write

$$z^4+4 z^3+10 z^2+4 z+1 = (z-a)(z-\bar{a})\left (z-\frac{1}{a}\right )\left (z-\frac{1}{\bar{a}}\right )$$

Write $a=r e^{i \theta}$ and get the following equations:

$$\left ( r+\frac{1}{r}\right ) \cos{\theta}=-2$$ $$\left (r^2+\frac{1}{r^2}\right) + 4 \cos^2{\theta}=10$$

From these equations, one may deduce that a solution is $r=\phi+\sqrt{\phi}$ and $\cos{\theta}=1/\phi$, where $\phi=(1+\sqrt{5})/2$ is the golden ratio. Thus the poles take the form

$$z_k = \pm \left (\phi\pm\sqrt{\phi}\right) e^{\pm i \arctan{\sqrt{\phi}}}$$

Now we have to find the residues of the integrand at these 8 poles. We can break this task up by computing:

$$\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1) \log^2{z}}{z^8+4 z^6+70z^4+4 z^2+1}\right ]=\sum_{k=1}^8 \operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] \log^2{z_k}$$

Here things got very messy, but the result is rather unbelievably simple:

$$\operatorname*{Res}_{z=z_k} \left [\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1}\right ] = \text{sgn}[\cos{(\arg{z_k})}]$$


Actually, this is a very simple computation. Inspired by @sos440, one may express the rational function of $z$ in a very simple form:

$$\frac{8 (z^2-1)(z^4-6 z^2+1)}{z^8+4 z^6+70z^4+4 z^2+1} = -\left [\frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right ]$$


$$p(z)=z^4+4 z^3+10 z^2+4 z+1$$

The residue of this function at the poles are then easily seen to be $\pm 1$ according to whether the pole is a zero of $p(z)$ or $p(-z)$.


That is, if the pole has a positive real part, the residue of the fraction is $+1$; if it has a negative real part, the residue is $-1$.

Now consider the log piece. Expanding the square, we get 3 terms:

$$\log^2{|z_k|} - (\arg{z_k})^2 + i 2 \log{|z_k|} \arg{z_k}$$

Summing over the residues, we find that because of the $\pm1$ contributions above, that the first and third terms sum to zero. This leaves the second term. For this, it is crucial that we get the arguments right, as $\arg{z_k} \in [0,2 \pi)$. Thus, we have

$$\begin{align}I= \int_0^{\infty} dv \frac{8 (v^2-1)(v^4-6 v^2+1)}{v^8+4 v^6+70v^4+4 v^2+1} \log{v} &= \frac12 \sum_{k=1}^8 \text{sgn}[\cos{(\arg{z_k})}] (\arg{z_k})^2 \\ &= \frac12 [2 (\arctan{\sqrt{\phi}})^2 + 2 (2 \pi - \arctan{\sqrt{\phi}})^2 \\ &- 2 (\pi - \arctan{\sqrt{\phi}})^2 - 2 (\pi + \arctan{\sqrt{\phi}})^2]\\ &= 2 \pi^2 -4 \pi \arctan{\sqrt{\phi}} \\ &= 4 \pi \, \text{arccot}{\sqrt{\phi}}\\\end{align}$$

  • 22,756
  • 140
  • 38
  • 83
Ron Gordon
  • 134,112
  • 16
  • 181
  • 296
  • 117
    +1. $\text{arccot}(\sqrt{\phi})$-definitely one of the most weirdest closed form solutions to have ever been obtained! –  Dec 02 '13 at 06:31
  • 14
    @ShikariShambu: which makes me wonder if we can generate a list of the oddest closed-form solutions to integrals, sums, products, diff eq'ns, and the like. What makes a closed form "odd"? What makes this one odd? The juxtaposition of the arccotangent and $\phi$? This could lead to...a weird discussion, but a fun one nonetheless. – Ron Gordon Dec 18 '13 at 23:41
  • 13
    I know that $\phi$ has some nice properties, but at the end of the day it's just another algebraic number: $$\phi = \frac{1+\sqrt{5}}{2},$$ and all we've done is taken the arc cotangent of the square root of it and multiplied by $4\pi$. - just wondering why this result would be seen as "odd". It most likely appears because it happens to be a root of some polynomial which has $(1+\sqrt{5})/2$ as one of its roots. – gone Feb 12 '14 at 18:11
  • 3
    @pbs: what is odd to one person is not to another. Just a matter of taste. In this case, a matter of expectations. Was the presence of $\phi$ or its square root obvious to you when you first approached this problem? The arctan or arccot, depending on how you expressed the result? It was far from obvious to me, or that we'd have such a nice, compact expression for a result. – Ron Gordon Feb 12 '14 at 18:20
  • 3
    @RonGordon True, and no it certainly was not expected! Then again, I don't expect any particular algebraic number to appear in a result until it does appear, so I guess the surprise here is that $\phi$ has some special significance. I like your expectations idea. – gone Feb 12 '14 at 18:24
  • 7
    @RonGordon When I see things like this, I don't consider it odd so much as a pleasant surprise. Phi is known as a number of beauty for many reasons, and to see it emerge from, to quote our friend dfeuer, "that horrid-looking thing", is quite a pleasant surprise indeed! No tin-foil hats, no cube-root-of-31-conspiracies, just running into an old friend at the grocery =) – corsiKa Mar 13 '14 at 19:28
  • 104
    @corsiKa: $\phi$ alone, sure. But now imagine your meeting your old friend, but dressed in drag with a Kaiser-era military helmet on, spike and all. That's sort of the feeling you get when you see, not regular old $\phi$, but ARCCOT SQRT $\phi$. – Ron Gordon Mar 14 '14 at 17:08
  • 3
    I hope you don't mind @RonGordon sir, I wish to give you +50 rep; the exemplary answer is both enlightening, as well as inspiring. It says I can give you the bounty in 23 hours; I will give the bounty when it's possible. – Panglossian Oporopolist Mar 02 '15 at 09:44
  • Sir, is there any chance that you can explain what you mean by "use the symmetry from the map $t \to 1/t$" ? – Our Jul 20 '18 at 07:16
  • 3
    @onurcanbektas In $\int_1^{\infty} dt \frac{t^{-1/2}}{1-t^2} \log{\left (\frac{5-2 t+t^2}{1-2 t +5 t^2} \right )}$, changing $t$ to $1/t$ in the argument to $\log$ has the effect of inverting numerator and denominator, so the overall $\log$ term simply picks up a minus sign. The integral, therefore, barely changes under the change of variables $t = 1/u$, $dt = -du/u^2$: you get $\int_0^1 du \frac{u^{1/2}}{1-u^2} \log{\left (\frac{5-2 u+u^2}{1-2 u +5 u^2} \right )}$. You can then restore $t$ as a variable of integration and recombine the integrals. – Connor Harris Jul 24 '18 at 13:59
  • @ConnorHarris Oh, I see. Thanks for explanation. – Our Jul 24 '18 at 16:20
  • I chanced on your afterthoughts on Reddit and wouldn't have known of it otherwise: https://old.reddit.com/r/math/comments/1qpus4/master_of_integration/cdfysit/. Did you link to it here? It'd be such a shame not to be aware of it! – NNOX Apps Apr 22 '19 at 08:25
  • 1
    This is very old, but PLEASE recommend me some study materials to help me become a master of integration like this. – CalebWilliamsUIC Dec 22 '20 at 19:44


