9

＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿

高校数学解説

積分メモ

735

0

\begin{array}{r} \int_{0}^{\frac{π}{4}} \frac{1 - \tan x}{\sqrt{\tan x}} (\frac{π}{4} - x) d x = \frac{π \ln 2}{2 \sqrt{2}} \end{array}

\begin{array}{r} \int_{0}^{\frac{π}{2}} \frac{x \ln \tan x}{\sqrt{\tan x}} d x = \frac{π}{8 \sqrt{2}} (16 β (2) - π^{2} - 2 π \ln 2) \end{array}

\begin{array}{r} \int_{0}^{\frac{π}{2}} \sinh^{- 1} \sqrt{\sin x} d x = \frac{π \ln 2}{2} \end{array}

\begin{array}{r} \int_{0}^{\frac{π}{3}} \frac{x (2 - \cos x)}{\cos x \sqrt{2 \cos x - 1}} d x = \frac{3}{2} ζ (2) \end{array}

tips

先ず，次の等式が成り立つ。

\begin{array}{r} \int_{0}^{1} \frac{(1 - x)^{n} \sqrt{x}}{1 - x^{2}} d x = \frac{2^{2 n}}{2 n (\binom{2 n}{n})} - 2^{n} \sum_{n < m} \frac{2^{m}}{2 m (\binom{2 m}{m})} \end{array}

両辺に

\frac{1}{n (\binom{2 n}{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。

\begin{array}{r} \int_{0}^{1} \frac{1 - x^{2}}{x (1 + x^{2})} \tanh^{- 1} x \tanh^{- 1} \sqrt{1 - x^{2}} d x = \frac{3}{2} ζ (2) \ln \frac{4}{1 + \sqrt{2}} \end{array}

\begin{array}{r} \int_{0}^{1} \frac{\sin^{- 1} \frac{x^{2}}{2}}{(1 + x^{2}) \sqrt{2 + x^{2}}} d x = \frac{ζ (2)}{24} \end{array}

\begin{array}{r} \int_{0}^{1} \frac{1 - x}{(1 + x) (1 + x + x^{2})} \ln \frac{1}{x} \ln (1 + x) d x = \frac{ζ (3)}{18} \end{array}

\begin{array}{r} \int_{0}^{1} \frac{(1 - x) (2 + x) (1 + 2 x)}{x (1 + x) (1 + x + x^{2})} \ln \frac{1}{x} \ln \frac{(1 + x)^{2}}{1 + x + x^{2}} d x = ζ (3) \end{array}

\begin{array}{r} \int_{0}^{1} \frac{(1 - x) (2 + x) (1 + 2 x)}{x (1 + x) (1 + x + x^{2})} \ln \frac{1}{x} \ln^{2} \frac{(1 + x)^{2}}{1 + x + x^{2}} d x = \frac{ζ (4)}{9} \end{array}

tips

先ず，次の等式が成り立つ。

\begin{array}{r} ζ (4) = 2 R (4) + 2 R (1, 3) + 3 R (1, 1, 2) \end{array}

ただし，

R (k_{1}, \dots, k_{r}) = \sum_{0 < n_{1} < \dots < n_{r}} n_{1}^{- k_{1}} \dots n_{r}^{- k_{r}} {(\binom{2 n_{r}}{n_{r}})}^{- 1}

。

\begin{aligned} R (4) & = \frac{17}{36} ζ (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} d x \\ 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} d x \end{aligned}

