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積分メモ

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$\BA\D\\ \int_0^\frac{\pi}{4} \frac{1-\tan x}{\sqrt{\tan x}}\L(\frac{\pi}{4}-x\R)dx=\frac{\pi\ln2}{2\sqrt{2}} \EA$

$\BA\D\\ \int_0^\frac{\pi}{2} \frac{x\ln\tan x}{\sqrt{\tan x}}\,dx =\frac{\pi}{8\sqrt{2}}(16\beta(2)-\pi^2-2\pi\ln2) \EA$

$\BA\D\\ \int_0^\frac{\pi}{2} \sinh^{-1}\sqrt{\sin x}\,\,dx=\frac{\pi\ln2}{2} \EA$

$\BA\D\\ \int_0^\frac{\pi}{3} \frac{x(2-\cos x)}{\cos x\sqrt{2\cos x-1}}\,dx=\frac{3}{2}\zeta(2) \EA$

tips
先ず,次の等式が成り立つ。

$\BA\D\\ \int_0^1 \frac{(1-x)^n\sqrt{x}}{1-x^2}\,dx=\frac{2^{2n}}{2n\binom{2n}{n}}-2^n\sum_{n< m}\frac{2^m}{2m\binom{2m}{m}} \EA$


両辺に$\D \frac{1}{n\binom{2n}{n}}$を掛けて$n$で和をとり,右辺第1項を積分で表し,第2項は級数として計算する。

$\BA\D\\ \int_0^1 \frac{1-x^2}{x(1+x^2)}\tanh^{-1}x\,\tanh^{-1}\sqrt{1-x^2}\,dx=\frac{3}{2}\zeta(2)\ln\frac{4}{1+\sqrt{2}} \EA$

$\BA\D\\ \int_0^1 \frac{\sin^{-1}\frac{x^2}{2}}{(1+x^2)\sqrt{2+x^2}}\,dx=\frac{\zeta(2)}{24} \EA$

$\BA\D\\ \int_0^1 \frac{1-x}{(1+x)(1+x+x^2)}\ln\frac{1}{x}\,\ln(1+x)\,dx=\frac{\zeta(3)}{18} \EA$

$\BA\D\\ \int_0^1 \frac{(1-x)(2+x)(1+2x)}{x(1+x)(1+x+x^2)}\ln\frac{1}{x}\,\ln\frac{(1+x)^2}{1+x+x^2}\,dx=\zeta(3) \EA$

$\BA\D\\ \int_0^1 \frac{(1-x)(2+x)(1+2x)}{x(1+x)(1+x+x^2)}\ln\frac{1}{x}\,\ln^2\frac{(1+x)^2}{1+x+x^2}\,dx=\frac{\zeta(4)}{9} \EA$

tips
先ず,次の等式が成り立つ。

$\BA\D\\ \zeta(4)=2R(4)+2R(1,3)+3R(1,1,2) \EA$


ただし,$\D R(k_1,\cdots,k_r)=\sum_{0< n_1<\cdots< n_r}n_1^{-k_1}\cdots n_r^{-k_r}\binom{2n_r}{n_r}^{-1}$

$\BA\D\\ R(4)&=\frac{17}{36}\zeta(4)\\ R(1,3)&=2\int_0^1 \frac{1}{x}\ln^2\frac{1}{1-\frac{x^2}{4}}\,\ln\frac{1+\sqrt{1-x^2}}{x}\,dx\\ R(1,1,2)&=\frac{1}{2}\int_0^1 \frac{x}{1-\frac{x^2}{4}}\ln^2\frac{1}{1-\frac{x^2}{4}}\,\ln\frac{1+\sqrt{1-x^2}}{x}\,dx \EA$


を代入し,$\D x=\frac{2t}{1+t^2}$と置換する。

$\BA\D\\ \int_0^1 \frac{(1-x^2)\ln\frac{1}{x}}{x(1+x+x^2)}(3{\rm Li}_2(x)+2{\rm Li}_2(-x)+2\ln^2(1+x))\,dx=\zeta(4) \EA$

