Integral $\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx$

Is there a closed form for the integral $$\int_0^1\frac{\ln\left(x+\sqrt2\right)}{\sqrt{2-x}\,\sqrt{1-x}\,\sqrt{\vphantom{1}x}}\mathrm dx.$$ I do not have a strong reason to be sure it exists, but I would be very interested to see an approach to find one if it does exist.

• The numerical calculation gives $1.71611436816290240644184900503$. – user64494 Nov 17 '13 at 21:46
• @FRida Mauer : Try wolframalpha.com All you need is an Internet connection. They seem to accept Maple syntax, since theyare run by Mathematica, I'm sure they accept Mathematica syntax, and I bet they accept just about any reasonable syntax. If woframalpha.com can't find a closed form, then probably none exists. It may give you a nasty answer involving lots of special functions, which you may or may not consider a closed form. – Stefan Smith Nov 17 '13 at 21:57
• @StefanSmith: false. There are loads of examples in this site in which we evaluate definite integrals that WA, Mathematica, Maple, etc., do not find simple closed forms for. – Ron Gordon Nov 17 '13 at 21:59
• @StefanSmith: My pleasure. Start with this math.stackexchange.com/questions/562694/… – Ron Gordon Nov 18 '13 at 22:16
• Perhaps the following intermediary result might prove itself useful : $$\int_0^1\frac{\ln(x+a)}{\sqrt{x(1-x)}}dx=\pi\cdot\left[2\cdot\ln\left(1+\sqrt{1+\tfrac1a}\right)+\ln\frac a4\right]$$ – Lucian Nov 18 '13 at 22:24

For $a > 0$, let $b = \frac12 + \frac1a$ and $I(a)$ be the integral $$I(a) = \int_0^1 \frac{\log(a+x)}{\sqrt{x(1-x)(2-x)}}dx$$ Substitute $x$ by $\frac{1}{p+\frac12}$, it is easy to check we can rewrite $I(a)$ as

$$I(a) = -\sqrt{2}\int_\infty^{\frac12}\frac{\log\left[a (p + b)/(p + \frac12)\right]}{\sqrt{4p^3 - p}} dp$$ Let $\wp(z), \zeta(z)$ and $\sigma(z)$ be the Weierstrass elliptic, zeta and sigma functions associated with the ODE:

$$\wp'(z)^2 = 4\wp(z)^3 - g_2 \wp(z) - g_3\quad\text{ for }\quad g_2 = 1 \;\text{ and }\; g_3 = 0.$$

In terms of $\wp(z)$, we can express $I(a)$ as

$$I(a) = \sqrt{2}\int_0^\omega \log\left[a \left(\frac{\wp(z) + b}{\wp(z) + \frac12}\right)\right] dz = \frac{1}{\sqrt{2}}\int_{-\omega}^\omega \log\left[a \left(\frac{\wp(z) + b}{\wp(z) + \frac12}\right)\right] dz$$

where $\;\displaystyle \omega = \int_\frac12^\infty \frac{dp}{\sqrt{4p^3 - p}} = \frac{\pi^{3/2}}{2\Gamma\left(\frac34\right)^2}\;$ is the half period for $\wp(z)$ lying on real axis. Since $g_3 = 0$, the double poles of $\wp(z)$ lies on a square lattice $\mathbb{L} = \{\; 2\omega ( m + i n ) : m, n \in \mathbb{Z} \;\}$ and and we can pick the other half period $\;\omega'$ as $\;i\omega$.

Notice $\wp(\pm i \omega) = -\frac12$. If we pick $u \in (0,\omega)$ such that $\wp(\pm i u) = -b$, the function inside the square brackets in above integral is an ellitpic function with zeros at $\pm i u + \mathbb{L}$ and poles at $\pm i \omega + \mathbb{L}$. We can express $I(a)$ in terms of $\sigma(z)$ as

$$I(a) = \frac{1}{\sqrt{2}}\int_{-\omega}^\omega \log\left[ C\frac{\sigma(z-iu)\sigma(z+iu)}{\sigma(z-i\omega)\sigma(z+i\omega)}\right] dz \quad\text{ where }\quad C = a\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-iu)\sigma(iu)}\right).$$

Let $\varphi_{\pm}(\tau)$ be the integral $\displaystyle \int_{-\omega}^\omega \log\sigma(z+\tau) dz$ for $\Im(\tau) > 0$ and $< 0$ respectively. Notice $\sigma(z)$ has a simple zero at $z = 0$. We will choose the branch cut of $\log \sigma(z)$ there to be the ray along the negative real axis.