を代入し，

x = \frac{2 t}{1 + t^{2}}

と置換する。

\begin{array}{r} \int_{0}^{1} \frac{(1 - x^{2}) \ln \frac{1}{x}}{x (1 + x + x^{2})} (3 {L i}_{2} (x) + 2 {L i}_{2} (- x) + 2 \ln^{2} (1 + x)) d x = ζ (4) \end{array}

\begin{array}{r} \int_{0}^{1} \frac{(1 - x^{2}) \tanh^{- 1} \sqrt{1 - x^{2}}}{(1 - 2 x \cos t + x^{2}) (1 + 2 x \cos t + x^{2})} d x = \frac{π}{2} \frac{\cos^{- 1} \sqrt{\sin t}}{\cos t} (0 \leq t < \frac{π}{2}) \end{array}

\begin{array}{r} \int_{0}^{1} x^{- 1 + α} (1 - x)^{- 1 + β} (\frac{1 + a x}{1 - a x} + \frac{1 + b x}{1 - b x})_{2} F_{1} [\begin{array}{c} 1, α \\ α + β \end{array}; a b x] d x = \frac{Γ (α) Γ (β)}{Γ (α + β)}_{2} F_{1} [\begin{array}{c} 1, α \\ α + β \end{array}; a]_{2} F_{1} [\begin{array}{c} 1, α \\ α + β \end{array}; b] \end{array}

\begin{array}{r} \int_{0}^{1} \frac{(\sin^{- 1} a b x)^{2}}{x} (\frac{1 + a^{2} x^{2}}{1 - a^{2} x^{2}} + \frac{1 + b^{2} x^{2}}{1 - b^{2} x^{2}}) \ln \frac{1 + \sqrt{1 - x^{2}}}{x} d x = (\sin^{- 1} a)^{2} (\sin^{- 1} b)^{2} \end{array}

\begin{array}{r} \int_{0}^{\frac{π}{2}} (\frac{x}{\sin x} - 1) \frac{1}{\tan \frac{x}{2}} \ln \frac{1}{\cos x} d x = 2 \int_{0}^{1} \frac{\sqrt{x}}{1 - x^{2}} \ln \frac{2}{1 + x} d x = \frac{π^{2}}{8} + \frac{π \ln 2}{2} - 2 β (2) \end{array}

$α + β + γ = 1, ω = {(\frac{μ}{1 - μ})}^{γ}$ に対して

\begin{array}{r} \int_{0}^{1} x^{- α} (1 - x)^{- β} (1 - μ x)^{- γ} d x = \frac{Γ (1 - α) Γ (1 - β)}{Γ (γ)} \int_{0}^{1} x^{- 1 + γ} (1 - μ x)^{- 1 + α} d x = \frac{Γ (1 - α) Γ (1 - β)}{Γ (1 + γ)} μ^{- γ} \int_{0}^{ω} (1 + x^{\frac{1}{γ}})^{- 1 + β} d x \end{array}

\begin{array}{r} \int_{0}^{\frac{1}{2}} \frac{d x}{\sqrt{x + x^{4}}} = \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{3}}} \end{array}

tips

x = \frac{1 - t}{2 + t}

と置換する。

\begin{array}{r} \int_{0}^{z} \frac{d x}{\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{d x}{\sqrt{1 - \frac{2 + \sqrt{3}}{4} \sin^{2} x}} \end{array}

tips

\begin{array}{r} \frac{1}{2} \int_{0}^{z^{2}} \frac{d x}{\sqrt{x (1 + x^{3})}} \end{array}

において

\frac{x}{1 + x} = \frac{\tan^{2} t}{\sqrt{3}}

と置換する。

\begin{array}{r} \int_{0}^{\frac{1}{\sqrt{\sqrt{3} - 1}}} \frac{d x}{\sqrt{1 + x^{6}}} = \frac{Γ (\frac{1}{3}) Γ (\frac{1}{6})}{8 Γ (\frac{1}{2})} \end{array}

\begin{array}{r} \frac{π}{2} \int_{\frac{1}{\sqrt{3}}}^{\sqrt{\frac{3}{5}}} \frac{\tan^{- 1} \frac{1}{\sqrt{2 - x^{2}}}}{1 + x^{2}} d x - \int_{\frac{1}{\sqrt{2}}}^{\sqrt{\frac{3}{5}}} \frac{\tan^{- 1} \frac{1}{\sqrt{2 - x^{2}}}}{1 + x^{2}} \tan^{- 1} \frac{(1 + x^{2}) \sqrt{2 x^{2} - 1}}{x (3 - 5 x^{2})} d x = \frac{π^{3}}{280} \end{array}

