Integrals and Miscellaneous 9

@oommemem told me this equation. Let us show this.$$\int_0^\infty\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac{\pi}{n}$$PROOF.
Substituting $\dfrac{1}{y}=1+x^n$ we get$$\int_0^\infty\frac{dx}{1+x^n}=\frac{1}{n}\int_0^1y^{-\frac{1}{n}}(1-y)^{\frac{1}{n}-1}dy$$We can see that is the beta function,namely$$\int_0^1y^{-\frac{1}{n}}(1-y)^{\frac{1}{n}-1}dy=B\left(1-\frac{1}{n},\frac{1}{n}\right)=\G\left(1-\frac{1}{n}\right)\G\left(\frac{1}{n}\right)$$where using the reflection formula$$\G\left(1-\frac{1}{n}\right)\G\left(\frac{1}{n}\right)=\pi\csc\frac{\pi}{n}$$
Hence,$$\int_0^\infty\frac{dx}{1+x^n}=\frac{\pi}{n}\csc\frac{\pi}{n}$$

\begin{eqnarray}\int_0^1 x^s\arcsin xdx&=&\frac{\pi}{2(s+1)}-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)\G(\frac{s+3}{2})}\\\int_0^1 x^s\ln x\arcsin xdx&=&-\frac{\pi}{2(s+1)^2}+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)^2\G(\frac{s+3}{2})}\\&&+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{4(s+1)\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]\\\int_0^1 x^s\ln^2 x\arcsin xdx&=&\frac{\pi}{(s+1)^3}-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{(s+1)^3\G(\frac{s+3}{2})}\\&&-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)^2\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]\\ && -\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{8(s+1)\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]^2\\ &&+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{8(s+1)\G(\frac{s+3}{2})}\left[\psi'\left(\frac{s+3}{2}\right)-\psi'\left(\frac{s}{2}+1\right)\right]\end{eqnarray}\begin{eqnarray}\int_0^1 x^s\arccos xdx&=&\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)\G(\frac{s+3}{2})}\\\int_0^1 x^s\ln x\arccos xdx&=&-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)^2\G(\frac{s+3}{2})}\\&&-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{4(s+1)\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]\\\int_0^1 x^s\ln^2 x\arccos xdx&=&\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{(s+1)^3\G(\frac{s+3}{2})}\\&&+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{2(s+1)^2\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]\\&&+\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{8(s+1)\G(\frac{s+3}{2})}\left[\psi\left(\frac{s+3}{2}\right)-\psi\left(\frac{s}{2}+1\right)\right]^2\\&&-\frac{\sqrt{\pi}\G(\frac{s}{2}+1)}{8(s+1)\G(\frac{s+3}{2})}\left[\psi'\left(\frac{s+3}{2}\right)-\psi'\left(\frac{s}{2}+1\right)\right]\end{eqnarray}

x^s・arcsin x の定積分

\begin{eqnarray}\int_0^1 x^s\arctan xdx &=& \frac{1}{4(s+1)}\left[\pi-\psi\left(\frac{s}{4}+1\right)+\psi\left(\frac{s}{4}+\frac{1}{2}\right)\right]\\ \int_0^1 x^s\ln x\arctan xdx &=& -\frac{1}{4(s+1)^2}\left[\pi-\psi\left(\frac{s}{4}+1\right)+\psi\left(\frac{s}{4}+\frac{1}{2}\right)\right]\\&&+\frac{1}{16(s+1)}\left[-\psi'\left(\frac{s}{4}+1\right)+\psi'\left(\frac{s}{4}+\frac{1}{2}\right)\right]\\ \int_0^1 x^s\ln^2 x\arctan xdx &=& \frac{1}{2(s+1)^3}\left[\pi-\psi\left(\frac{s}{4}+1\right)+\psi\left(\frac{s}{4}+\frac{1}{2}\right)\right]\\&&+\frac{1}{8(s+1)^2}\left[\psi'\left(\frac{s}{4}+1\right)-\psi'\left(\frac{s}{4}+\frac{1}{2}\right)\right]\\&&-\frac{1}{64(s+1)}\left[\psi''\left(\frac{s}{4}+1\right)-\psi''\left(\frac{s}{4}+\frac{1}{2}\right)\right]\end{eqnarray}