$\BA\D\\ \int_0^1 \frac{(1-x^2)\tanh^{-1}\sqrt{1-x^2}}{(1-2x\cos t+x^2)(1+2x\cos t+x^2)}\,dx=\frac{\pi}{2}\frac{\cos^{-1}\sqrt{\sin t}}{\cos t}\qquad \L(0\le t<\frac{\pi}{2}\R) \EA$

$\BA\D\\ \int_0^1 x^{-1+\alpha}(1-x)^{-1+\beta}\left(\frac{1+ax}{1-ax}+\frac{1+bx}{1-bx}\right) {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};abx \right]dx =\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};a\right] {_2}F_1\left[\begin{matrix} 1,\alpha\\ \alpha+\beta \end{matrix};b\right] \EA$

$\BA\D\\ \int_0^1 \frac{(\sin^{-1}abx)^2}{x}\left(\frac{1+a^2x^2}{1-a^2x^2}+\frac{1+b^2x^2}{1-b^2x^2} \right)\ln\frac{1+\sqrt{1-x^2}}{x}\,dx =(\sin^{-1}a)^2(\sin^{-1}b)^2 \EA$

$\BA\D\\ \int_0^\frac{\pi}{2} \L(\frac{x}{\sin x}-1\R)\frac{1}{\tan\frac{x}{2}}\ln\frac{1}{\cos x}\,dx =2\int_0^1 \frac{\sqrt{x}}{1-x^2}\ln\frac{2}{1+x}\,dx =\frac{\pi^2}{8}+\frac{\pi\ln2}{2}-2\beta(2) \EA$

$\D \alpha+\beta+\gamma=1,\quad \omega=\L(\frac{\mu}{1-\mu}\R)^{\gamma}$に対して

$\BA\D\\ \int_0^1 x^{-\alpha}(1-x)^{-\beta}(1-\mu x)^{-\gamma}dx =\frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(\gamma)}\int_0^1 x^{-1+\gamma}(1-\mu x)^{-1+\alpha}dx =\frac{\Gamma(1-\alpha)\Gamma(1-\beta)}{\Gamma(1+\gamma)}\mu^{-\gamma}\int_0^\omega (1+x^{\frac{1}{\gamma}})^{-1+\beta}dx \EA$

$\BA\D\\ \int_0^\frac{1}{2} \frac{dx}{\sqrt{x+x^4}}=\int_0^1 \frac{dx}{\sqrt{1-x^3}} \EA$

tips
$\D x=\frac{1-t}{2+t}~$と置換する。

$\BA\D\\ \int_0^z \frac{dx}{\sqrt{1+x^6}}=\frac{1}{2\sqrt[4]{3}}\int_0^{2\tan^{-1}\sqrt{\frac{\sqrt{3}z^2}{1+z^2}}}\frac{dx}{\sqrt{1-\frac{2+\sqrt{3}}{4}\sin^2 x}} \EA$

tips
$\BA\D\\ \frac{1}{2}\int_0^{z^2} \frac{dx}{\sqrt{x(1+x^3)}} \EA$


において$\D \frac{x}{1+x}=\frac{\tan^2 t}{\sqrt{3}}$と置換する。

$\BA\D\\ \int_0^{\frac{1}{\sqrt{\sqrt{3}-1}}} \frac{dx}{\sqrt{1+x^6}} =\frac{\Gamma(\frac{1}{3})\Gamma(\frac{1}{6})}{8\Gamma(\frac{1}{2})} \EA$

$\BA\D\\ \frac{\pi}{2}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{\frac{3}{5}}}\frac{\tan^{-1}\frac{1}{\sqrt{2-x^{2}}}}{1+x^{2}}dx-\int_{\frac{1}{\sqrt{2}}}^{\sqrt{\frac{3}{5}}}\frac{\tan^{-1}\frac{1}{\sqrt{2-x^{2}}}}{1+x^{2}}\tan^{-1}\frac{\left(1+x^{2}\right)\sqrt{2x^{2}-1}}{x\left(3-5x^{2}\right)}dx=\frac{\pi^3}{280} \EA$