When we move $\tau$ around, as long as we don't cross the real axis, the line segment $[\tau-\omega,\tau+\omega]$ won't touch the branch cut and everything will be well behaved. We have

\begin{align} & \varphi_{\pm}(\tau)''' = -\wp(\tau+\omega) + \wp(\tau-\omega) = 0\\ \implies & \varphi_{\pm}(\tau)'' = \zeta(\tau+\omega) - \zeta(\tau-\omega) \quad\text{ is a constant}\\ \implies & \varphi_{\pm}(\tau)'' = 2 \zeta(\omega)\\ \implies & \varphi_{\pm}(\tau) = \zeta(\omega) \tau^2 + A_{\pm} \tau + B_{\pm} \quad\text{ for some constants } A_{\pm}, B_{\pm} \end{align}

Let $\eta = \zeta(\omega)$ and $\eta' = \zeta(\omega')$. For elliptic functions with general $g_2, g_3$, there is always an identity $$\eta \omega' - \omega \eta' = \frac{\pi i}{2}$$ as long as $\omega'$ is chosen to satisfy $\Im(\frac{\omega'}{\omega}) > 0$. In our case, $\omega' = i\omega$ and the symmetric of $\mathbb{L}$ forces $\eta = \frac{\pi}{4\omega}$. This implies

$$\varphi_{\pm}(\tau) = \frac{\pi}{4\omega}\tau^2 + A_{\pm}\tau + B_{\pm}$$

Because of the branch cut, $A_{+} \ne A_{-}$ and $B_{+} \ne B_{+}$. In fact, we can evaluate their differences as

\begin{align} A_{+} - A_{-} &= \lim_{\epsilon\to 0} \left( -\log\sigma(i\epsilon-\omega) + \log\sigma(-i\epsilon-\omega) \right) = - 2 \pi i\\ B_{+} - B_{-} &= \lim_{\epsilon\to 0} \int_{-\omega}^0 \left( \log\sigma(i\epsilon+z) - \log\sigma(-i\epsilon+z) \right) dz = 2\pi i\omega \end{align} Apply this to our expression of $I(a)$, we get

\begin{align} I(a) &= \frac{1}{\sqrt{2}}\left(2\omega\log C + \varphi_{-}(-iu)+\varphi_{+}(iu)-\varphi_{-}(-i\omega)-\varphi_{+}(i\omega)\right)\\ &= \frac{1}{\sqrt{2}}\left\{ 2\omega\log\left[a\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-iu)\sigma(iu)}\right)\right] + \frac{\pi}{2\omega}(\omega^2 - u^2) + 2\pi(u-\omega) \right\} \end{align}

Back to our original problem where $a = \sqrt{2} \iff b = \frac{1+\sqrt{2}}{2}$. One can use the duplication formula for $\wp(z)$ to vertify $u = \frac{\omega}{2}$. From this, we find: $$I(\sqrt{2}) = \sqrt{2}\omega\left\{ \log\left[\sqrt{2}\left(\frac{\sigma(-i\omega)\sigma(i\omega)}{\sigma(-i\frac{\omega}{2})\sigma(i\frac{\omega}{2})}\right)\right] - \frac{5\pi}{16}\right\}$$

It is known that $| \sigma(\pm i\omega) | = e^{\pi/8}\sqrt[4]{2}$. Furthermore, we have the identity:

$$\wp'(z) = - \frac{\sigma(2z)}{\sigma(z)^4} \quad\implies\quad \left|\sigma\left( \pm i\frac{\omega}{2} \right)\right| = \left|\frac{\sigma(\pm i \omega)}{\wp'\left(\pm i\frac{\omega}{2}\right)}\right|^{1/4} = \left(\frac{\sigma(\omega)}{1+\sqrt{2}}\right)^{1/4}$$ Combine all these, we get a result matching other answer.