x^s・arctan x の定積分

From @infseriesbot$$I(a):=\int_0^\frac{\pi}{2}\frac{\arctan(a\sin x)}{\sin x}dx=\frac{\pi}{2}\ln(a+\sqrt{a^2+1})$$PROOF.
The derivative with respect to $a$ is$$I'(a)=\int_0^\frac{\pi}{2}\frac{dx}{1+a^2\sin^2x}$$Setting $\tan x=t$ we get$$I'(a)=\int^\infty_0\frac{dt}{1+(a^2+1)t^2}=\frac{\pi}{2\sqrt{a^2+1}}$$It follows from $I(0)=0$ that$$I(a)=\frac{\pi}{2}\int_0^a\frac{dt}{\sqrt{t^2+1}}$$Substituting $t+\sqrt{t^2+1}=u$ we find$$I(a)=\frac{\pi}{2}\int_1^{a+\sqrt{a^2+1}}\frac{du}{u}$$Hence,$$I(a)=\frac{\pi}{2}\ln(a+\sqrt{a^2+1})$$

$$\int_0^1\ln\G(x)dx=\frac{\ln 2\pi}{2}$$PROOF.
We can obtain that imediately with the help of Kummer's formula:\begin{eqnarray}\log\G(x)&=&\frac{1}{2}\log2\pi+\frac{1}{2}\sum_{n=1}^\infty\frac{\cos2n\pi x}{n}\\&&+\frac{1}{\pi}\sum_{n=1}^\infty\frac{1}{n}(\g+\log2n\pi)\sin2n\pi x\tag{1}\end{eqnarray}Proof of this formula here:

【γ18】対数ガンマ関数のフーリエ級数表示（ガンマ関数の基礎18）

$$\int_0^1\ln^2\G(x)dx=\frac{\pi^2}{48}+\frac{\ln^22\pi}{4}+\frac{(\g+\ln2\pi)^2}{12}-\frac{\g+\ln2\pi}{\pi^2}\zeta'(2)+\frac{\zeta''(2)}{2\pi^2}$$PROOF.
We Square (1) and integrate from $0$ to $1$, where the terms of $\cos2n\pi x$ , $\sin2n\pi x$ , $\sin2m\pi x\cos2n\pi x$ , $\sin2m\pi x\sin2n\pi x(m\neq n)$ , $\cos2m\pi x\cos2n\pi x(m\neq n)$ vanish. Therefore,$$\int_0^1\ln^2\G(x)dx=\int_0^1\left(\frac{\ln^22\pi}{4}+\sum_{n=1}^\infty\frac{\cos^22n\pi x}{4n^2}+\sum_{n=1}^\infty\frac{(\g+\ln2n\pi)^2}{n^2\pi^2}\sin^22n\pi x\right)dx$$\begin{eqnarray*}&=& \frac{\ln^22\pi}{4}+\frac{\zeta(2)}{8}+\frac{1}{2\pi^2}\sum_{n=1}^\infty\frac{(\g+\ln2n\pi)^2}{n^2} \\ &=& \frac{\pi^2}{48}+\frac{\ln^22\pi}{4}+\frac{1}{2\pi^2}\sum_{n=1}^\infty\frac{(\g+\ln2\pi)^2+2(\g+\ln2\pi)\ln n+\ln^2n}{n^2}\end{eqnarray*}The zeta function and its derivatives are$$\zeta(s)=\sum_{n=1}^\infty\dfrac{1}{n^s}\;,\;\zeta'(s)=-\sum_{n=1}^\infty\frac{\ln n}{n^s}\;,\;\zeta''(s)=\sum_{n=1}^\infty\frac{\ln^2 n}{n^s}$$Hence,$$\int_0^1\ln^2\G(x)dx=\frac{\pi^2}{48}+\frac{\ln^22\pi}{4}+\frac{(\g+\ln2\pi)^2}{12}-\frac{\g+\ln2\pi}{\pi^2}\zeta'(2)+\frac{\zeta''(2)}{2\pi^2}$$