tips
$\BA\D\\ \int_0^1\int_0^1\int_0^1 \frac{dxdydz}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2}} &=\frac{2}{\sqrt{\pi}}\int_0^1\int_0^1\int_0^1 \frac{1}{(1+x^2)(1+y^2)(1+z^2)}\int_0^\infty e^{-t^2(4+x^2+y^2+z^2)}dtdxdydz\\ &=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-4t^2}\L(\int_0^1 \frac{e^{-t^2x^2}}{1+x^2}dx\R)^3dt\\ &=\frac{2}{\sqrt{\pi}}\int_0^\infty e^{-4t^2}\L(\frac{\pi}{4}e^{t^2}\big(1-{\rm erf}(t)^2\big)\R)^3dt\\ &=\frac{\pi^3}{32\sqrt{\pi}}\int_0^\infty e^{-t^2}\big(1-{\rm erf}(t)^2\big)^3dt\\ &=\frac{\pi^3}{32\sqrt{\pi}}\frac{8\sqrt{\pi}}{35}\\ &=\frac{\pi^3}{140} \EA$


また,

$\BA\D\\ \int_0^1\int_0^1\int_0^1 \frac{dxdydz}{(1+x^2)(1+y^2)(1+z^2)\sqrt{4+x^2+y^2+z^2}} &=\int_0^1\int_0^1 \frac{\tan^{-1}\sqrt{\frac{3+x^2+y^2}{5+x^2+y^2}}}{(1+x^2)(1+y^2)\sqrt{3+x^2+y^2}}\,dxdy\\ &=\frac{1}{4}\int_0^1\int_0^1 \frac{\tan^{-1}\sqrt{\frac{3+x+y}{5+x+y}}}{(1+x)(1+y)\sqrt{xy}\sqrt{3+x+y}}\,dxdy\\ &=\int_0^\frac{1}{2}\int_0^p \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{((1+p)^2-q^2)\sqrt{p^2-q^2}\sqrt{3+2p}}dqdp +\int_\frac{1}{2}^1\int_0^{1-p} \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{((1+p)^2-q^2)\sqrt{p^2-q^2}\sqrt{3+2p}}dqdp\\ &=\frac{\pi}{2}\int_0^\frac{1}{2} \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp +\int_\frac{1}{2}^1 \frac{\tan^{-1}\L(\frac{1-p}{1+p}\sqrt{\frac{2p+1}{2p-1}}\R)\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp\\ &=\frac{\pi}{2}\int_0^1 \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp -\frac{\pi}{2}\int_\frac{1}{2}^1 \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp +\int_\frac{1}{2}^1 \frac{\tan^{-1}\L(\frac{1-p}{1+p}\sqrt{\frac{2p+1}{2p-1}}\R)\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp\\ &=\frac{\pi}{2}\int_0^1 \frac{\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp -\int_\frac{1}{2}^1 \frac{\tan^{-1}\L(\frac{1+p}{1-p}\sqrt{\frac{2p-1}{2p+1}}\R)\tan^{-1}\sqrt{\frac{3+2p}{5+2p}}}{(1+p)\sqrt{1+2p}\sqrt{3+2p}}dp \EA$


$\D\frac{1}{2-x^2}=\frac{3+2p}{5+2p}$と置換する。

$\BA\D\\ \int_0^1\int_0^1\int_0^1 \frac{dxdydz}{(2+x^2+y^2+z^2)^2}+\frac{1}{2}\int_0^1 \frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{(1+x^2)(2+x^2)\sqrt{3+x^2}}\,dx=\frac{\pi}{8\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}} \EA$

