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Please help me to evaluate this integral: $$ I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\, \,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1 $$ Mathematica could not evaluate it in a closed form. A numerical integration returned $$I \approx 4.25314982536869548103063\ldots\,,\tag2$$ but neither WolframAlpha nor ISC+ could find a plausible closed form for this.

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    $\begingroup$ Maybe it just doesn't exist? $\endgroup$
    – Kaster
    Jul 24, 2014 at 18:15
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    $\begingroup$ It can be rewritten as $\quad\displaystyle\int_0^\infty\frac{\ln(\sinh x)}{1+\sinh x}\cdot e^{x/2}~dx,\quad$ for which Mathematica is able to return a hideous expression. $\endgroup$
    – Lucian
    Jul 24, 2014 at 19:02
  • $\begingroup$ @Lucian I simplified the result to 3 lines with only 1 PolyLog term, but it still looks bad... $\endgroup$ Jul 24, 2014 at 19:53
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    $\begingroup$ Is there anymore background information on the integral. Eg. From a problem? If you created it yourself, try and play around with the bounds and see if there are nicer expressions. $\endgroup$ Jul 24, 2014 at 20:57
  • $\begingroup$ @Lucian Your observation already suffices to obtain an answer. After the substitution $t=e^{x/2}$ the integral becomes $\int R(t)\ln t\,dt$ with rational $R(t)$, and thus can always be computed in terms of dilogarithms. $\endgroup$ Jul 25, 2014 at 9:26

3 Answers 3

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With some help from a CAS I got this result with only one dilogarithm term: $$I=\frac{\sqrt[4]2\sqrt{2-\sqrt2}}{16}\\\left\{\left[16\ln2+\left(1+\sqrt2\right)\left(4\pi-16\arctan\sqrt{\sqrt2-1}\right)\right]\ln\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\-\Big(32\left(1+\sqrt2\right)\ln2+4\pi\Big)\arctan\sqrt{\sqrt2-1}-3\pi\arctan\frac{2\sqrt{2+10\sqrt2}}7\\-16\arctan^2\sqrt{\sqrt2-1}+4\ln^2\!\left(1+\sqrt2+\sqrt{2+2\sqrt2}\right)\\+3\pi^2-\frac{32}{\sqrt{2-\sqrt2}}\,\Re\left[\left(-1\right)^{5/8}\operatorname{Li}_2\!\left(2i\left( 1+\sqrt{1+i}\right)\right)\right]\right\}.$$


Update: It was suggested in comments that this expression gives a numerically incorrect result when evaluated with Maple. Here is what I got with Maple 18:

Maple screenshot

This value numerically matches the integral. I also checked it with Mathematica 10 and it gave the same numeric result.

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    $\begingroup$ How did you do this? $\endgroup$
    – Kirill
    Aug 29, 2014 at 23:21
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    $\begingroup$ I started with a large expression similar to one given in another answer, then simplified it using Mathematica, then using Maple, then manually, and again, and again..., and finally I got this :) I do not recall any particularly remarkable idea that I used during this process. $\endgroup$ Aug 30, 2014 at 0:52
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    $\begingroup$ Compared with the other answer, this is positively elegant. $\endgroup$
    – Kirill
    Aug 30, 2014 at 9:22
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    $\begingroup$ Forgive my obscure reference to Spartacus, but: The bringer of results!... The slayer of integrals!... The champion of Stack Exchange!... VLADIMIR !!! :-) $\endgroup$
    – Lucian
    Aug 31, 2014 at 15:11
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    $\begingroup$ It looks like Cleo's style answer $\endgroup$ Sep 3, 2014 at 11:07
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\begin{align} -\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x &=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\ &=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\ &=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\ \end{align} For simplicity's sake, let $\displaystyle\frac{1}{x^4-2x^2-1}=\sum^4_{k=1}\frac{c_k}{x-r_k}$. The integral becomes $$I=-4\sum^4_{k=1}\int^1_0\frac{c_k\ln(1-x^4)-c_k\ln{2}-2c_k\ln{x}}{x-r_k}{\rm d}x$$ The second integral is \begin{align} -c_k\ln{2}\int^1_0\frac{1}{x-r_k}{\rm d}x=-c_k\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right) \end{align} The third integral is \begin{align} -2c_k\int^1_0\frac{\ln{x}}{x-r_k}{\rm d}x &=2c_k\int^{1/r_k}_0\frac{\ln(r_kx)}{1-x}{\rm d}x\\ &=-2c_k{\rm Li}_2\left(\frac{1}{r_k}\right) \end{align} Pluck these results back in. $$I=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)-\int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\right]$$ The remaining integral is \begin{align} &\ \ \ \ \ \int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\\ &=\sum_{j=1,-1,i,-i}\int^1_0\frac{\ln(1+jx)}{x-r_k}{\rm d}x\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\ln\left(1+\frac{j\lambda}{y}+jr_k\right)\frac{{\rm d}y}{y}\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\left[\ln\left(1+\frac{1+jr_k}{j\lambda}y\right)-\ln\left(\frac{y}{j\lambda}\right)\right]\frac{{\rm d}y}{y}\\ &=-\sum_{j=1,-1,i,-i}\int^{\frac{1+jr_k}{j-jr_k}}_{-\frac{1+jr_k}{jr_k}}\frac{\ln(1+y)}{y}-\frac{1}{y}\ln\left(\frac{y}{1+jr_k}\right){\rm d}y\\ &=-\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right] \end{align} Final Result: \begin{align} \color\purple{\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x =4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)+\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]\right]} \end{align} where \begin{align} \frac{c_1}{x-r_1}&=\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x-\sqrt{1+\sqrt{2}}\right)}\\ \frac{c_2}{x-r_2}&=-\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x+\sqrt{1+\sqrt{2}}\right)}\\ \frac{c_3}{x-r_3}&=-\frac{1}{4\sqrt{2}\left(x+i\sqrt{\sqrt{2}-1}\right)}\\ \frac{c_4}{x-r_4}&=-\frac{1}{4\sqrt{2}\left(x-i\sqrt{\sqrt{2}-1}\right)}\\ \end{align}