From @integralsbot$$\int_0^\infty\frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\G^2(\frac{1}{4})}{6\sqrt{\pi}}\frac{a^3-b^3}{a^4-b^4}$$PROOF.
Rationalizing the dominator and decomposing into two integrals, we get$$LHS=\frac{1}{a^4-b^4}\left[\int_0^\infty\left(\sqrt{x^4+a^4}-x^2\right)dx-\int_0^\infty\left(\sqrt{x^4+b^4}-x^2\right)dx\right]$$Now we consider the first term. Substituting $\sqrt{x^4+a^4}-x^2=a^2\sqrt{y}$, we find$$x=\frac{a}{\sqrt{2}}y^{-\frac{1}{4}}(1-y)^{\frac{1}{2}}\;,\;dx=-\frac{a}{\sqrt{2}}\left[\frac{1}{4}y^{-\frac{5}{4}}(1-y)^{\frac{1}{2}}+\frac{1}{2}y^{-\frac{1}{4}}(1-y)^{-\frac{1}{2}}\right]dy$$Hence\begin{eqnarray*}\int_0^\infty\left(\sqrt{x^4+a^4}-x^2\right)dx &=& \frac{a^3}{2\sqrt{2}}\left[\frac{1}{2}B\left(\frac{1}{4},\frac{3}{2}\right)+B\left(\frac{5}{4},\frac{1}{2}\right)\right] \\ &=& \frac{a^3\G^2(\frac{1}{4})}{6\sqrt{\pi}}\end{eqnarray*}Obviously,$$\int_0^\infty\left(\sqrt{x^4+b^4}-x^2\right)dx=\frac{b^3\G^2(\frac{1}{4})}{6\sqrt{\pi}}$$Therefore,$$\int_0^\infty\frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}=\frac{\G^2(\frac{1}{4})}{6\sqrt{\pi}}\frac{a^3-b^3}{a^4-b^4}$$

From @integralsbot$$\int_0^\pi\frac{\ln(1+\frac{\cos x}{2})}{\cos x}dx=\zeta(2)$$PROOF.
\begin{eqnarray*}\int_0^\pi\frac{\ln(1+\frac{\cos x}{2})}{\cos x}dx &=& \int_0^\frac{\pi}{2}\frac{\ln(1+\frac{\cos x}{2})}{\cos x}dx-\int_0^\frac{\pi}{2}\frac{\ln(1-\frac{\cos x}{2})}{\cos x}dx \\ &=& \int_0^\frac{\pi}{2}\ln\frac{1+\frac{\cos x}{2}}{1-\frac{\cos x}{2}}\frac{dx}{\cos x}\\ &=& \int_0^\frac{\pi}{2}\frac{2}{\cos x}\sum_{n=0}^\infty\frac{1}{2n+1}\left(\frac{\cos x}{2}\right)^{2n+1}dx \\ &=& 2\sum_{n=0}^\infty\frac{1}{2n+1}\left(\frac{1}{2}\right)^{2n+1}\int_0^\frac{\pi}{2}\cos^{2n}xdx \\ &=& \pi\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{1}{2n+1}\left(\frac{1}{2}\right)^{2n+1} \\ &=& \pi\arcsin\frac{1}{2} \\ &=& \frac{\pi^2}{6}\end{eqnarray*}

From @integralsbot$$\int_0^\infty\frac{dx}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}}=\frac{\G^2(\frac{1}{8})}{2^\frac{11}{4}\G(\frac{1}{4})}$$PROOF.
Substituting $8x^2+8x+1=y$, it follows from $dx=\dfrac{\sqrt{2}}{8}(1+y)^{-\frac{1}{2}}$ that$$\int_0^\infty\frac{dx}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}}=\frac{1}{2}\int_1^\infty y^{-\frac{1}{4}}(y^2-1)^{-\frac{1}{2}}dy$$Substituting $y^2=\dfrac{1}{t}$ we get\begin{eqnarray*}\int_0^\infty\frac{dx}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}} &=& \frac{1}{4}\int_0^1t^{-\frac{7}{8}}(1-t)^{-\frac{1}{2}} dt\\ &=& \frac{1}{4}B\left(\frac{1}{8},\frac{1}{2}\right) \\ &=& \frac{\sqrt{\pi}}{4}\frac{\G(\frac{1}{8})}{\G(\frac{5}{8})}\end{eqnarray*}Legendre duplication formula gives $$\G\left(\frac{1}{4}\right)=\frac{1}{2^{\frac{3}{4}}\sqrt{\pi}}\G\left(\frac{1}{8}\right)\G\left(\frac{5}{8}\right)$$Hence,$$\int_0^\infty\frac{dx}{\sqrt{x^2+x}\sqrt[4]{8x^2+8x+1}}=\frac{\G^2(\frac{1}{8})}{2^\frac{11}{4}\G(\frac{1}{4})}$$

