積分botを解けるだけ解く, その6

半径$1$の半円形の積分路を考えると,
$$\begin{eqnarray*} &&\int_0^{\pi/2}\frac{x\ln\sin x}{\sin 2x}\,dx\\ &=&\frac 14\int_0^{\pi}\frac{x\ln\sin\frac x2}{\sin x}\,dx\\ &=&-\frac 12\int_0^{\pi}\frac{e^{ix}\ln e^{ix}}{1-e^{2ix}}\left(\ln\left(\frac{1-e^{ix}}2\right)-\frac 12\ln e^{ix}+\frac{i\pi}{2}\right)\,dx\\ &=&-\frac i2\int_{-1}^1\frac{\ln x}{1-x^2}\left(\ln\left(\frac{1-x}2\right)-\frac 12\ln x+\frac{i\pi}2\right)\,dx\\ &=&\Re\left(-\frac i2\int_0^1\frac{\ln x}{1-x^2}\left(\ln\left(\frac{1-x}2\right)-\frac 12\ln x+\frac{i\pi}2\right)\,dx\right)\\ &+&\Re\left(-\frac i2\int_0^1\frac{\ln x+i\pi}{1-x^2}\left(\ln\left(\frac{1+x}2\right)-\frac 12\ln x\right)\,dx\right)\\ &=&\frac{\pi}{4}\int_0^1\frac{\ln x}{1-x^2}\,dx+\frac{\pi}{2}\int_0^1\frac{1}{1-x^2}\left(\ln\left(\frac{1+x}2\right)-\frac 12\ln x\right)\,dx\\ &=&\frac{\pi}2\int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x^2}\,dx\\ &=&\frac{\pi}4\int_0^1\left(\frac 1{1-x}+\frac 1{1+x}\right)\ln\left(\frac{1+x}2\right)\,dx \end{eqnarray*}$$
ここで, $x\to 1-2x$の置換により,
$$\begin{eqnarray*} \int_0^1\frac{\ln\left(\frac{1+x}{2}\right)}{1-x}\,dx&=&\int_0^{1/2}\frac{\ln(1-x)}{x}\,dx\\ &=&-\Li_2\left(\frac 12\right)\\ &=&\frac{\ln^2 2}{2}-\frac{\pi^2}{12} \end{eqnarray*}$$
また,
$$\begin{eqnarray*} \int_0^1\frac{\ln\left(\frac{1+x}2\right)}{1+x}\,dx&=&\left[\frac 12\ln^2(1+x)-\ln 2\ln(1+x)\right]_0^1\\ &=&-\frac{\ln^2 2}{2} \end{eqnarray*}$$
より, これを先ほどの式に代入して,
$$\begin{eqnarray*} \int_0^{\pi/2}\frac{x\ln\sin x}{\sin 2x}\,dx=-\frac{\pi^3}{48} \end{eqnarray*}$$
が得られる.