  • 17,008
  • 5
  • 44
  • 62
  • 91
    @OliverBel IMHO, if the question is not asking explicitly for a proof, and there is no obvious conjectured closed form, I think it's OK to post the result first and add a proof later, when time permits. The result itself can be useful for the person asking. It may be downvoted if not useful, but it still qualifies as an answer. The guideline for comments say "Avoid answering questions in comments", and I see no reason to withhold the result until the full proof is ready to be posted. – Vladimir Reshetnikov Nov 11 '13 at 22:20
  • 45
    @VladimirReshetnikov: I defer to Hamming: "The purpose of computing is insight, not numbers." Unless the result itself is particularly illuminating, I do not agree that it is an answer. There is a user who continues to insist on putting Maple output as an answer, and I think that degrades the site. I'd prefer if Cleo gave some insight as to where this answer came from. – Ron Gordon Nov 11 '13 at 22:53
  • 44
    @RonGordon Of course, I would prefer to see the proof too. I hope that Cleo posts it eventually, or somebody else does it, inspired by the fact the closed form exists (and maybe also guided by its shape). My point is that writing a proof clearly and typesetting it can take hours (at least for me) and I can imagine that not everybody can allocate the required time promptly. I would prefer to at least see the result posted promptly in such cases. (And from my experience being a PhD student in theoretical physics I can say that sometimes the result is the only thing a person cares of :) – Vladimir Reshetnikov Nov 11 '13 at 23:17
  • 10
    @VladimirReshetnikov: fine, but we should ask Cleo to at least say something about how he got this and that a proof is in the works. Maybe he manipulated Mathematica to produce the answer in a way that the OP could not. (That still is not good enough for an answer, but it is at least good enough for a comment.) But just sticking a result in that agrees with the numerics is not good enough for this site IMO. – Ron Gordon Nov 11 '13 at 23:20
  • 7
    I agree with both of you, the answer is beautiful in itself - I would almost say incredibly beautiful given the ratio of simplicity of the result to complexity of the problem, but we don't *learn* anything from this answer, manipulation of a CAS is a useful skill to have when tackling these problems, and I'd like to see *something* that tells me how I should tackle things like this. – Bennett Gardiner Nov 12 '13 at 07:10
  • 4
    Assuming this answer is correct (which it certainly is to many decimal places), it seems very strange. $\pi$ times an arccotangent seems odd enough. Tossing the square root of $\phi$ into there seems even stranger. – dfeuer Nov 12 '13 at 07:19
  • 4
    @dfeuer: I can tell you where the sqrt of $\phi$ comes from: it is the phase of the roots of the octic (word?) in the integrand I posted in the comments to the question. Still working out the rest. – Ron Gordon Nov 12 '13 at 09:28
  • Hey @RonGordon, what do you mean by "phase of the roots"? I see the roots are of the form $$u = (\pm 1\pm i)\pm\sqrt{-1\pm 2 i}$$(close enough?) – Bennett Gardiner Nov 12 '13 at 11:54
  • @Bennett: that's right as far as I can tell from my mobile phone. – Ron Gordon Nov 12 '13 at 13:35
  • So what is this "phase"? I'm not familiar. – Bennett Gardiner Nov 12 '13 at 13:36
  • 3
    @BennettGardiner I suppose, the phase is [$\arg z$](http://mathworld.wolfram.com/ComplexArgument.html). – Vladimir Reshetnikov Nov 12 '13 at 21:27
  • 1
    @BennettGardiner: Vladimir is correct. More to follow. – Ron Gordon Nov 13 '13 at 06:02
  • 3
    @VladimirReshetnikov: another thought. I think you should care about how we get the answers here. I hope that you were able to profit from the process I described to get to the destination. Without the road, how will we ever understand how to get to the destination? And how will we ever find a better road? Thus, I do not consider Cleo's "answers" as such. Cleo has posted four such answers without any hint as to a thought process. For all I know, he may just have found a way to manipulate Maple to output a plausible answer. But I have no idea. – Ron Gordon Nov 15 '13 at 12:29
  • 80
    This style of answer is complete disrespect. This situation seems for me like this: Cleo found interesting problem, and solved it. He is lazy to write the solution but want to show how clever he is, so decided to post only the final result. The reference to the definition of golden ratio made me laugh. If OP asks question of such level he definitely familiar with this constant. Note that this is not a single example. ALL Cleo's answer are of this style, and even after polite ephasis that these answers is not what OP's wanted he continues to post only final results! – Norbert Nov 17 '13 at 09:14
  • 9
    @Norbert: I completely agree. What drives people to upvote this answer, I have no idea (perhaps a slavish devotion to upvoting the first answer, which I know runs rampant among certain users). These answers are not such, and should not only be vigorously downvoted, but also flagged as non-answers. – Ron Gordon Nov 17 '13 at 18:51
  • 14
    @Norbert I would not take it too personally. There is definitely nothing offensive in the answer. Certainly, it is of low quality and may be downvoted by those who deem it not useful. – Liu Jin Tsai Nov 18 '13 at 21:06
  • 6
    I will not change my mind, cause I know the style of this user. – Norbert Nov 18 '13 at 21:14
  • 5
    Can you give us any hint at how you arrived at your solution? Which program did you use? – Abdulh Khazzak Gustav ElFakiri Nov 21 '13 at 21:32
  • 5
    @Norbert to my opinion this is not a clever answer, this is an arrogant answer and I don't understand why it has so many up votes. An explanation to this solution of the integral is needed. – Bman72 Apr 06 '14 at 06:33
  • 144
    «Disrespectful answer» and «arrogant answer»? Really?! I find the answer pretty useless, as I did not learn anything from it, but disrespectful and arrogant are judgements which seem totally inappropriate to this answer! I **really** wish commenters would limit the dramatic charge of their comments to ¡the value of an integral! – Mariano Suárez-Álvarez Apr 26 '14 at 02:16
  • 16