$$\mathfrak{R}\Li_2(e^{\pm\frac{\pi}{3}i})=\frac{\pi^2}{36}$$PROOF.
The dilogarithm identity$$\Li_2(1-x)+\Li_2\left(1-\frac{1}{x}\right)=-\frac{\ln^2x}{2}$$which we obtained in

多重対数関数（ポリログ）の関係式一覧・証明付き

Let $x=e^{\frac{\pi}{3}i}$ we get$$\Li_2(e^{-\frac{\pi}{3}i})+\Li_2(e^{\frac{\pi}{3}i})=\frac{\pi^2}{18}$$Hence,$$\mathfrak{R}\Li_2(e^{\pm\frac{\pi}{3}i})=\frac{\pi^2}{36}$$

\begin{eqnarray}\int_0^\frac{\pi}{2} \ln^2(2\sin x)dx &=& \frac{\pi^3}{24} \\ \int_0^\frac{\pi}{4} \ln^2(2\sin x)dx &=& \frac{\pi^3}{192}-\frac{\ln2}{2}G-\frac{\pi}{16}\ln^22-\mathfrak{I}\Li_3(1-i) \\ \int_0^\frac{\pi}{6} \ln^2(2\sin x)dx &=& \frac{7\pi^3}{216}\end{eqnarray}\begin{eqnarray*}\int_0^1\frac{\ln^2x}{\sqrt{4-x^2}}dx &=& \frac{7\pi^3}{216}\\ {}_4F_3\left[\begin{matrix}\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2},\dfrac{1}{2}\\\dfrac{3}{2},\dfrac{3}{2},\dfrac{3}{2}\end{matrix};\frac{1}{4}\right]&=&\frac{7\pi^3}{216}\end{eqnarray*}\begin{eqnarray}\int_0^\frac{\pi}{2} x\ln^2(2\sin x)dx &=& \Li_4\left(\frac{1}{2}\right)-\frac{19\pi^4}{2880}+\frac{7}{8}\zeta(3)\ln2\\&&-\frac{\pi^2}{24}\ln^22+\frac{\ln^42}{24}\end{eqnarray}\begin{eqnarray}\int_0^\frac{\pi}{4} x\ln^2(2\sin x)dx&=&\frac{5}{16}\Li_4\left(\frac{1}{2}\right)-\frac{11\pi^4}{1152}-\frac{\pi G}{8}\ln2\\&&+\frac{35}{128}\zeta(3)\ln2-\frac{11\pi^2}{384}\ln^22\\&&+\frac{5\ln^42}{384}-\frac{\pi}{4}\mathfrak{I}\Li_3(1-i)\end{eqnarray}PROOF.