$$\begin{eqnarray*} \int_0^2\frac{\tan^{-1}\frac{2-x}{1+2x}}{x^2-4x-1}\,dx&=&\int_0^2\frac{\tan^{-1}2-\tan^{-1}x}{(x-2)^2-5}\,dx\\ &=&\int_0^2\int_x^2\frac 1{(1+y^2)((x-2)^2-5)}\,dydx\\ &=&\frac 45\int_{0\lt x,y,\,x+y\lt 1}\frac{1}{\left(1-\frac{4}{5}x^2\right)(1+4y^2)}\,dxdy \end{eqnarray*}$$
ここで, Euler積分により,
$$\begin{eqnarray*} \int_{0\lt x,y,\,x+y\lt 1}\frac{1}{(1-a^2x^2)(1-b^2x^2)}\,dxdy&=&\sum_{0\leq n,m}a^{2n}b^{2m}\int_{0\lt x,y\,x+y\lt 1}x^{2n}y^{2m}\,dxdy\\ &=&\sum_{0\leq n,m}\frac{a^{2n}b^{2m}}{(2n+1)(2m+1)}\frac{(2n+1)!(2m+1)!}{(2n+2m+2)!} \end{eqnarray*}$$
ここで, Multivariable connected sums and multiple polylogarithms( https://arxiv.org/abs/2103.05492 ) のExample 5.6より,
$$\begin{eqnarray*} f(a,b):=\sum_{0\lt a,b}\frac{a^nb^m}{nm}\frac{n!m!}{(n+m)!}=\Li_2(a)+\Li_2(b)-\Li_2(a+b-ab) \end{eqnarray*}$$
だから,
$$\begin{eqnarray*} &&\sum_{0\leq n,m}\frac{a^{2n}b^{2m}}{(2n+1)(2m+1)}\frac{(2n+1)!(2m+1)!}{(2n+2m+2)!}\\ &=&\frac 1{4ab}(f(a,b)-f(-a,b)-f(a,-b)+f(-a,-b))\\ &=&-\frac 1{4ab}\left(\Li_2(a+b-ab)-\Li_2(-a+b+ab)-\Li_2(a-b+ab)+\Li_2(-a-b-ab)\right) \end{eqnarray*}$$
よって, 解析接続により, $a=\frac 2{\sqrt{5}}, b=2i$として,
$$\begin{eqnarray*} &&\frac 45\int_{0\lt x,y,\,x+y\lt 1}\frac{1}{\left(1-\frac{4}{5}x^2\right)(1+4y^2)}\,dxdy\\ &=&-\frac 1{5\cdot\frac 2{\sqrt{5}}\cdot 2i}\left(\Li_2\left(2i+\frac{2}{\sqrt{5}}(1-2i)\right)-\Li_2\left(-2i+\frac{2}{\sqrt{5}}(1+2i)\right)-\Li_2\left(2i-\frac{2}{\sqrt{5}}(1-2i)\right)+\Li_2\left(-2i-\frac{2}{\sqrt{5}}(1+2i)\right)\right)\\ &=&-\frac 1{4i\sqrt{5}}\left(\Li_2\left(2i+\frac{2}{\sqrt{5}}(1-2i)\right)+\Li_2\left(-2i-\frac{2}{\sqrt{5}}(1+2i)\right)-\Li_2\left(-2i+\frac{2}{\sqrt{5}}(1+2i)\right)-\Li_2\left(2i-\frac{2}{\sqrt{5}}(1-2i)\right)\right) \end{eqnarray*}$$
ここで,
$$\begin{eqnarray*} \Li_2(z)+\Li_2\left(\frac{z}{z-1}\right)=-\frac 12\ln^2(1-z) \end{eqnarray*}$$
に,
$$\begin{eqnarray*} z=2i+\frac{2}{\sqrt{5}}(1-2i),\quad -2i+\frac{2}{\sqrt{5}}(1+2i) \end{eqnarray*}$$
を代入すると,
$$\begin{eqnarray*} \Li_2\left(2i+\frac{2}{\sqrt{5}}(1-2i)\right)+\Li_2\left(-2i-\frac{2}{\sqrt{5}}(1+2i)\right)&=&-\frac 12\ln^2\left(1-2i-\frac{2}{\sqrt{5}}(1-2i)\right)\\ &=&-\frac 12\ln^2\left((1-2i)\left(1-\frac 2{\sqrt{5}}\right)\right)\\ \Li_2\left(-2i+\frac{2}{\sqrt{5}}(1+2i)\right)+\Li_2\left(2i-\frac{2}{\sqrt{5}}(1-2i)\right)&=&-\frac 12\ln^2\left(1+2i-\frac{2}{\sqrt{5}}(1+2i)\right)\\ &=&-\frac 12\ln^2\left((1+2i)\left(1-\frac 2{\sqrt{5}}\right)\right) \end{eqnarray*}$$