    @Norbert: From Cleo's user profile: "I have a medical condition that makes it very difficult for me to engage in conversations, or post long answers, sorry for that." Would you like to reconsider your judgement? – Edward Kmett Oct 10 '15 at 08:52
  • 13
    @EdwardKMETT, you are newbie here. Believe me, you don't know the situation in detail. As for the medical condition... this is a lie, because first explanation Cleo gave a long time ago was.... Guess what? This is her religion to give naked answers and if someone doubt in correctness then one should add that answer to his system of axioms. This is ridiculous! – Norbert Oct 10 '15 at 10:21
  • 6
    @Norbert :'( Well I do hope we did not cause the leave of this user. He/she was extremely valuable and I do love much of her work. – Simply Beautiful Art Feb 17 '17 at 19:40
  • 8
    @SimplyBeautifulArt: Sorry to burst your bubble. True mathematics is not about being selfish and keeping one's methods to oneself. Also, he/she **lies intentionally to deceive people** into pitying him/her and allowing mathematically empty answers to remain, which is contrary to the spirit of mathematics in seeking truth (or at least insight). How do you know that all his/her answers are true, since he/she lies about motives?? See http://meta.math.stackexchange.com/q/11723 and http://meta.math.stackexchange.com/questions/11759/what-to-do-with-users-posting-unhelpful-answers#comment45802_11760. – user21820 Feb 21 '17 at 06:26
  • 2
    @user21820 of course I do not believe her answers are the best for the site, but I also believe that the talented di benefit the site, selfish or not, in some ways. I also do not like the thought of decently high rep users leaving us. It makes me sad to see. – Simply Beautiful Art Feb 21 '17 at 12:01
  • 3
    @SimplyBeautifulArt: Note the following: (1) One-line answers should in almost all cases be just a comment. (2) High reputation means the ability to gain high reputation in the SE environment, and does not at all imply quality of their contributions. There are many users with high reputation gained by answering as many basic calculus questions as they can every day, who have such a poor command of logic that they spout nonsense as well. (3) This user will have zero reputation if all of his/her such mathematically empty posts are removed! – user21820 Feb 21 '17 at 12:22
  • 3
    68 votes for just posting an answer.... – imranfat Apr 17 '17 at 15:43
  • 15
    @imranfat, do you think someone would dare to look at that integral without knowing there is a nice closed form for it ? This answer is half of the way of solving the integral. This question gained popularity of the interesting closed form that his user provided. – Zaid Alyafeai Apr 25 '17 at 03:25
  • @ZaidAlyafeai. Ok, let's wait for "the other half", then I rest my case :) – imranfat Apr 25 '17 at 17:17
  • 8
    @ZaidAlyafeai I agree whole-heartedly. There are a ton of integrals on this site without answers because no closed form is known, and so people give up when they can't find one. While having an answer isn't everything, it is fairly vital in knowing that an integral IS solvable, and thus worthy of time spent solving it. Moreover, knowing the solution can often be a step in the direction of solving it. Are proofs better in 99% of cases? Yes. However, good luck getting those proofs on a bunch of impossible looking integrals without knowing the integrals are solvable first :/ – Brevan Ellefsen May 15 '17 at 22:53
  • 1
    @BrevanEllefsen take this for example https://math.stackexchange.com/questions/2242141/an-integral-of-a-rational-function-of-logarithm-and-nonlinear-arguments. – Zaid Alyafeai May 15 '17 at 22:59
  • 2
    Gee guys, no need for all of this hand-wringing; just downvote useless answers without saying anything and move on! – J. M. ain't a mathematician Apr 12 '20 at 01:21
  • 1
    @user21820 Are you claiming that Cleo is lying in his/her bio about the medical condition? How did you reach this conclusion? – Blue Jun 04 '20 at 20:07
  • 4
    @Blue: Indeed. I reached this conclusion after investing sufficient effort to verify the claims by numerous other users who have witnessed Cleo's earlier behaviour, such as what Norbert said above. At first, she said she got her answers from Namagiri, and then when questioned further said "There are many ways to prove this result. The easiest one is to work in an axiomatic system that accepts it as an axiom. I prefer this approach when I know the result. Therefore, the full proof is given here." to justify her proof-less answer!! Later, she put the lie in her profile and deleted her comments. – user21820 Jun 05 '20 at 15:52
  • @user21820 Oh I see. – Blue Jun 05 '20 at 18:07
  • 1
    Why do so many users assume, at the start of this comment thread, @RonGordon, Norbert ?? I agree with what you post here ++++, but I'm confused as to your reasoning about referring to "Cleo" as "he" or "his". Especially given "their" gravatar. – amWhy Dec 11 '20 at 23:48
  • I should have used “they” because I did not know Cleo’s gender. I apologize for that. – Ron Gordon Dec 11 '20 at 23:51
  • 1
    @user21820 The sad part of the situation is, if she came out and admitted that she was solving these via computer assistance (which seems likely), even the method through which she finessed a computer program to output these results would be immensely useful. Instead, she chooses to remain mysterious, for reasons unbeknownst to us. – Rushabh Mehta Jun 16 '21 at 20:42
  • 1
    @DonThousand: Exactly. This kind of behaviour is really disgusting. – user21820 Jun 18 '21 at 10:40
  • 7
    Years later, and we're still talking about this. However one feels about the ordeal, bickering about it 8 years later is just impractical. Cleo is evidently nowhere *near* as active on the site as she used to be. Besides, her answer does not endorse the notion that "posting answers without proof is acceptable". Nobody is going to start frequently posting such naked answers and justifying their actions with "Cleo did it, so I can too", it's a slippery slope. I can understand users' shared frustration from years ago, but *8 years later*, the point does not anymore meet the same moral conviction. – Mr Pie Aug 25 '21 at 10:31
  • 2
    Cheers, Cleo! It is 2022 and you have become a legend for me. May your answer last forever. – coudy Feb 17 '22 at 18:40
  • Sometimes I miss Cleo's answers. – Wylnorr Apr 08 '22 at 13:06