log(sin x)の2乗の対数正弦積分（調和数・ポリログ）

\begin{equation}\int_0^x\ln(2\sinh x)dx=\frac{x^2}{2}-\frac{\pi^2}{12}+\frac{1}{2}\Li_2\left(e^{-2x}\right)\end{equation}\begin{equation}\int_0^x x\ln(2\sinh x)dx=\frac{x^3}{3}+\frac{x}{2}\Li_2\left(e^{-2x}\right)+\frac{1}{4}\Li_3\left(e^{-2x}\right)-\frac{\zeta(3)}{4}\end{equation}\begin{eqnarray}\int_0^x \ln^2(2\sinh x)dx &=& \frac{x^3}{3}-\frac{\zeta(3)}{2}+\frac{1}{2}\Li_3(e^{-2x})+\Li_3(1-e^{-2x})\\&&+x\Li_2(e^{-2x})-\ln(1-e^{-2x})\Li_2(1-e^{-2x})+x\ln^2(1-e^{-2x})\end{eqnarray}\begin{eqnarray}\int_0^x x^2\ln(2\sinh x)dx &=& \frac{x^4}{4}-\frac{\pi^4}{360}+\frac{x^2}{2}\Li_2(e^{-2x})\\&&+\frac{x}{2}\Li_3(e^{-2x})+\frac{1}{4}\Li_4(e^{-2x})\end{eqnarray}\begin{eqnarray}\int_0^x &&x\ln^2(2\sinh x)dx\\ =&& \frac{x^4}{4}-\frac{7\pi^4}{720}+\frac{1}{2}\left[\Li_4(1-e^{-2x})-\Li_4\left(\frac{1}{1-e^{2x}}\right)\right] \\ && +x\left[\Li_3(e^{-2x})+\Li_3(1-e^{-2x})-\zeta(3)\right]+\frac{\ln(1-e^{-2x})}{2}\Li_3(e^{-2x}) \\&& +x^2\Li_2(e^{-2x})-x\ln(1-e^{-2x})\Li_2(1-e^{-2x}) -\frac{\ln^4(1-e^{-2x})}{48}\\ && -\frac{x}{6}\ln^3(1-e^{-2x})+\left(x^2-\frac{\pi^2}{24}\right)\ln^2(1-e^{-2x})-\frac{\zeta(3)}{2}\ln(1-e^{-2x})\end{eqnarray}

log(sinh x)の対数正弦積分（調和数・ポリログ）

$$\int_0^1\frac{x^{-\frac{3}{4}}(1-x)^{-\frac{1}{4}}}{1+a\sqrt{x(1-x)}}dx=\frac{2\pi}{\sqrt{2+a}}$$PROOF.
\begin{eqnarray*}\int_0^1\frac{x^{-\frac{3}{4}}(1-x)^{-\frac{1}{4}}}{1+a\sqrt{x(1-x)}}dx &=& \int_0^1x^{-\frac{3}{4}}(1-x)^{-\frac{1}{4}}\sum_{n=0}^\infty\left(-ax^\frac{1}{2}(1-x)^\frac{1}{2}\right)^n dx \\ &=& \sum_{n=0}^\infty(-a)^n\int_0^1x^{\frac{n}{2}-\frac{3}{4}}(1-x)^{\frac{n}{2}-\frac{1}{4}}dx\\ &=& \sum_{n=0}^\infty(-a)^n\frac{1}{n!}\G\left(\frac{n}{2}+\frac{1}{4}\right)\G\left(\frac{n}{2}+\frac{3}{4}\right)\\&&\hskip 10em\quad(\because \mathrm{Beta\: function}) \\ &=& \sum_{n=0}^\infty(-a)^n\frac{1}{n!}2^{\frac{1}{2}-n}\sqrt{\pi}\G\left(n+\frac{1}{2}\right)\\&&\hskip 10em\quad(\because\mathrm{duplication\: formula})\\ &=& \sqrt{2}\pi\sum_{n=0}^\infty \frac{(\frac{1}{2})_n}{n!}\left(-\frac{a}{2}\right)^n \\ &=& \sqrt{2}\pi\cdot \frac{1}{\sqrt{1+\frac{a}{2}}} \\ &=&\frac{2\pi}{\sqrt{2+a}}\end{eqnarray*}