これより,
$$\begin{eqnarray*} &&-\frac 1{4i\sqrt{5}}\left(\Li_2\left(2i+\frac{2}{\sqrt{5}}(1-2i)\right)+\Li_2\left(-2i-\frac{2}{\sqrt{5}}(1+2i)\right)-\Li_2\left(-2i+\frac{2}{\sqrt{5}}(1+2i)\right)-\Li_2\left(2i-\frac{2}{\sqrt{5}}(1-2i)\right)\right)\\ &=&\frac 1{8i\sqrt{5}}\left(\ln^2\left((1-2i)\left(1-\frac 2{\sqrt{5}}\right)\right)-\ln^2\left((1+2i)\left(1-\frac 2{\sqrt{5}}\right)\right)\right)\\ &=&\frac 1{8i\sqrt{5}}\left(\ln\left((1-2i)\left(1-\frac 2{\sqrt{5}}\right)\right)-\ln\left((1+2i)\left(1-\frac 2{\sqrt{5}}\right)\right)\right)\left(\ln\left((1-2i)\left(1-\frac 2{\sqrt{5}}\right)\right)+\ln\left((1+2i)\left(1-\frac 2{\sqrt{5}}\right)\right)\right)\\ &=&\frac 1{8i\sqrt{5}}\ln\left(\frac{1-2i}{1+2i}\right)\ln\left(5\left(1-\frac{2}{\sqrt{5}}\right)\right)\\ &=&-\frac 1{8i\sqrt{5}}\cdot 2i\tan^{-1}2\cdot 2\tanh^{-1}\frac 2{\sqrt{5}}\\ &=&-\frac 1{2\sqrt{5}}\tan^{-1}2\tanh^{-1}\frac{2}{\sqrt{5}} \end{eqnarray*}$$

まず,
$$\begin{eqnarray*} \int_0^{\infty}\frac{x\ln(1+x^2)}{\sinh\pi x}\,dx&=&\int_0^{\infty}\frac{x\ln\left(1+\frac 1{x^2}\right)}{\sinh\pi x}\,dx+2\int_0^{\infty}\frac{x\ln x}{\sinh\pi x}\,dx \end{eqnarray*}$$
第1項は前回の記事( https://mathlog.info/articles/1929 ) の命題5の証明の最後の式,
$$\begin{eqnarray*} \int_0^{\infty}\frac{x\ln\left(1+\frac{a^2}{x^2}\right)}{e^{2\pi x}-1}\,dx&=&-\frac {a^2}4(1-2\ln a)-\frac a2(1+2\ln\Gamma(a))+\ln K(a)+\frac a2\ln 2\pi+\frac{a(1-a)}2 \end{eqnarray*}$$
と$K\left(\frac 12\right)=\frac{A^{3/2}}{2^{1/24}e^{1/8}}$をもちいて,
$$\begin{eqnarray*} \int_0^{\infty}\frac{x\ln\left(1+\frac 1{x^2}\right)}{\sinh\pi x}\,dx&=&2\int_0^{\infty}\frac{x\ln\left(1+\frac 1{x^2}\right)}{e^{\pi x}-1}\,dx-2\int_0^{\infty}\frac{x\ln\left(1+\frac 1{x^2}\right)}{e^{2\pi x}-1}\,dx\\ &=&8\int_0^{\infty}\frac{x\ln\left(1+\frac 1{4x^2}\right)}{e^{2\pi x}-1}\,dx-2\left(-\frac 14-\frac 12+\frac 12\ln 2\pi\right)\\ &=&8\left(-\frac 1{16}(1+2\ln 2)-\frac 14(1+\ln\pi)+\ln K\left(\frac 12\right)+\frac 14\ln 2\pi+\frac 18\right)+\frac 32-\ln 2\pi\\ &=&-\frac {3}{2}+8\ln K\left(\frac 12\right)+\ln 2+\frac 32-\ln 2\pi\\ &=&8\ln K\left(\frac 12\right)-\ln\pi\\ &=&\ln\frac{A^{12}}{\pi e\sqrt[3]{2}} \end{eqnarray*}$$