NEW ANSWER. I found yet another way of solving this problem. My new solution does not use contour integration, and is based on the following observation: for $|z| \leq 1$,

$$ - \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log(1 - zx) \, dz= \pi \sin^{-1} z - \pi \log \left( \tfrac{1}{2}+\tfrac{1}{2}\sqrt{1-z^{2}} \right) . $$

As I want to keep both the old answer and the new answer, I posted my new solution to other page. You can check it here.

OLD ANSWER. Okay here is another solution. It is also related to my generalization.

We claim the following proposition:

Proposition. If $0 < r < 1$ and $r < s$, then $$ I(r, s) := \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx = 4\pi \arcsin r. \tag{1} $$

Assuming this proposition, all that we have to do is to solve the non-linear system of equations

$$ 2rs = 2 \quad \text{and} \quad r^{2} + s^{2} - 1 = 2. $$

The unique solution satisfying the condition of the proposition is $r = \phi - 1$ and $s = \phi$. So by $\text{(1)}$ we have

\begin{align*} \int_{-1}^{1} \frac{1}{x} \sqrt{\frac{1+x}{1-x}} \log \left( \frac{1 + 2x + 2x^{2}}{1 - 2x + 2x^{2}} \right) \, dx & = I(\phi-1, \phi) \\ &= 4\pi \arcsin (\phi - 1) = 4\pi \operatorname{arccot} \sqrt{\phi}. \end{align*}

Thus it remains to prove the proposition.

Proof of Proposition. We divide the proof into several steps.

Step 1. (Case reduction by analytic continuation) We first remark that given $r$ and $s$, we always have

$$ \min_{-1 \leq x \leq 1} \{ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} \} > 0. \tag{2} $$

Indeed, it is not hard to check if we utilize the following equality

$$ 1 \pm 2rsx + (r^{2} + s^{2} - 1)x^{2} = (1 \pm rsx)^{2} - (1 - r^{2})(1 - s^{2}) x^{2}. $$

Then $\text{(2)}$ shows that the integrand of $I(r, s)$ remains holomoprhic under small perturbation of $s$ in $\Bbb{C}$. So it allows us to extend $s \mapsto I(r, s)$ as a holomorphic function on some open set containing the line segment $(r, \infty) \subset \Bbb{C}$. Then by the principle of analytic continuation, it is sufficient to prove that $\text{(1)}$ holds for $r < s < 1$.

Step 2. (Integral representation of $I$) Put $r = \sin \alpha$ and $s = \sin \beta$, where $ 0 < \alpha < \beta < \frac{\pi}{2}$. Then

\begin{align*} I(r, s) &= \int_{-1}^{1} \frac{1+x}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \\ &= \int_{0}^{1} \frac{2}{x\sqrt{1-x^{2}}} \log \left( \frac{1 + 2rsx + (r^{2} + s^{2} - 1)x^{2}}{1 - 2rsx + (r^{2} + s^{2} - 1)x^{2}} \right) \, dx \qquad (\because \text{ parity}) \\ &= \int_{1}^{\infty} \frac{2}{\sqrt{x^{2}-1}} \log \left( \frac{x^{2} + 2rsx + (r^{2} + s^{2} - 1)}{x^{2} - 2rsx + (r^{2} + s^{2} - 1)} \right) \, dx \qquad (x \mapsto x^{-1}) \\ &= \int_{0}^{1} \frac{2}{t} \log \left( \frac{\left(t+t^{-1}\right)^{2} + 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)}{\left(t+t^{-1}\right)^{2} - 4rs\left(t+t^{-1}\right) + 4(r^{2} + s^{2} - 1)} \right) \, dt, \end{align*}

where in the last line we utilized the substitution $x = \frac{1}{2}(t + t^{-1})$. If we introduce the quartic polynomial \begin{align*} p(t) = t^{4} + 4rst^{3} + (4r^{2}+4s^{2}-2)t^{2} + 4rst + 1, \end{align*}

then by the property $p(1/t) = t^{-4}p(t)$, we can simplify

\begin{align*} I(r, s) &= 2 \int_{0}^{1} \frac{\log p(t) - \log p(-t)}{t} \, dt = \int_{0}^{\infty} \frac{\log p(t) - \log p(-t)}{t} \, dt \\ &= - \int_{0}^{\infty} \left( \frac{p'(t)}{p(t)} + \frac{p'(-t)}{p(-t)} \right) \log t \, dt = - \frac{1}{2} \Re \int_{-\infty}^{\infty} \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z \, dz, \end{align*}

where we choose the branch cut of $\log$ in such a way that it avoids the upper-half plane

$$\Bbb{H} = \{ z \in \Bbb{C} : \Im z > 0 \}.$$

Step 3. (Residue calculation) Since

$$ f(z) := \left( \frac{p'(z)}{p(z)} + \frac{p'(-z)}{p(-z)} \right) \log z = O\left(\frac{\log z}{z^{2}} \right) \quad \text{as } z \to \infty, $$

by replacing the contour of integration by a semicircle of sufficiently large radius, it follows that

\begin{align*} I(r, s) = - \frac{1}{2} \Re \left\{ 2 \pi i \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z) \right\} = \pi \Im \sum_{z_{0} \in \Bbb{H}} \operatorname{Res}_{z = z_{0}} f(z). \end{align*}

(It turns out that $f(z)$ has only logarithmic singularity at the origin. So it does not account for the value of $I(r, s)$.) But by a simple calculation, together with the condition $ 0 < \alpha < \beta < \frac{\pi}{2}$, we easily notice that the zeros of $p(z)$ are exactly

$$ e^{\pm i(\alpha + \beta)} \quad \text{and} \quad -e^{\pm i(\alpha - \beta)}. $$

Now let $Z_{+}$ be the set of zeros of $p(z)$ in $\Bbb{H}$ and $Z_{-}$ be the set of zeros of $p(z)$ in $-\Bbb{H}$. Then

$$ Z_{+} = \{ e^{i(\beta+\alpha)}, -e^{-i(\beta - \alpha)} \} \quad \text{and} \quad Z_{-} = \{ e^{-i(\beta+\alpha)}, -e^{i(\beta- \alpha)} \}. $$

