Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ [closed]

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.

• Maybe it just doesn't exist? Jul 24, 2014 at 18:15
• 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. Jul 24, 2014 at 19:02
• @Lucian I simplified the result to 3 lines with only 1 PolyLog term, but it still looks bad... Jul 24, 2014 at 19:53
• 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. Jul 24, 2014 at 20:57
• @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. Jul 25, 2014 at 9:26

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:

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

• How did you do this? Aug 29, 2014 at 23:21
• 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. Aug 30, 2014 at 0:52
• Compared with the other answer, this is positively elegant. Aug 30, 2014 at 9:22
• Forgive my obscure reference to Spartacus, but: The bringer of results!... The slayer of integrals!... The champion of Stack Exchange!... VLADIMIR !!! :-) Aug 31, 2014 at 15:11
• It looks like Cleo's style answer Sep 3, 2014 at 11:07

\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}

• Do you think these sums can be evaluated? Probably not, but would like your comments @SuperAbound
– apg
Aug 28, 2014 at 15:17
• @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. Aug 28, 2014 at 15:20
• OK nice. Interesting work.
– apg
Aug 28, 2014 at 17:31
• The numerical value doesn't match the other answer. At least the $2$ in $2\log2\log(\cdots)$ looks wrong. Aug 29, 2014 at 9:17
• @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. Aug 29, 2014 at 23:10

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:

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

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.

• 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. Sep 2, 2014 at 8:05