第2項は,

$$\begin{eqnarray*} 2\int_0^{\infty}\frac{x^{s-1}}{\sinh\pi x}\,dx&=&4\pi^{-s}(1-2^{-s})\Gamma(s)\zeta(s) \end{eqnarray*}$$
と, $\frac{\zeta'(2)}{\zeta(2)}=\gamma+\ln 2\pi-12\ln A$より,

$$\begin{eqnarray*} 2\int_0^{\infty}\frac{x\ln x}{\sinh\pi x}\,dx&=&4\left.\frac{d}{ds}\left(\pi^{-s}(1-2^{-s})\Gamma(s)\zeta(s)\right)\right|_{s=2}\\ &=&4\pi^{-2}(1-2^{-2})\zeta(2)\left(\frac{\zeta'(2)}{\zeta(2)}-\ln\pi+\frac{2^{-2}\ln 2}{1-2^{-2}}+\psi(2)\right)\\ &=&\frac 12\left(1+\frac 43\ln 2-12\ln A\right)\\ &=&\ln\frac{\sqrt[3]{4}\sqrt{e}}{A^6} \end{eqnarray*}$$
よって, 求める積分は,

$$\begin{eqnarray*} \ln\frac{A^{12}}{\pi e\sqrt[3]{2}}+\ln\frac{\sqrt[3]{4}\sqrt{e}}{A^6}=\ln\frac{\sqrt[3]{2}A^6}{\pi\sqrt{e}} \end{eqnarray*}$$

まず,
$$\begin{eqnarray*} \frac 1{\sqrt{1+z}}=\sum_{0\leq n}\frac{(-1)^n}{2^{2n}}\binom{2n}nz^n \end{eqnarray*}$$
に, $z=e^{2ix}$を代入して, 虚部を比較することにより,
$$\begin{eqnarray*} \frac{\sin\frac x2}{\sqrt{\cos x}}=\sqrt{2}\sum_{0\leq n}\frac{(-1)^{n-1}}{2^{2n}}\binom{2n}n\sin 2nx \end{eqnarray*}$$
となることと, $|x|\leq 1$に対し,
$$\begin{eqnarray*} \sum_{0\lt n}\frac{1}{2^{2n}n}\binom{2n}nx^n&=&\int_0^x\frac 1{t}\left(\frac 1{\sqrt{1-t}}-1\right)\,dt\\ &=&2\ln\frac{2}{1+\sqrt{1-x}} \end{eqnarray*}$$
であることを確認しておく.
$$\begin{eqnarray*} &&\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\tan^{-1}\frac{2\sin x}{2\cos x-1}\,dx\\ &=&\int_0^{\pi/3}\frac{\sin\frac x2}{\sqrt{\cos x}}\tan^{-1}\frac{2\sin x}{2\cos x-1}\,dx-\int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\tan^{-1}\frac{2\sin x}{1-2\cos x}\,dx\\ &=&\Im\left(\int_0^{\pi/3}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln(2e^{ix}-1)\,dx-\int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln(1-2e^{-ix})\,dx\right)\\ &=&\Im\left(\int_0^{\pi/3}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln(2e^{ix}-1)\,dx+\int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}(\ln(2e^{ix}-1)-i\pi)\,dx\right)\\ &=&\Im\left(\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln(2e^{ix}-1)\,dx\right)-\pi\int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\,dx\\ &=&\Im\left(\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln\left(1-\frac{e^{-ix}}{2}\right)\,dx\right)+\int_0^{\pi/2}\frac{x\sin\frac x2}{\sqrt{\cos x}}\,dx-\pi\int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\,dx\\ \end{eqnarray*}$$