This in particular shows that

$$ \frac{p'(z)}{p(z)}\log z = \sum_{z_{0} \in Z_{+}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H} $$


$$ \frac{p'(-z)}{p(-z)}\log z = -\sum_{z_{0} \in -Z_{-}} \frac{\log z}{z - z_{0}} + \text{holomorphic function on } \Bbb{H}. $$

So it follows that

\begin{align*} I(r, s) &= \pi \Im \left\{ \sum_{z_{0} \in Z_{+}} \log z_{0} - \sum_{z_{0} \in -Z_{-}} \log z_{0} \right\} \\ &= \pi \Im \left\{ \log e^{i(\beta+\alpha)} + \log e^{i(\pi-\beta+\alpha)} - \log e^{i(\pi-\beta-\alpha)} - \log e^{i(\beta-\alpha)} \right\} \\ &= \pi \Im \left\{ i(\beta+\alpha) + i(\pi-\beta+\alpha) - i(\pi-\beta-\alpha) - i(\beta-\alpha) \right\} \\ &= 4\pi \alpha = 4\pi \arcsin r. \end{align*}

This completes the proof.

Sangchul Lee
  • 141,630
  • 16
  • 242
  • 385
  • 10
    Very nice, and it helps me simplify a piece of my proof as well (I think our solutions have more in common than not). One question, though: what about the branch point of the log in the residue calculation? I know it doesn't seem to matter as you do end up with the correct solution, but you may want to say something about avoiding the branch point at the origin and defining a branch of the log (which I think you do anyway with your restrictions on $\alpha$ and $\beta$) in the complex plane. – Ron Gordon Nov 17 '13 at 08:45
  • 4
    @RonGordon, As written in the solution, the branch cut of the log is chosen so that it avoids the upper half plane. So it would be safe if we choose it as the negative y-axis, but I think the standard branch cut $(-\infty, 0)$ would also works. – Sangchul Lee Nov 17 '13 at 15:44
  • 3
    Oh my, there it is. My bad, so sorry. In any case, reading your solution helped me simplify a small part of mine, so thanks. – Ron Gordon Nov 17 '13 at 15:53

Our aim is to give an elementary proof of Proposition formula (1) in the answer of @sos440. We first note that $$ \min_{-1\leq x\leq1}\{1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\}>0. $$ Indeed, if $x=\pm1$ then $$ 1\pm2rsx+(r^{2}+s^{2}-1)x^{2}\geq(r-s)^{2}>0, $$ if $x=0$ then $$ 1\pm2rsx+(r^{2}+s^{2}-1)x^{2}=1>0, $$ if $-1<x<1$, $x\neq0$ then the equations \begin{eqnarray*} \frac{\partial}{\partial s}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0,\\ \frac{\partial}{\partial r}(1\pm2rsx+(r^{2}+s^{2}-1)x^{2}) & = & 0, \end{eqnarray*} give $\pm r=sx$, $\pm s=rx$, which is impossible.

In the second step we show that $I(r,s)$ is independent of $s$. $$ \frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx. $$ Substituting $x:=-x$ and adding them we obtain $$ 2\frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{2}{\sqrt{1-x^{2}}}\cdot\frac{4r(1+(r^{2}-s^{2}-1)x^{2})}{(1-2rsx+(r^{2}+s^{2}-1)x^{2})(1+2rsx+(r^{2}+s^{2}-1)x^{2}}\, dx, $$ that is, $$ \frac{\partial}{\partial s}I(r,s)=\int_{-1}^{1}\frac{1}{\sqrt{1-x^{2}}}\cdot\frac{4r(-s^{2}+r^{2}-1)x^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}x^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)x^{2}}\, dx. $$ Substituting $x:=\sin(t)$ we have $$ \frac{\partial}{\partial s}I(r,s) = \int_{-\pi/2}^{\pi/2}\frac{4r(-s^{2}+r^{2}-1)\sin(t)^{2}+4r}{1+(r^{2}+s^{2}-1)^{2}\sin(t)^{4}+(2s^{2}-4r^{2}s^{2}+2r^{2}-2)\sin(t)^{2}}\, dt $$ $$ =\int_{-\pi/2}^{\pi/2}-\frac{8r((-s^{2}+r^{2}-1)\cos(2t)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(2t)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(2t)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dt $$ $$ = \int_{-\pi}^{\pi}-\frac{4r((-s^{2}+r^{2}-1)\cos(y)+s^{2}-r^{2}-1)}{(r^{2}+s^{2}-1)^{2}\cos(y)^{2}-2(r^{2}-s^{2}-1)(r^{2}+1-s^{2})\cos(y)+r^{4}+(2-6s^{2})r^{2}+(s^{2}+1)^{2}}\, dy. $$ Introducing the new variable $T:=\tan\frac{y}{2}$ we obtain \begin{eqnarray*} \frac{\partial}{\partial s}I(r,s) & = & \int_{-\infty}^{\infty}-\frac{4r(s^{2}-r^{2})T^{2}-4r}{(r-s)^{2}(r+s)^{2}T^{4}+((2-4s^{2})r^{2}+2s^{2})T^{2}+1}\, dT\\ & = & -\frac{4r(s^{2}-r^{2})}{(r-s)^{2}(r+s)^{2}}\int_{-\infty}^{\infty}\frac{T^{2}+a}{T^{4}+bT^{2}+b^{2}/4+d}\, dT\\ & = & -\frac{4r(-s^{2}+r^{2})}{(r-s)^{2}(r+s)^{2}}\cdot\frac{(2a(b^{2}+4d)+(b^{2}+4d)^{3/2})\pi}{(b^{2}+4d)^{3/2}\sqrt{\sqrt{b^{2}+4d}+b}}, \end{eqnarray*} where $$ a=-\frac{1}{s^{2}-r^{2}}, $$ $$ b=\frac{(2-4s^{2})r^{2}+2s^{2}}{(r-s)^{2}(r+s)^{2}}, $$ $$ b^{2}+4d=\frac{4}{(r-s)^{2}(r+s)^{2}}. $$ It gives $2ab^{2}+8da+(b^{2}+4d)^{3/2}=0$.

