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

Question

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.

Answer 1

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.

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

Answer 2

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

Answer 3

I know I'm late answering this, but here's an alternative solution that relies on the Beta function and some algebric work. It's by no means as elegant as the solution presented by @achille hui.

$$\begin{align}I&=\int_{0}^{1}\underbrace{\frac{\log\left(\sqrt{2}+x\right)}{\sqrt{(2-x)(1-x)x}}dx}_{x\rightarrow 1-x}=\int_{0}^{1}\frac{\log\left(\sqrt{2}+1-x\right)}{\sqrt{x(1-x^2)}}dx\end{align}$$

In one hand, we've got that $I$ is equal to: $$\int_{0}^{1}\frac{\log\left(1-\frac{x}{\sqrt{2}+1}\right)+\log\left(\sqrt{2}+1\right)}{\sqrt{x(1-x^2)}}dx$$

On the other hand, $I$ can be expressed as: $$\int_{0}^{1}\underbrace{\frac{\log\left(\sqrt{2}+1-x\right)}{\sqrt{x(1-x^2)}}dx}_{x\rightarrow \frac{1-x}{1+x}}=\int_{0}^{1}\frac{\log\left(\sqrt{2}+\frac{2x}{1+x}\right)}{\sqrt{x(1-x^2)}}dx\\=\int_{0}^{1}\frac{\log\left(\sqrt{2}\right)-\log\left(1+x\right)+\log\left(1+\left(\sqrt{2}+1\right)x\right)}{\sqrt{x(1-x^2)}}dx$$

Combining both representations: $$\begin{align}2I&=\int_{0}^{1}\frac{\log\left(\sqrt{2}\right)+\log\left(\sqrt{2}+1\right)-\log\left(1+x\right)+\log\left(1+2x-x^2\right)}{\sqrt{x(1-x^2)}}dx\\I&=\frac{1}{2}\int_{0}^{1}\frac{\frac{3}{2}\log\left(2\right)+\log\left(\sqrt{2}+1\right)+\log(x)-\log\left(1+x\right)+\log\left(\frac{x^{-1}-x}{2}+1\right)}{\sqrt{x(1-x^2)}}dx\end{align}$$

Now, to compute these integrals, let's split them according to the substitutions that we need to apply.

$$\begin{align} I_1&=\frac{1}{2}\int_{0}^{1}\underbrace{\frac{\frac{3}{2}\log\left(2\right)+\log\left(\sqrt{2}+1\right)+\log(x)}{\sqrt{x(1-x^2)}}dx}_{x\rightarrow x^{1/2}}\\ &=\frac{1}{4}\int_{0}^{1}\left(\frac{3}{2}\log\left(2\right)+\log\left(\sqrt{2}+1\right)+\frac{\log(x)}{2}\right)x^{1/4-1}\left(1-x\right)^{1/2-1}dx\\ &=\left(\frac{3}{8}\log\left(2\right)+\frac{\log\left(\sqrt{2}+1\right)}{4}\right)\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)+\frac{1}{8}\lim_{t\to 1/4}\frac{d}{dt}\mathfrak{B}\left(t,\frac{1}{2}\right)\\ &=\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{8}\left(3\log(2)+2\log\left(\sqrt{2}+1\right)+\psi^{(0)}\left(\frac{1}{4}\right)-\psi^{(0)}\left(\frac{3}{4}\right)\right)\\ &=\boxed{\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{8}\left(3\log(2)+2\log\left(\sqrt{2}+1\right)-\pi\right)} \end{align}$$

$$\begin{align} I_2&=-\frac{1}{2}\int_{0}^{1}\underbrace{\frac{\log(1+x)}{\sqrt{x(1-x^2)}}dx}_{x\rightarrow \frac{1-x}{1+x}}=-\frac{1}{2}\int_{0}^{1}\frac{\log\left(\frac{2}{1+x}\right)}{\sqrt{x(1-x^2)}}dx\\ &=-\frac{\log(2)}{4}\int_{0}^{1}\frac{dx}{\sqrt{x(1-x^2)}}\\ &=\boxed{-\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{8}\log(2)} \end{align}$$

$$\begin{align} I_3&=\frac{1}{2}\int_{0}^{1}\frac{\log\left(\frac{x^{-1}-x}{2}+1\right)}{\sqrt{x(1-x^2)}}dx=\int_{0}^{1}\underbrace{\frac{\log\left(\sinh\left(-\log(x)\right)+1\right)}{x\sqrt{2(\sinh\left(-\log(x)\right)}}dx}_{x\rightarrow \exp\left(-\sinh^{-1}(\tan^2(x))\right)}\\ &=-2\int_{0}^{\pi/2}\underbrace{\frac{\log\left(\cos(x)\right)}{\sqrt{1+\cos^2(2x)}}dx}_{x\rightarrow \pi/2-x}=-2\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(2x)}}dx\\ &=-\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\cos(x)\right)}{\sqrt{1+\cos^2(2x)}}dx=-\int_{0}^{\pi/2}\underbrace{\frac{\log\left(\sin(2x)\right)-\log(2)}{\sqrt{1+\cos^2(2x)}}dx}_{2x\rightarrow x}\\ &=\frac{1}{2}\int_{0}^{\color{red}{\pi}}\frac{\log(2)-\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx=\frac{\color{red}{2}}{2}\int_{0}^{\color{red}{\pi/2}}\frac{\log(2)-\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx\\ &=\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{4}\log(2)-\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{32}\left(\psi^{(0)}\left(\frac{1}{4}\right)+2\psi^{(0)}\left(\frac{2}{4}\right)-3\psi^{(0)}\left(\frac{3}{4}\right)\right)\\ &=\boxed{\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{16}\left(3\log(2)+\pi\right)} \end{align}$$

Note: the first integral just requires the substitution $\cos(x)\rightarrow x$, then it will be the standard Beta function application; I recently solved the second integral [here][1].

Gathering all results: $$\frac{\mathfrak{B}\left(\frac{1}{4},\frac{1}{2}\right)}{16}\left(7\log(2)+4\log(\sqrt{2}+1)-\pi\right)\\ \frac{\Gamma^2\left(\frac{1}{4}\right)}{\sqrt{2\pi}}\left(\frac{7}{16}\log(2)+\frac{\log(\sqrt{2}+1)}{4}-\frac{\pi}{16}\right)$$ [1]: https://math.stackexchange.com/a/4491145/705845

Answer 4

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{1} {\ln\pars{x + \root{2}} \over \root{2 - x}\root{1 - x}\root{x}}\,\dd x:\ {\Large ?}}$

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 !!!

Answer 5

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.