第1項は
$$\begin{eqnarray*} &&\Im\left(\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\ln\left(1-\frac{e^{-ix}}{2}\right)\,dx\right)\\ &=&\int_0^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\sum_{0\lt n}\frac{1}{2^nn}\sin nx\,dx\\ &=&\sqrt{2}\int_0^{\pi/2}\sum_{0\lt n,m}\frac{(-1)^{m-1}}{2^{n+2m}n}\binom{2m}m\sin nx\sin 2mx\,dx\\ &=&\sqrt{2}\sum_{0\lt n,m}\frac{(-1)^{m-1}}{2^{n+2m}n}\binom{2m}m\int_0^{\pi/2}\sin nx\sin 2mx\,dx \end{eqnarray*}$$
ここで,
$n=2m$のとき,
$$\begin{eqnarray*} \int_0^{\pi/2}\sin nx\sin 2mx\,dx &=& \int_0^{\pi/2}\sin^2 2mx\,dx\\ &=&\frac 12\int_0^{\pi/2}(1-\cos 4mx)\,dx\\ &=&\frac{\pi}4 \end{eqnarray*}$$
$n\neq 2m$のとき,
$$\begin{eqnarray*} \int_0^{\pi/2}\sin nx\sin 2mx\,dx &=& -\frac 12\int_0^{\pi/2}(\cos(n+2m)x-\cos(n-2m)x)\,dx\\ &=&-\frac 12\left[\frac{\sin(n+2m)x}{n+2m}-\frac{\sin(n-2m)x}{n-2m}\right]_0^{\pi/2}\\ &=&\begin{cases} 0\quad(n=2r)\\ \frac{(-1)^{m+r}}{2}\left(\frac 1{2r-2m+1}-\frac 1{2r+2m+1}\right)\quad(n=2r+1) \end{cases} \end{eqnarray*}$$
よって,
$$\begin{eqnarray*} &&\sqrt{2}\sum_{0\lt n,m}\frac{(-1)^{m-1}}{2^{n+2m}n}\binom{2m}m\int_0^{\pi/2}\sin nx\sin 2mx\,dx\\ &=&\frac{\sqrt{2}}4\pi\sum_{0\lt m}\frac{(-1)^{m-1}}{2^{4m}(2m)}\binom{2m}m-\frac{1}{\sqrt{2}}\sum_{0\leq r,0\lt m}\frac{(-1)^r}{2^{2r+2m+1}(2r+1)}\binom{2m}m\left(\frac 1{2r-2m+1}-\frac{1}{2r+2m+1}\right)\\ &=&-\frac{\sqrt{2}}{8}\pi\sum_{0\leq m}\frac 1{2^{2m}m}\binom{2m}{m}\left(-\frac 14\right)^m+\frac{1}{\sqrt{2}}\sum_{0\leq r}\frac{(-1)^r}{2^{2r+1}(2r+1)}\sum_{0\leq m}\frac 1{2^{2m}}\binom{2m}m\left(\frac 1{2m-2r-1}+\frac{1}{2m+2r+1}\right)\\ &=&-\frac{\sqrt{2}}{8}\pi\cdot 2\ln\frac{2}{1+\frac{\sqrt{5}}2}+\frac{1}{\sqrt{2}}\sum_{0\leq r}\frac{(-1)^r}{2^{2r+1}(2r+1)}\left(\frac 1{-2r-1}\,{}_2F_1\left[\begin{matrix}\frac 12,-\frac 12-r\\\frac 12-r\end{matrix};1\right]+\frac 1{2r+1}\,{}_2F_1\left[\begin{matrix}\frac 12,\frac 12+r\\\frac 32+r\end{matrix};1\right]\right)\\ &=&\frac{\sqrt{2}}{4}\pi\ln\frac{2+\sqrt{5}}4+\frac 1{\sqrt{2}}\sum_{0\leq r}\frac{(-1)^r}{2^{2r+1}(2r+1)}\left(0+\frac{1}{2r+1}\frac{\Gamma\left(\frac 12\right)\Gamma\left(r+\frac 32\right)}{\Gamma(r+1)}\right)\\ &=&\frac{\sqrt{2}}{4}\pi\ln\frac{\phi^3}{4}+\frac {\sqrt{2}}{4}\pi\sum_{0\leq r}\frac{(-1)^r}{2^{2r+1}(2r+1)}\frac{\left(\frac 12\right)_r}{r!}\\ &=&\frac{\sqrt{2}}{4}\pi\ln\frac{\phi^3}{4}+\frac {\sqrt{2}}{4}\pi\sinh^{-1}\frac{1}{2}\\ &=&\frac{\sqrt{2}}{4}\pi\ln\frac{\phi^3}{4}+\frac {\sqrt{2}}{4}\pi\ln\phi\\ &=&\sqrt{2}\pi\ln\phi-\frac{\pi}{\sqrt{2}}\ln 2 \end{eqnarray*}$$