tips
$\BA\D\\ &\int_0^1\int_0^1\int_0^1 \frac{dxdydz}{(2+x^2+y^2+z^2)^2}\\ =&2\int_0^1\int_0^1\int_0^1 \int_0^\infty t^3e^{-t^2(2+x^2+y^2+z^2)}\,dt dxdydz\\ =&2\int_0^\infty e^{-2t^2}\L(t\int_0^1 e^{-t^2x^2}dx\R)^3\,dt\\ =&2\L(\frac{\sqrt{\pi}}{2}\R)^3\int_0^\infty e^{-2t^2}{\rm erf}(t)^3\,dt\\ =&\L(\frac{\sqrt{\pi}}{2}\R)^4\int_0^\infty te^{-t^2}{\rm erf}(t)^4\,dt\\ =&\frac{\pi^2}{16}\int_0^\infty te^{-t^2}\L(1-\frac{4}{\pi}\int_0^1 \frac{e^{-t^2(1+x^2)}}{1+x^2}\,dx\R)^2\,dt\\ =&\frac{\pi^2}{16}\int_0^\infty te^{-t^2}\,dt-\frac{\pi}{2}\int_0^\infty t\int_0^1 \frac{e^{-t^2(2+x^2)}}{1+x^2}\,dxdt +\int_0^\infty t\int_0^1\int_0^1 \frac{e^{-t^2(3+x^2+y^2)}}{(1+x^2)(1+y^2)}\,dxdydt\\ =&\frac{\pi^2}{32}-\frac{\pi}{4}\int_0^1 \frac{dx}{(1+x^2)(2+x^2)}+\frac{1}{2}\int_0^1 \int_0^1 \frac{dxdy}{(1+x^2)(1+y^2)(3+x^2+y^2)}\\ =&\frac{\pi^2}{32}-\frac{\pi}{4}\int_0^1 \frac{dx}{(1+x^2)(2+x^2)}+\frac{1}{2}\int_0^1 \frac{1}{(1+x^2)(2+x^2)}\L(\frac{\pi}{4}-\frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{\sqrt{3+x^2}}\R)\,dx\\ =&\frac{\pi^2}{32}-\frac{\pi}{8}\int_0^1 \frac{dx}{(1+x^2)(2+x^2)}-\frac{1}{2}\int_0^1 \frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{(1+x^2)(2+x^2)\sqrt{3+x^2}}\,dx\\ =&\frac{\pi^2}{32}-\frac{\pi}{8}\L(\frac{\pi}{4}-\frac{1}{\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}}\R)-\frac{1}{2}\int_0^1 \frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{(1+x^2)(2+x^2)\sqrt{3+x^2}}\,dx\\ =&\frac{\pi}{8\sqrt{2}}\tan^{-1}\frac{1}{\sqrt{2}}-\frac{1}{2}\int_0^1 \frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{(1+x^2)(2+x^2)\sqrt{3+x^2}}\,dx \EA$

$\D a\ge 0,~R>0$に対して

$\BA\D\\ \int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_{2m-1}}{{(R+1+x_1^2+\cdots+x_{2m-1}^2)}^{m+a}} =\frac{\Gamma(1+a)}{m\Gamma(m+a)}\sum_{k=1}^m(-1)^{k-1}\binom{m}{k}\L(\frac{\pi}{4}\R)^{m-k}\int_0^1\cdots\int_0^1 \L(1-\frac{R}{R+k+x_1^2+\cdots+x_k^2}\R) \,\frac{dx_1\cdots dx_k}{(1+x_1^2)\cdots(1+x_k^2){(R+k+x_1^2+\cdots+x_k^2)}^{a}} \EA$