From Yury Brychkov, Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas. But the result in this book is incorrect. Let us derive the true result.$$I:=\int_0^ax^\frac{1}{2}\left(1-b\sqrt{x(a-x)}\right)^\frac{1}{2}dx=?$$Substituting $x=ay$ we have$$\int_0^ax^\frac{1}{2}\left(1-b\sqrt{x(a-x)}\right)^\frac{1}{2}dx=a^\frac{3}{2}\int_0^1y^\frac{1}{2}\left(1-ab\sqrt{y(1-y)}\right)^\frac{1}{2}dy$$Considering the Maclaurin series of $(1-X)^\frac{1}{2}$, we find the beta function appears and obtain that $$I=a^\frac{3}{2}\left[\frac{2}{3}-\sum_{n=0}^\infty\frac{(2n-1)!!}{(n+1)!}\left(\frac{ab}{2}\right)^{n+1}B\left(\frac{n}{2}+2,\frac{n+3}{2}\right)\right]$$Applying the duplication formula of the gamma function to the numerator\begin{eqnarray*}I &=& a^\frac{3}{2}\left[\frac{2}{3}-\sum_{n=0}^\infty\left(\frac{ab}{2}\right)^{n+1}\frac{2n+4}{(2n+5)(2n+3)(2n+1)}\right] \\ &=& -a^\frac{3}{2}\sum_{n=0}^\infty\left(\frac{ab}{2}\right)^{n}\frac{2(n+1)}{(2n+3)(2n+1)(2n-1)}\\ &=& a^\frac{3}{2}\sum_{n=0}^\infty\left(\frac{ab}{2}\right)^{n}\left[\frac{1/4}{2n+1}+\frac{1/8}{2n+3}-\frac{3/8}{2n-1}\right]\end{eqnarray*}It follows from $$\sum_{n=0}^\infty\frac{x^n}{2n+1}=\frac{1}{2\sqrt{x}}\ln\frac{1+\sqrt{x}}{1-\sqrt{x}}\;,\;\sum_{n=0}^\infty\frac{x^n}{2n+3}=\frac{1}{2x^\frac{3}{2}}\ln\frac{1+\sqrt{x}}{1-\sqrt{x}}-\frac{1}{x}$$$$\sum_{n=0}^\infty\frac{x^n}{2n-1}=\frac{\sqrt{x}}{2}\ln\frac{1+\sqrt{x}}{1-\sqrt{x}}-1$$that$$I=\frac{\sqrt{a}}{4b}\left[\frac{(1+\frac{3ab}{2})(1-\frac{ab}{2})}{\sqrt{2ab}}\ln\frac{\sqrt{2}+\sqrt{ab}}{\sqrt{2}-\sqrt{ab}}+\frac{3}{2}ab-1\right]$$

From Yury Brychkov, Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas.
$$I:=\int_0^a\frac{\ln\left(1-bx(a-x)\right)}{x}dx=-2\arcsin^2\frac{a\sqrt{b}}{2}$$PROOF.
Substituting $x=ay$, we get$$I=\int_0^1\frac{\ln\left(1-a^2 b y(1-y)\right)}{y}dy$$The Maclaurin expansion of $\ln(1-X)$ yields$$I=-\sum_{n=1}^\infty\frac{(a^2b)^n}{n}\int_0^1y^{n-1}(1-y)^ndy$$The integral on the right side is the Beta function.$$I=-\sum_{n=1}^\infty\frac{(a^2b)^n}{n}B(n,n+1)$$ Hence,$$I=-\sum_{n=1}^\infty\frac{(a\sqrt{b})^{2n}}{n^2\binom{2n}{n}}$$The series representation of $\arcsin^2 X$ is $$2\arcsin^2x=\sum_{n=1}^\infty\frac{1}{n^2\binom{2n}{n}}(2X)^{2n}$$Therefore,$$I=-2\arcsin^2\frac{a\sqrt{b}}{2}$$

$$\int_0^1\frac{\sqrt{x}\ln(1-x)}{1+x^3}dx=-\frac{2}{9}G-\frac{\pi}{6}\ln2$$$$\int_0^1\frac{\sqrt{x}\ln x}{1+x^3}dx=-\frac{4}{9}G$$$$\int_0^1\frac{\sqrt{x}}{1+x^3}\ln\frac{1-x}{x}dx=\frac{2}{9}G-\frac{\pi}{6}\ln2$$

\begin{eqnarray}\sum_{n=0}^\infty\frac{(-1)^nO_{3n}}{2n+1} &=& -\frac{G}{6}+\frac{3}{8}\pi\ln2-\frac{\pi}{8}\\ &&-\frac{\sqrt{3}}{2}\ln(\sqrt{3}+1)+\frac{\sqrt{3}}{4}\ln2\end{eqnarray}\begin{eqnarray*}&&\mathfrak{I}\left[\Li_2\left(\frac{\sqrt{6}-\sqrt{2}}{4}e^{-\frac{5}{12}\pi i}\right)-\Li_2\left(\frac{\sqrt{6}+\sqrt{2}}{4}e^{\frac{1}{12}\pi i}\right)\right]\\ &&\quad\quad\quad=-\frac{4}{3}G+\frac{2}{3}\pi\ln2-\frac{\pi}{3}\ln(\sqrt{3}+1)\end{eqnarray*}

Proof here

複素積分演習（対数と２つの分岐点）