Since $\frac{\partial}{\partial s}I(r,s)=0$ we have $$ I(r,s)=I(r,1)=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{(1+rx)^{2}}{(1-rx)^{2}}\right)dx. $$ From this $$ \frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}\frac{4}{1-r^{2}x^{2}}\, dx. $$ Similarly as above we get $$ \frac{\partial}{\partial r}I(r,1)=\int_{-1}^{1}\frac{4}{\sqrt{1-x^{2}}(1-r^{2}x^{2})}\, dx=\frac{4\pi}{\sqrt{1-r^{2}}}=4\pi(\arcsin r)'. $$ It implies $$ I(r,1)=4\pi\arcsin r+C. $$ Taking the limit $\lim_{r\to0+}$ we obtain $C=0$, that is, $I(r,s)=4\pi\arcsin r$.

  • 3,271
  • 18
  • 26
  • 3
    @sos440 Thanks. It was not a trivial task. How could you discover your general Proposition? – vesszabo Mar 09 '14 at 19:28
  • 1
    I first tried variants of the original integral by changing coefficients. Using inverse symbolic calculators, I found [some patterns](http://math.stackexchange.com/questions/568807/integral-int-11-frac1x-sqrt-frac1x1-x-log-left-fracr-1). Then I tried choose a nice parameters that makes the (conjectured) result look simple. – Sangchul Lee Mar 09 '14 at 20:22

For the purposes of alternative methods, it may be of interest to note that the integrand

$$f(x)=\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)$$ may be rewritten in terms of hyperbolic trigonometric functions. Using $$\tanh^{-1}(z) = \frac{1}{2}\log\left(\frac{1+z}{1-z}\right),$$ and we obtain

$$f(x)=\frac{1}{x}e^{\tanh^{-1}x}\log\left(\frac{1+\frac{2x}{1+2x^2}}{1-\frac{2x}{1+2x^2}}\right) = e^{\tanh^{-1} x}\left(\frac{2\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)}{x}\right).$$

The rational function in the bracket, which we will denote $s(x)$, is symmetric about $x=0$.

The desired integral is

$$I=\int_{-1}^1 f(x)dx = \int_{-1}^1e^{\tanh^{-1}x}s(x)dx,$$

which, by adding the indicated useful definite integral to both side, gives

$$I + \int_{-1}^1 e^{-\tanh^{-1}x}s(x)dx = 2\int_{-1}^1 \frac{s(x)dx}{\sqrt{1-x^2}}.$$

Now using the change of variable $x=-y$ we have $$\int_{-1}^1 e^{-\tanh^{-1} x}s(x)dx = -\int_1^{-1} e^{\tanh y}s(-y)dy = \int_{-1}^1 e^{\tanh y}s(y)dy = I,$$ by the symmetry of $s(x)$. Hence, we finally obtain

$$I = \int_{-1}^1\frac{s(x)dx}{\sqrt{1-x^2}} = 2\int_{-1}^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$

This integral is symmetric about $x=0$, so we have

$$I=4\int_0^1\frac{1}{x\sqrt{1-x^2}}\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx,$$ which can be rewritten $$I=-4\int_0^1\left(\frac{d}{dx}\text{sech}^{-1}x\right)\tanh^{-1}\left(\frac{2x}{1+2x^2}\right)dx.$$

Using integration by parts this results in


We could also make the change of variable $y=\text{sech}^{-1}x$ to obtain

$$I=8\int_0^\infty\frac{y(\cosh^2(y)-2)\sinh y}{\cosh^4(y)+4}dy= 8\int_0^\infty\frac{y\sinh^3 y}{\cosh^4y+4}dy-8\int_0^\infty\frac{y\sinh y}{\cosh^4 y+4}dy.$$

  • 9,831
  • 5
  • 40
  • 73

This answer provides a way to find $I=\displaystyle\int_0^1\dfrac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\ dx$ (which @RonGordon obtained above) with differentiating under the integral sign. A $u$-substitution of $u=\dfrac{1+x^2}{1-x^2}$ yields this.

$$I=\dfrac{1}{2}\displaystyle\int_1^\infty\dfrac{\ln\left(\frac{u^2+2u+2}{u^2-2u+2}\right)}{\sqrt{u^2-1}}\ du.$$ Now integrate by parts with $a=\ln\left(\frac{u^2+2u+2}{u^2-2u+2}\right)$ and $db=\dfrac{du}{\sqrt{u^2-1}}.$ $$I=\left.\ln\left(\dfrac{u^2+2u+2}{u^2-2u+2}\right)\ln(u+\sqrt{u^2-1})\right]^\infty_1+2\displaystyle\int_1^\infty\dfrac{u^2-2}{u^4+4}\ln\left(u+\sqrt{u^2-1}\right)\ du$$ The first term is equal to $0$, so we are left with this. $$I=2\displaystyle\int_1^\infty\dfrac{u^2-2}{u^4+4}\ln\left(u+\sqrt{u^2-1}\right)\ du$$ We now begin the step of differentiating under the integral. Consider the following integral: $$f(a)=a\displaystyle\int_1^\infty\dfrac{x^2-a^2}{x^4+a^4}\ln\left(x+\sqrt{x^2-1}\right)\ dx$$ Note that trivially, $f(0)=0.$ A quick $u=\dfrac{x}{a}$ yields this. $$f(a)=\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{u^2-1}{u^4+1}\ln\left(au+\sqrt{(au)^2-1}\right)\ du$$ Differentiating with respect to $a$ and using the Chain Rule, we get this. $$f'(a)=-1\times\dfrac{-1}{a^2}\times\dfrac{\left(\frac{1}{a}\right)^2-1}{\left(\frac{1}{a}\right)^4+1}\ln\left(a\left(\dfrac{1}{a}\right)+\sqrt{\left(a\left(\dfrac{1}{a}\right)\right)^2-1}\right)+\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{x^2-1}{x^4+1}\times\dfrac{x}{\sqrt{(ax)^2-1}}\ dx$$ Luckily, the first term cancels, so we are left with this. $$f'(a)=\displaystyle\int_{\frac{1}{a}}^\infty\dfrac{x^2-1}{x^4+1}\times\dfrac{x}{\sqrt{(ax)^2-1}}\ dx$$ A $u$-substitution of $u=\sqrt{(ax)^2-1}$ yields this. $$f'(a)=\displaystyle\int_0^\infty\dfrac{u^2+1-a^2}{(u^2+1)^2+a^4}\ du$$ Consider the integral with $u\mapsto\dfrac{\sqrt{a^4+1}}{u}$ $$f'(a)=\dfrac{1}{\sqrt{a^4+1}}\displaystyle\int_0^\infty\dfrac{(1-a^2)u^2+(a^4+1)}{u^4+2u^2+(a^2+1)}\ du$$ If we add these two versions of the integral and divided the numerator and denominator of the integrand by $u^2$, we get the following. $$f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\displaystyle\int_0^\infty\dfrac{1+\frac{\sqrt{a^4+1}}{u^2}}{\left(u-\frac{\sqrt{a^4+1}}{u}\right)^2+2\left(1+\sqrt{a^4+1}\right)}\ du$$ We can finally perform a very nice substitution of $w=u-\dfrac{\sqrt{a^4+1}}{u}$ to solve this integral. $$f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\displaystyle\int_{-\infty}^\infty\dfrac{dw}{w^2+2\left(1+\sqrt{a^4+1}\right)}\ dw$$ Thus, we can finally say that $f'(a)=\dfrac{(1-a^2)+\sqrt{a^4+1}}{2\sqrt{a^4+1}}\times\dfrac{\pi}{\sqrt{2\left(1+\sqrt{a^4+1}\right)}}.$ After a bit of considerable algebra, we can simply that to obtain this. $$f'(a)=\dfrac{\pi}{2}\sqrt{\dfrac{\sqrt{a^4+1}-a^2}{a^4+1}}$$ Integrating, we can now say this about the value of $f(a).$ $$f(a)=\dfrac{\pi}{2}\displaystyle\int_0^a\sqrt{\dfrac{\sqrt{x^4+1}-x^2}{x^4+1}}\ dx$$ Only one $u$-substitution of $u=\sqrt{x^4+1}-x^2$ is required here to obtain this. $$f(a)=\dfrac{\pi}{2\sqrt{2}}\displaystyle\int_{\sqrt{a^4+1}-a^2}^1\dfrac{du}{\sqrt{1-u^2}}$$ This, of course, is equal to $\dfrac{\pi\arccos\left(\sqrt{a^4+1}-a^2\right)}{2\sqrt{2}}.$