tips
$\BA\D\\ &\frac{\Gamma(m+a)}{2}\int_0^1\cdots\int_0^1 \frac{dx_1\cdots dx_{2m-1}}{{(R+1+x_1^2+\cdots+x_{2m-1}^2)}^{m+a}}\\ =&\int_0^1\cdots\int_0^1\int_0^\infty t^{2a+2m-1} e^{-t^2(R+1+x_1^2+\cdots+x_{2m-1}^2)} \,dtdx_1\cdots dx_{2m-1}\\ =&\int_0^\infty t^{2a+2m-1}e^{-t^2(R+1)}\L(\int_0^1 e^{-t^2x^2}\,dx\R)^{2m-1}\,dt\\ =&\int_0^\infty t^{2a+2m-1}e^{-t^2(R+1)}\L(\frac{\sqrt{\pi}}{2t}{\rm erf}(t)\R)^{2m-1}\,dt\\ =&\L(\frac{\sqrt{\pi}}{2}\R)^{2m-1}\int_0^\infty t^{2a}e^{-t^2(R+1)}{\rm erf}(t)^{2m-1}\,dt\\ =&\frac{1}{m}\L(\frac{\sqrt{\pi}}{2}\R)^{2m}\int_0^\infty (Rt^2-a)t^{2a-1}e^{-Rt^2}{\rm erf}(t)^{2m}\,dt\\ =&\frac{1}{m}\L(\frac{\pi}{4}\R)^{m}\int_0^\infty (Rt^2-a)t^{2a-1}e^{-Rt^2}\L(1-\frac{4}{\pi}\int_0^1 \frac{e^{-t^2(1+x^2)}}{1+x^2}\,dx\R)^{m}\,dt\\ =&\frac{1}{m}\L(\frac{\pi}{4}\R)^{m}\sum_{k=0}^m(-1)^k\binom{m}{k}\L(\frac{4}{\pi}\R)^{k}\int_0^\infty (Rt^2-a)t^{2a-1}e^{-Rt^2}\L(\int_0^1 \frac{e^{-t^2(1+x^2)}}{1+x^2}\,dx\R)^{k}\,dt\\ =&\frac{1}{m}\sum_{k=0}^m(-1)^k\binom{m}{k}\L(\frac{\pi}{4}\R)^{m-k}\int_0^\infty (Rt^2-a)t^{2a-1}e^{-Rt^2}\L(\int_0^1\cdots\int_0^1 \frac{e^{-t^2(k+x_1^2+\cdots+x_k^2)}}{(1+x_1^2)\cdots(1+x_k^2)}\,dx_1\cdots dx_k\R)\,dt\\ =&\frac{1}{m}\sum_{k=0}^m(-1)^k\binom{m}{k}\L(\frac{\pi}{4}\R)^{m-k}\int_0^1\cdots\int_0^1\int_0^\infty (Rt^2-a)t^{2a-1}e^{-t^2(R+k+x_1^2+\cdots+x_k^2)}\,dt\,\frac{dx_1\cdots dx_k}{(1+x_1^2)\cdots(1+x_k^2)}\\ =&\frac{\Gamma(1+a)}{2m}\sum_{k=0}^m(-1)^k\binom{m}{k}\L(\frac{\pi}{4}\R)^{m-k}\int_0^1\cdots\int_0^1 \L(\frac{R}{R+k+x_1^2+\cdots+x_k^2}-1\R) \,\frac{dx_1\cdots dx_k}{(1+x_1^2)\cdots(1+x_k^2){(R+k+x_1^2+\cdots+x_k^2)}^{a}}\\ =&\frac{\Gamma(1+a)}{2m}\sum_{k=1}^m(-1)^{k-1}\binom{m}{k}\L(\frac{\pi}{4}\R)^{m-k}\int_0^1\cdots\int_0^1 \L(1-\frac{R}{R+k+x_1^2+\cdots+x_k^2}\R) \,\frac{dx_1\cdots dx_k}{(1+x_1^2)\cdots(1+x_k^2){(R+k+x_1^2+\cdots+x_k^2)}^{a}} \EA$


結果的には実数の範囲では$a\notin -{\mathbb N},~ R>-1$で成り立つか。

$\BA\D\\ \int_0^1 \L(\frac{1}{1+x^2}+\frac{1}{2+x^2}\R)\frac{\tan^{-1}\frac{1}{\sqrt{3+x^2}}}{\sqrt{3+x^2}}\,dx=\frac{\pi\tan^{-1}\frac{1}{\sqrt{2}}}{4\sqrt{2}} \EA$

tips
$\BA\D\\ \int_0^\infty e^{-2t^2}{\rm erf}(t)(1-{\rm erf}(t)^2)\,dt \EA$


について,➀カッコを分解してそれぞれ部分積分した後式$A$を用いる。➁直接式$A$を用いる。➀=➁

$\BA\D\\ \int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}\,dt=\frac{\pi}{4}(1-{\rm erf}(x)^2) \quad \cdots A \EA$