第2項は,
$$\begin{eqnarray*} \int_0^{\pi/2}\frac{x\sin\frac x2}{\sqrt{\cos x}}\,dx&=&\sqrt{2}\sum_{0\lt n}\frac{(-1)^{n-1}}{2^{2n}}\binom{2n}n\int_0^{\pi/2}x\sin 2nx\,dx\\ &=&\sqrt{2}\sum_{0\lt n}\frac{(-1)^{n-1}}{2^{2n}}\binom{2n}n\frac{(-1)^{n-1}\pi}{4n}\\ &=&\frac{\sqrt{2}}{4}\pi\sum_{0\lt n}\frac{1}{2^{2n}n}\binom{2n}n\\ &=&\frac{\sqrt{2}}{4}\cdot 2\ln 2\\ &=&\frac{\pi}{\sqrt{2}}\ln 2 \end{eqnarray*}$$
第3項は,
$$\begin{eqnarray*} \int_{\pi/3}^{\pi/2}\frac{\sin\frac x2}{\sqrt{\cos x}}\,dx &=&2\int_{\pi/6}^{\pi/4}\frac{\sin x}{\sqrt{2\cos^2 x-1}}\,dx\\ &=&-\sqrt{2}\left[\cosh^{-1}(\sqrt{2}\cos x)\right]_{\pi/6}^{\pi/4}\\ &=&\sqrt{2}\cosh^{-1}\frac{\sqrt{6}}2\\ &=&\sqrt{2}\ln\frac{\sqrt{6}+\sqrt{2}}2\\ &=&\frac 1{\sqrt{2}}\ln(2+\sqrt{3}) \end{eqnarray*}$$
よって, もとの積分は,
$$\begin{eqnarray*} &&\sqrt{2}\pi\ln\phi-\frac{\pi}{\sqrt{2}}\ln 2+\frac{\pi}{\sqrt{2}}\ln 2-\frac{\pi}{\sqrt{2}}\ln(2+\sqrt{3})\\ &=&\sqrt{2}\pi\ln\phi-\frac{\pi}{\sqrt{2}}\ln(2+\sqrt{3})\\ \end{eqnarray*}$$

$0\lt \Re(s)$のとき,
$$\begin{eqnarray*} \int_0^{\infty}x^{s-1}\frac{x}{\sinh x}\,dx&=&2(1-2^{s+1})\Gamma(s+1)\zeta(s+1) \end{eqnarray*}$$
より, Mellin inversion theoremより, $-1\lt \Re(s)\lt 0$のとき,
$$\begin{eqnarray*} \int_0^{\infty}x^{s-1}\left(1-\frac{x}{\sinh x}\right)\,dx&=&-2(1-2^{-s-1})\Gamma(s+1)\zeta(s+1) \end{eqnarray*}$$
$s\to -1$として,
$$\begin{eqnarray*} \int_0^{\infty}\frac{\sinh x-x}{x^2\sinh x}\,dx&=&-2\lim_{s\to -1}(1-2^{-s-1})\Gamma(s+1)\zeta(s+1)\\ &=&\ln 2 \end{eqnarray*}$$

$$\begin{eqnarray*} \int_0^{\infty}\frac{dx}{\sqrt{x^4+a^4}+\sqrt{x^4+b^4}}&=&\frac 1{a^4-b^4}\int_0^{\infty}(\sqrt{x^4+a^4}-\sqrt{x^4+b^4})\,dx\\ &=&\frac 1{a^4-b^4}\left(\int_0^{\infty}(\sqrt{x^4+a^4}-x^2)\,dx-\int_0^{\infty}(\sqrt{x^4+b^4}-x^2)\,dx\right)\\ &=&\frac {a^3-b^3}{a^4-b^4}\int_0^{\infty}(\sqrt{x^4+1}-x^2)\,dx\\ &=&\frac {a^3-b^3}{a^4-b^4}\int_0^{\infty}\frac 1{\sqrt{1+x^4}+x^2}\,dx \end{eqnarray*}$$
よって,
$$\begin{eqnarray*} \int_0^{\infty}\frac 1{\sqrt{1+x^4}+x^2}\,dx&=&\frac{\Gamma\left(\frac 14\right)^2}{6\sqrt{\pi}} \end{eqnarray*}$$