We will now manipulate this result to a function with $\arctan$ in it.


Our desired value for our original integral is $\sqrt{2}f\left(\sqrt{2}\right).$

$$\boxed{\displaystyle\int_0^1\dfrac{\ln\left(x^4-2x^2+5\right)-\ln\left(5x^4-2x^2+1\right)}{1-x^2}\ dx=\pi\arctan\left(\sqrt{\dfrac{\sqrt{5}-1}{2}}\right)=\pi\text{arccot}\sqrt{\phi}}$$

So the final answer to the original problem is $4\pi\text{arccot}\sqrt{\phi}.$

  • 789
  • 4
  • 14
  • 1
    Credit goes to [Ronak's answer](https://brilliant.org/problems/a-calculus-problem-by-tunk-fey-ariawan/) on Brilliant.org (Pulished on Dec 31, 2015). – MathGod Dec 27 '16 at 06:56
  • 3
    @IshanSingh I didn't copy from that answer; I was given the original integral and subsequently the initial $u=\dfrac{1+x^2}{1-x^2}$ by a friend who posed this question to me. – Arcturus Dec 28 '16 at 07:58
  • You can remove the square root with $4\pi\,\text{arccosec}\,{\phi}$ – Henry Sep 17 '20 at 01:22

Noteworthy, RIES (http://mrob.com/pub/ries/index.html) finds closed form from numerical value in the form of an equation: $$ \cos{\left( \frac{x}{\pi} \right)}+1=\frac{2}{\phi^6}. $$

Simplifying above, we get another form of the result: $$ I = \pi \arccos{(17-8\sqrt{5})}. $$

  • 3
    I used RIES also but made some simple adjustments to get a simpler form of the answer $I=8\pi \arcsin(\phi-1)$. – Somos Jun 25 '17 at 19:09
  • @Somos Your answer (about $16.74$) seems to be double that of Andrzej Odrzywolek and of Arcturus (about $8.37$) – Henry Sep 17 '20 at 01:19
  • @Henry Oops! Thanks. The correct answer is $\,I=4\pi \arcsin(\phi-1)$. Somehow I doubled it. – Somos Sep 17 '20 at 02:19

This is not really an answer, but grossly too long for an comment. I didn't know how to simplify it beyond the final solution.

$$I=\int_{-1}^1 \frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right)\text{d}{x}$$

Begin with the substitution of $x=-\cos2a$ $$I=\int_{-1}^1 \frac{1}{-\cos2a}\sqrt{\frac{1-\cos2a}{1+\cos2a}}\ln\left(\frac{2\cos^2 2a-2\cos 2a+1}{2\cos^2 2a-2\cos2a+1}\right)\text{d}{x}$$

By the tangent and cos double angle properties

$$I=\int_{-1}^1 -\sec2a|\tan a|\ln\left(\frac{-2\cos^22a+\cos 4a+2}{2\cos2a+\cos4a+2}\right)\text{d}{a}$$

Were just getting started. Now replace $a=\frac{1}{2}\text{gd}(b)$ where $\text{gd}$ is the Gudermannian function.

$$I=\int_{-1}^1 -\sec(\text{gd}(b))|\tan(\text{gd}(\frac{b}{2}))|\ln\left(\frac{-2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}{2\cos^2(\text{gd}(b))+\cos (2\text{gd}(b))+2}\right)\text{d}{a}$$

Hehe. Now we get to simplify a bit. This is under the definition of Gudermannian properties.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{-2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}{2\text{sech}^2 b+(\text{sech}^2b+\tanh^2b)+2}\right)$$

Now, use properties of $\tanh$ and $\text{sech} $ to simplify even further

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{(1-\text{sech}^2 b)+2}{(1+\text{sech}^2 b)+2}\right)$$

Our goal is to create an $\text{arctanh}$ function, but that will obviously take some serious effort. Factor out a $3$ to generate that $1$ needed even if it makes an ugly factoring.

$$I=\int_{-1}^1 -\text{cosh}\space b|\sinh\frac{b}{2}|\ln\left(\frac{3(1-\frac{\text{sech}^2 b}{3})}{3(1+\frac{\text{sech}^2 b}{3})}\right)$$

And now cut out all of the 3's. After this cut, use a property of $\ln$'s to reciprocate the argument of $\ln$. And multiply 2 and 1/2

$$I=\int_{-1}^1 2\text{cosh}\space b|\sinh\frac{b}{2}|\frac{1}{2}\ln\left(\frac{(1+\frac{\text{sech}^2 b}{3})}{(1-\frac{\text{sech}^2 b}{3})}\right)$$

And what do you know! You're there! Use a property of $\ln$ and $\text{arctanh}$ to generate a much CLEANER form (also by throwing the 2 in front).

$$I=2\int_{-1}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This function is even, and we can know that because all parts of what is above, $\cosh b,|\sinh b|, $ etc. all even. So we can do the following.

$$I=4\int_{0}^1 \text{cosh}\space b|\sinh\frac{b}{2}|\text{arctanh}(\frac{\text{sech}^2b}{3})$$

This is just an idea, and like I said not a real solution. I have no idea where to continue beyond this, but I thought it may help to come up with a new idea to solve.