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  • $\begingroup$ Do you think these sums can be evaluated? Probably not, but would like your comments @SuperAbound $\endgroup$
    – apg
    Aug 28, 2014 at 15:17
  • $\begingroup$ @AlexanderGiles Technically, they can be, but it is certainly not possible by hand, as would get an ugly barrage of terms as shown in user153012's answer. $\endgroup$ Aug 28, 2014 at 15:20
  • $\begingroup$ OK nice. Interesting work. $\endgroup$
    – apg
    Aug 28, 2014 at 17:31
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    $\begingroup$ The numerical value doesn't match the other answer. At least the $2$ in $2\log2\log(\cdots)$ looks wrong. $\endgroup$
    – Kirill
    Aug 29, 2014 at 9:17
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    $\begingroup$ @SuperAbound The answer's numerical value doesn't match the integral's numerical value, that's the main issue. What I meant was: you have $-4\sum_k\int\frac{-c_k \log 2}{x-r_k}$ in one line, evaluate the integral to $-c_k\log2\log\frac{1-r_k}{-r_k}$, and then in the answer you have $4\sum_k c_k(2\log2\log\frac{1-r_k}{-r_k})$, where the $2$ came out of nowhere. I also think the evaluation of $\int\frac{\log(1-x^4)}{x-r_k}$ is incorrect, it doesn't match the integral numerically, according to my computer. $\endgroup$
    – Kirill
    Aug 29, 2014 at 23:10
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Okay, first of all: it is not easy to answer this question, and I'm not able to show the truth in an elegant way. I did the job with Maple computer algebra system (CAS).

In this form there is no chance to evaulate to solution even with CAS, so we transform into the form, what @Lucian offered. Let $f$ be $$f:=\frac{\ln(\sinh x)}{1+ \sinh x} \cdot e^{x/2}$$

I calculated for you the indefinite integral $\int f(x) \ dx$. It is too long to render with $\LaTeX$, so I paste here the raw Maple output:

indefinite integral

After that I calculated the definite integral $\int_0^{\infty} f(x) \ dx$. Here is the closed form:

definite integral

Which is by 100 digits precision: $4.253149825368695481030630266176679795147658282 \\ 387249226057297960587772473334190569825379100016048097$

A good approximation for this integral for example $$\frac{5}{9} \cdot 3^{4/5} \cdot 5^{2/3} \cdot \ln(3)^{8/9}.$$ The value of this one is $4.2531498232523271$ which is correct for $9$ digits, and also has quite an easy form.

If you need the closed form in raw text I can paste here. For a more accurate answer I think we need here @Cleo or @Tunk-Fey.

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    $\begingroup$ After about 4 days I tried to crack this integral, I decided to gave up. Of course I have a partial answer, but as a partial answer it turns out somewhat too lengthy and messy. I doubt it can be solved analytically. Maybe @Cleo can do it. $\endgroup$
    – Tunk-Fey
    Sep 2, 2014 at 8:05

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