を示せば十分である. これは, $x^2\to x, x\to \frac 12\left(\frac 1x-x\right)$の置換により,
$$\begin{eqnarray*} \int_0^{\infty}\frac{1}{\sqrt{1+x^4}+x^2}\,dx&=&\frac 12\int_0^{\infty}\frac 1{\sqrt{x}(\sqrt{1+x^2}+x)}\,dx\\ &=&\frac 12\int_0^1\frac 1{\sqrt{\frac 12\left(\frac 1x-x\right)}\left(\frac 12\left(\frac 1x+x\right)+\frac 12\left(\frac 1x-x\right)\right)}\frac 12\left(\frac 1{x^2}+1\right)\,dx\\ &=&\frac{\sqrt{2}}4\int_0^1\frac{1+x^2}{\sqrt{x}\sqrt{1-x^2}}\,dx\\ &=&\frac{\sqrt{2}}{8}\left(\int_0^1x^{-3/4}(1-x)^{-1/2}\,dx+\int_0^1x^{1/4}(1-x)^{-1/2}\,dx\right)\\ &=&\frac{\sqrt{2}}{8}\left(\frac{\Gamma\left(\frac 14\right)\Gamma\left(\frac 12\right)}{\Gamma\left(\frac 34\right)}+\frac{\Gamma\left(\frac 54\right)\Gamma\left(\frac 12\right)}{\Gamma\left(\frac 74\right)}\right)\\ &=&\frac {\sqrt{2}}6\frac{\sqrt{\pi}\Gamma\left(\frac 14\right)}{\Gamma\left(\frac 34\right)}\\ &=&\frac{\Gamma\left(\frac 14\right)^2}{6\sqrt{\pi}}\\ \end{eqnarray*}$$
よって, 示された.

$$\begin{eqnarray*} \int_0^1\frac{dx}{\sqrt[3]{x(1-x)}\left(x-1+\frac 1x\right)}&=&\int_0^1{x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\,dx \end{eqnarray*}$$
より,
$$\begin{eqnarray*} I:=\int_0^1\frac{x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\,dx=\int_0^1\frac{1-x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\, \end{eqnarray*}$$
とする. Euler積分より, $x\to\frac{x}{1-x+x^2}$と置換して,
$$\begin{eqnarray*} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}&=&\int_0^1x^{a-1}(1-x)^{b-1}\,dx\\ &=&\int_0^1\left(\frac{x}{1-x+x^2}\right)^{a-1}\left(1-\frac{x}{1-x+x^2}\right)^{b-1}\frac{1-x^2}{(1-x+x^2)^2}\,dx\\ &=&\int_0^1\frac{x^{a-1}(1-x)^{2b-1}(1+x)}{(1-x+x^2)^{a+b}}\,dx \end{eqnarray*}$$
ここで, $a=\frac 23, b=\frac 13$とすると,
$$\begin{eqnarray*} 3I&=&2\int_0^1\frac{x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\,dx+\int_0^1\frac{1-x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\,dx\\ &=&\int_0^1\frac{1+x}{\sqrt[3]{x(1-x)}(1-x+x^2)}\,dx\\ &=&\Gamma\left(\frac 13\right)\Gamma\left(\frac 23\right)\\ &=&\frac{2\pi}{\sqrt{3}} \end{eqnarray*}$$
よって,
$$\begin{eqnarray*} I=\frac{2\pi}{3\sqrt{3}} \end{eqnarray*}$$

$x\to \tanh x$の置換により,
$$\begin{eqnarray*} \int_0^1\frac{\ln x}{1-x^2}\ln\ln\frac{1+x}{1-x}\,dx&=&\int_0^{\infty}\ln 2x\ln\tanh x\,dx \end{eqnarray*}$$
ここで, Mellin変換を考える.