\begin{align} I(\sqrt{2}) &= \sqrt{2}\omega\left\{\log\left[\sqrt{2}\sigma(\omega)^{3/2}\sqrt{1+\sqrt{2}}\right] - \frac{5\pi}{16}\right\}\\ &= \frac{\pi^{3/2}}{\sqrt{2}\Gamma\left(\frac34\right)^2}\left\{\frac78\log 2 + \frac12\log(\sqrt{2}+1) - \frac{\pi}{8} \right\} \end{align}

• This is just great!! I am going to award a bounty of +500 to your answer. – Shobhit Bhatnagar Mar 15 '14 at 10:22
• @achillehui Dear Teacher, Can I ask you to look at my problem, Please..If you have few minutes? You are Professional Mathematician. Thank you so much. Best Regards.. math.stackexchange.com/questions/2559814/… – MathLover Dec 12 '17 at 21:44

$$\frac{\pi^{3/2}}{8\,\sqrt2}\cdot\frac{7\ln2-\ln\left(17-12\,\sqrt2\right)-\pi}{\Gamma\left(\frac34\right)^2}$$

• "...I would be very interested to see an approach to find one if it does exist." Apparently alchemy + silence is an accepted approach with some voters here. – Ron Gordon Nov 18 '13 at 19:07
• @Cleo I have to say, this is a very unusual approach, and much less useful for most people than, say, Ron Gordon's detailed answers, but I do not see why I shouldn't accept the position of yours a valid one (at least, formally). And, well, for some questions your answer is the only one we get. Having about 15 years of experience with Mathematica and Maple, I was not able to manipulate them to get any of your answers, and I'm pretty sure they are not generated by a computer. I think, Namagiri is a more likely source :) – Vladimir Reshetnikov Nov 18 '13 at 20:25
• @Cleo Do you even know what an "axiomatic system" is or are you just playing with words? – Did Nov 20 '13 at 21:50
• I have the feeling this Namagiri's magic consists in answering its own questions. Not sure that's Ramanujan's Namagiri. I dare to say that because I was advised not to take Cleo's posts and comments too seriously... – Julien Nov 21 '13 at 4:28
• (-1): This answer isn't at all useful for those trying to understand the solution. – user61527 Feb 19 '14 at 22:37


A '$\large\tt partial$' answer:

\begin{align} &\int_{0}^{1} {\ln\pars{x + \root{2}} \over \root{2 - x}\root{1 - x}\root{x}}\,\dd x \\[3mm]&= \int_{0}^{1/2} {\ln\pars{x + \root{2}} \over \root{2 - x}\root{1 - x}\root{x}}\,\dd x + \int_{1/2}^{1} {\ln\pars{x + \root{2}} \over \root{2 - x}\root{1 - x}\root{x}}\,\dd x \\[3mm]&= 2\int_{0}^{\root{2}/2} {\ln\pars{x^{2} + \root{2}} \over \root{2 - x^{2}}\root{1 - x^{2}}}\,\dd x - 2\int_{\root{2}/2}^{0} {\ln\pars{1 - x^{2} + \root{2}} \over \root{1 + x^{2}}\root{1 - x^{2}}}\,\dd x \\[3mm]&= 2\int_{0}^{\pi/4} {\ln\pars{\sin^{2}\pars{\theta} + \root{2}} \over \root{2 - \sin^{2}\pars{\theta}}}\,\dd\theta + 2\int_{0}^{\pi/4} {\ln\pars{1 + \root{2} - \sin^{2}\pars{\theta}} \over \root{1 + \sin^{2}\pars{\theta}}}\,\dd\theta \\[3mm]&= 2\int_{0}^{\pi/4} {\ln\pars{\sin^{2}\pars{\theta} + \root{2}} \over \root{1 + \cos^{2}\pars{\theta}}}\,\dd\theta + 2\int_{-\pi/2}^{-\pi/4} {\ln\pars{1 + \root{2} - \cos^{2}\pars{\theta}} \over \root{1 + \cos^{2}\pars{\theta}}}\,\dd\theta \\[3mm]&=2\int_{0}^{\pi/2} {\ln\pars{\sin^{2}\pars{\theta} + \root{2}} \over \root{1 + \cos^{2}\pars{\theta}}}\,\dd\theta =2\int_{0}^{\pi/2} {\ln\pars{\bracks{1 - \cos\pars{2\theta}}/2 + \root{2}} \over \root{1 + \bracks{1 + \cos\pars{2\theta}}/2}}\,\dd\theta \\[3mm]&=\int_{0}^{\pi} {\ln\pars{\bracks{2\root{2} + 1 - \cos\pars{\theta}}/2} \over \root{\bracks{3 + \cos\pars{\theta}}/2}}\,\dd\theta =\root{2}\int_{0}^{\pi} {\ln\pars{\root{2} + 1/2 - \cos\pars{\theta}/2} \over \root{3 + \cos\pars{\theta}}}\,\dd\theta \end{align} Mathematica can evaluate this integral ( it can not calculate the original one ):