  • 1
    After further inspection, I messed up my work here. I will leave this post here howver becUse the purpose of the post still holds (ideas to solve) –  Nov 22 '15 at 17:42
  • Don't you need to change the limits after you make the first change of variable $x = -\cos 2a$? – r9m Dec 02 '15 at 16:59
  • @user23055 not really. There are lots of mistakes and only consists of only substitution –  Apr 02 '16 at 04:48

Eight years later.

Starting from @Ron Gordon's substitution $$8 \int_0^{\infty} \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du$$ since the roots of the polynomials in $\color{red}{ u^2}$ are simple, we can use partial fraction decomposition (I shall not type the formulae) and we face four integrals $$I=\int \frac \alpha {\beta x^2+\gamma}\log(x)\,dx$$ where all coefficients are complex numbers. Then $$I=\frac{i \alpha \left(\text{Li}_2\left(\frac{i x \sqrt{\beta }}{\sqrt{\gamma }}\right)-\text{Li}_2\left(-\frac{i x \sqrt{\beta }}{\sqrt{\gamma }}\right)+\log (x) \left(\log \left(1-\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)-\log \left(1+\frac{i \sqrt{\beta } x}{\sqrt{\gamma }}\right)\right)\right)}{2 \sqrt{\beta\gamma }}$$ which make $$J=\int_0^\infty \frac \alpha {\beta x^2+\gamma}\log(x)\,dx=\frac{i \alpha \left(\log ^2\left(\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)-\log ^2\left(-\frac{i \sqrt{\beta }}{\sqrt{\gamma }}\right)\right)}{4 \sqrt{\beta\gamma }}=-\frac{\pi \alpha \log \left(\frac{\beta }{\gamma }\right)}{4 \sqrt{\beta\gamma }}$$ This gives as a result $$8 \int_0^{\infty} \frac{(u^2-1)(u^4-6 u^2+1)}{u^8+4 u^6+70 u^4+4 u^2+1} \log(u)\,du=\pi \left(\pi -\cot ^{-1}\left(\frac{1}{4} \sqrt{22+17 \sqrt{5}}\right)\right)$$ I have not been able to simplify further.


If you look at this question of mine, @Jyrki Lahtonen made the simplification I was not able to do.

Claude Leibovici
  • 214,262
  • 52
  • 92
  • 197

Figured I would contribute and add a self-contained real analytic method:

I will use the following representations that are fairly straight-forward to prove: $$2\int_{0}^{\infty}\frac{\cos(x)-\cos(t \, x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\ln \left((t^2 -a)^2 + b^2\right)-\ln \left((1-a)^2+b^2\right)$$ $$\int_{0}^{\infty}\frac{\sin(x)}{x}\left(e^{-x\sqrt{-a-bi}}+e^{-x\sqrt{-a+bi}}\right) \, dx=\arctan\left(\frac{1}{\sqrt{-a-bi}}\right)+\arctan\left(\frac{1}{\sqrt{-a+bi}}\right)$$ $$\int_{0}^{\infty}\frac{\cos(t)-\cos(t \, x)}{x^2-1} \, dx=\frac{\pi}{2}\sin(t),\> \text{for} \, \> t\geq 0$$

Now we can begin evaluating $I$: $$I=\int_{-1}^{1}\frac{1}{x}\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2x^2+2x+1}{2x^2-2x+1}\right) \, dx$$

Enforce the substitution $\sqrt{\frac{1+x}{1-x}} = u$: $$\implies I = \int_{0}^{\infty}\frac{4u^2}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right)\, du \stackrel{u \, \mapsto \frac{1}{u}}{=}\int_{0}^{\infty}\frac{4}{u^4-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$

From adding the two, we can deduce: $$I = \int_{0}^{\infty}\frac{2}{u^2-1}\ln\left(\frac{5u^4-2u^2+1}{u^4-2u^2+5}\right) \, du$$

Since $$\ln\left(5 u^4-2 u^2+1\right)-\ln(4)=\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)$$ $$\ln(u^4-2 u^2+5)-\ln(4)=\ln((u^2-1)^2+4)-\ln(4)$$

We can write $$A=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln\left(\left(u^2-\frac{1}{5}\right)^2+\frac{4}{25}\right)-\ln\left(\frac{4}{5}\right)\right)\,du\\ B=\int_{0}^{\infty}\frac{2}{u^2-1}\left(\ln((u^2-1)^2+4)-\ln(4)\right)\, du$$ such that $I = A-B$

Now from the starting representations we deduce: $$\begin{align} \implies A &= \int_{0}^{\infty}\frac{4}{u^2-1}\int_{0}^{\infty}\frac{\cos(t)-\cos(u \, t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\,dt\,du \\ &= 2\pi\int_{0}^{\infty}\frac{\sin(t)}{t}\left(e^{-t\sqrt{(-1-2i)/5}}+e^{-t\sqrt{(-1+2i)/5}}\right)\, dt \\ &= 2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)+2\pi\arctan\left(\frac{\sqrt{5}}{\sqrt{-1+2i}}\right) \\ &= \Re\left(4\pi \arctan\left(\frac{\sqrt{5}}{\sqrt{-1-2i}}\right)\right)\end{align}$$

Similarly, one finds $$B=\Re\left(4\pi\arctan\left(\frac{1}{\sqrt{-1-2i}}\right)\right)$$

By using the identity: $$\arctan (x)+\arctan (y)=\arctan \left(\frac{x+y}{1-x y}\right)$$

One deduces $$\boxed{I=4\pi \operatorname{arccot} \left(\sqrt{\phi}\right)}$$

  • 3,134
  • 1
  • 8
  • 35