$$\begin{eqnarray*} \int_0^{\infty}x^{s-1}\ln\tanh x\,dx&=&-\frac 2s\int_0^{\infty}\frac{x^s}{\sinh 2x}\,dx\\ &=&-\frac{2^{-s}}s\int_0^{\infty}\frac{x^s}{\sinh x}\,dx\\ &=&-\frac{2^{1-s}}s(1-2^{-1-s})\Gamma(s+1)\zeta(s+1)\\ &=&-2^{1-s}(1-2^{-1-s})\Gamma(s)\zeta(s+1) \end{eqnarray*}$$
よって,
$$\begin{eqnarray*} \int_0^{\infty}(2x)^{s-1}\ln\tanh x\,dx=-(1-2^{-1-s})\Gamma(s)\zeta(s+1) \end{eqnarray*}$$
よって, この両辺を$s$について微分して, $s=1$とすると,
$$\begin{eqnarray*} \int_0^{\infty}\ln 2x\ln\tanh x\,dx&=&-\left.\frac{d}{ds}(1-2^{-1-s})\Gamma(s)\zeta(s+1)\right|_{s=1}\\ &=&-\frac 34\zeta(2)\left(\frac{\ln 2}2-\gamma+\frac{\zeta'(2)}{\zeta(2)}\right)\\ &=&-\frac{\pi^2}{8}\left(\frac{\ln 2}{3}+\ln 2\pi-12\ln A\right)\\ &=&\frac{\pi^2}{24}\ln\frac{A^{36}}{16\pi^3} \end{eqnarray*}$$
ここで, $\frac{\zeta'(2)}{\zeta(2)}=\gamma+\ln2\pi -12\ln A$を用いた.

解析接続により, $0\lt a$の場合を示せば十分である. そのとき, $x\to ax$の置換により, $a=1$の場合を示せば十分である. $x\to\cosh x$の置換により,
$$\begin{eqnarray*} &&\int_1^{\infty}(x-1)^m\left(x-\sqrt{x^2-1}\right)^n\,dx\\ &=&\int_0^{\infty}e^{-nx}(\cosh x-1)^m\sinh x\,dx\\ &=&\frac 1{2^m}\int_0^{\infty}e^{-nx}\left(2\sinh\frac x2\right)^{2m}\sinh x\,dx\\ &=&\frac 1{2^{m+1}}\int_0^{\infty}e^{-(n-m)x}(1-e^{-x})^{2m}(e^x-e^{-x})\,dx\\ &=&\frac 1{2^{m+1}}\int_0^1x^{n-m-1}(1-x)^{2m}\left(\frac 1x-x\right)\,dx\\ &=&\frac 1{2^{m+1}}\left(\int_0^1x^{n-m-2}(1-x)^{2m}\,dx-\int_0^1x^{n-m}(1-x)^{2m}\,dx\right)\\ &=&\frac 1{2^{m+1}}\left(\frac{(n-m-2)!(2m)!}{(n+m-1)!}-\frac{(n-m)!(2m)!}{(n+m+1)!}\right)\\ &=&\frac 1{2^{m+1}}\frac{\left((n+m)(n+m-1)-(n-m)(n-m-1)\right)(n-m-2)!(2m)!}{(n+m+1)!}\\ &=&\frac 1{2^m}\frac{n(n-m-2)!(2m+1)!}{(n+m+1)!} \end{eqnarray*}$$
となって, 示すことができた.