I'm still struggling with the integral !!!

• If we let $\alpha$ be the first hypergeometric number (1,1,5/4, etc.) and $\beta$ be the second one, and let $G = \Gamma\left(\frac{3}{4}\right)$, we can simplify that result to: $$\frac{\pi^{3/2}}{8\sqrt{2}}\cdot\frac{-\frac{4}{3} \left(\alpha\left(\sqrt{2}-1\right)^2 +6 \log \left(\sqrt{2}-1\right)\right)+\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(7 \sqrt{2}-10\right) \beta +5 \left(\sqrt{2}-2\right)\right)}{G^2}$$ – Eugene Bulkin Mar 9 '14 at 7:25
• Now the weirdest thing? $\frac{8\sqrt{2}G^4}{5\pi^2} \left(\left(7 \sqrt{2}-10\right) \beta +5 \left(\sqrt{2}-2\right)\right) = -\pi/2$ and $-\frac{4}{3} \left(\alpha\left(\sqrt{2}-1\right)^2 +6 \log \left(\sqrt{2}-1\right)\right) = 7\ln 2 - \ln(17-12\sqrt{2})-\pi/2$. So those are the pieces of the known answer. – Eugene Bulkin Mar 9 '14 at 7:26
• Note for my above comments: when I say "log" I mean "ln" interchangeably; I was copy/pasting from Mathematica. – Eugene Bulkin Mar 9 '14 at 7:42

More of a sketchy draft, rather than an answer, but perhaps better than nothing in terms of actual approach: Please do not upvote ! As I said, this is NOT an actual answer. Thank you.

OK, I think I got it: Make the following simple substitution: $t=1-x$. Then the integral becomes $$\int_0^1\frac{\ln(a-t)}{\sqrt{1-t^2}\sqrt t}dt\qquad,\qquad a=1+\sqrt2$$ which, unlike its predecessor, can be expressed, by machines and/or people far smarter than I will ever be, in terms of hypergeometric functions of the form $_3F_2(a^{-2})$, as follows:

Perhaps by factoring a inside the log, then using the properties of the logarithm $\ln ab=\ln a+\ln b$ to break up the integral into a sum of two, then recognizing the expression of the Beta function of arguments $\frac12$ and $\frac14$ in the first, and using the Taylor expansion of the natural logarithm and or integration by parts for the second, then ultimately making use of the wealth of information on hypergeometric functions freely available at the NIST DLMF math data base or elsewhere in order to simplify those intermediary hypergeometric expressions. Hope this helps, and that it will serve as a starting point or source of inspiration for (more complete) future answers.

• So it doesn't look like you can do anything from those hypergeometric functions, really. You can transform them into a $_2F_1$ and another $_3F_2$ but the fact that you have $1/\alpha^2$ and not 1 throws a wrench into it for sure. – Eugene Bulkin Feb 23 '14 at 22:47