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.

Answer

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.

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

Attribution
Source : Link , Question Author : Frida Mauer , Answer Author : achille hui

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