$\BA\D\\ \int_0^\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1} \frac{\ln\frac{1}{x}}{(1-x)\sqrt{1-2(15+8\sqrt{3})x+x^2}}\,dx=\frac{2(2-\sqrt{3})}{3}\beta(2) \EA$

tips
$\BA\D\\ \frac{3}{2(2-\sqrt{3})}\int_0^\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1} \frac{\ln\frac{1}{x}}{(1-x)\sqrt{1-2(15+8\sqrt{3})x+x^2}}\,dx &=\frac{3(1+\sqrt{3})}{2}\int_0^\frac{\pi}{24} \frac{\tan2\theta \ln\frac{1}{\tan \theta}}{\sqrt{\cos 4\theta-\frac{\sqrt{3}}{2}}}\,d\theta\\ &=\frac{3(1+\sqrt{3})}{4}\int_\frac{\sqrt{3}}{2}^1 \frac{\ln\frac{\sqrt{1-x}}{\sqrt{2}-\sqrt{1+x}}}{(1+x)\sqrt{x-\frac{\sqrt{3}}{2}}}\,dx\\ &=\frac{3(1+\sqrt{3})}{8}\int_0^{\frac{1}{2}\L(1-\frac{\sqrt{3}}{2}\R)} \frac{\tanh^{-1}\sqrt{1-x}}{(1-x)\sqrt{1-\frac{\sqrt{3}}{2}-2x}}\,dx\\ &=\frac{3(1+\sqrt{3})}{4}\int_0^\frac{-1+\sqrt{3}}{2} \frac{\tanh^{-1}\sqrt{\frac{1}{2}\L(\L(\frac{1+\sqrt{3}}{2}\R)^2+x^2\R)}}{\L(\frac{1+\sqrt{3}}{2}\R)^2+x^2}\,dx\\ &=\frac{3}{2}\int_0^\frac{-1+\sqrt{3}}{1+\sqrt{3}}\frac{\tanh^{-1}\frac{(1+\sqrt{3})\sqrt{1+x^2}}{2\sqrt{2}}}{1+x^2}\,dx\\ &=\frac{3}{2}\int_0^\frac{\pi}{12} \tanh^{-1}\frac{\cos\frac{\pi}{12}}{\cos \theta}\,d\theta\\ &=\frac{\pi}{8}\int_0^1 \tanh^{-1}\frac{\cos\frac{\pi}{12}}{\cos \frac{\pi v}{12}}\,dv \EA$


$\D f(t)=\tanh^{-1}\frac{\cos t}{\cos tv}$とすれば,$\D f'(t)=-\frac{1}{2}\L(\frac{1+v}{\sin(1+v)t}+\frac{1-v}{\sin(1-v)t}\R)$となるので

$\BA\D\\ \int_0^1 f'(t)\,dv &=-\frac{1}{2}\int_0^1 \L(\frac{1+v}{\sin(1+v)t}+\frac{1-v}{\sin(1-v)t}\R)dv\\ &=-\frac{1}{2}\int_0^2 \frac{v}{\sin tv}\,dv \EA$


また,$\D f\L(\frac{\pi}{2}\R)=0$なので

$\BA\D\\ \frac{\pi}{8}\int_0^1 \tanh^{-1}\frac{\cos\frac{\pi}{12}}{\cos \frac{\pi v}{12}}\,dv &=\frac{\pi}{16} \int_\frac{\pi}{12}^\frac{\pi}{2}\int_0^2 \frac{v}{\sin tv}\,dv\,dt\\ &=\frac{\pi}{16} \int_0^2\int_\frac{\pi}{12}^\frac{\pi}{2} \frac{v}{\sin tv}\,dt\,dv\\ &=\frac{\pi}{16} \int_0^2 \ln\frac{\tan\frac{\pi v}{4}}{\tan \frac{\pi v}{24}}\,dv\\ &=\frac{\pi}{8} \int_0^1 \ln\frac{\tan\frac{\pi v}{2}}{\tan \frac{\pi v}{12}}\,dv\\ &=\frac{\pi}{8} \sum_{n=0}^\infty \frac{2}{2n+1}\int_0^1 \L(\cos \frac{(2n+1)\pi v}{6}-\cos(2n+1)\pi v\R)\,dv\\ &=\frac{3}{2}\sum_{n=0}^\infty \frac{\sin\frac{(2n+1)\pi}{6}}{(2n+1)^2}\\ &=\beta(2) \EA$

$\BA\D\\ \EA$

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投稿日:2023330

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