tips

\begin{aligned} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{d x d y d z}{(1 + x^{2}) (1 + y^{2}) (1 + z^{2}) \sqrt{4 + x^{2} + y^{2} + z^{2}}} & = \frac{2}{\sqrt{π}} \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})} d t d x d y d z \\ = \frac{2}{\sqrt{π}} \int_{0}^{\infty} e^{- 4 t^{2}} {(\int_{0}^{1} \frac{e^{- t^{2} x^{2}}}{1 + x^{2}} d x)}^{3} d t \\ = \frac{2}{\sqrt{π}} \int_{0}^{\infty} e^{- 4 t^{2}} {(\frac{π}{4} e^{t^{2}} (1 - e r f (t)^{2}))}^{3} d t \\ = \frac{π^{3}}{32 \sqrt{π}} \int_{0}^{\infty} e^{- t^{2}} (1 - e r f (t)^{2})^{3} d t \\ = \frac{π^{3}}{32 \sqrt{π}} \frac{8 \sqrt{π}}{35} \\ = \frac{π^{3}}{140} \end{aligned}

また，

\begin{aligned} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{d x d y d z}{(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}}} d x d y \\ = \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{x y} \sqrt{3 + x + y}} d x d y \\ = \int_{0}^{\frac{1}{2}} \int_{0}^{p} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{((1 + p)^{2} - q^{2}) \sqrt{p^{2} - q^{2}} \sqrt{3 + 2 p}} d q d p + \int_{\frac{1}{2}}^{1} \int_{0}^{1 - p} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{((1 + p)^{2} - q^{2}) \sqrt{p^{2} - q^{2}} \sqrt{3 + 2 p}} d q d p \\ = \frac{π}{2} \int_{0}^{\frac{1}{2}} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p + \int_{\frac{1}{2}}^{1} \frac{\tan^{- 1} (\frac{1 - p}{1 + p} \sqrt{\frac{2 p + 1}{2 p - 1}}) \tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p \\ = \frac{π}{2} \int_{0}^{1} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p - \frac{π}{2} \int_{\frac{1}{2}}^{1} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p + \int_{\frac{1}{2}}^{1} \frac{\tan^{- 1} (\frac{1 - p}{1 + p} \sqrt{\frac{2 p + 1}{2 p - 1}}) \tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p \\ = \frac{π}{2} \int_{0}^{1} \frac{\tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p - \int_{\frac{1}{2}}^{1} \frac{\tan^{- 1} (\frac{1 + p}{1 - p} \sqrt{\frac{2 p - 1}{2 p + 1}}) \tan^{- 1} \sqrt{\frac{3 + 2 p}{5 + 2 p}}}{(1 + p) \sqrt{1 + 2 p} \sqrt{3 + 2 p}} d p \end{aligned}

\frac{1}{2 - x^{2}} = \frac{3 + 2 p}{5 + 2 p}

と置換する。

\begin{array}{r} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{d x d y d z}{(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}}} d x = \frac{π}{8 \sqrt{2}} \tan^{- 1} \frac{1}{\sqrt{2}} \end{array}

tips

\begin{aligned} \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{d x d y d z}{(2 + x^{2} + y^{2} + z^{2})^{2}} \\ = & 2 \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \int_{0}^{\infty} t^{3} e^{- t^{2} (2 + x^{2} + y^{2} + z^{2})} d t d x d y d z \\ = & 2 \int_{0}^{\infty} e^{- 2 t^{2}} {(t \int_{0}^{1} e^{- t^{2} x^{2}} d x)}^{3} d t \\ = & 2 {(\frac{\sqrt{π}}{2})}^{3} \int_{0}^{\infty} e^{- 2 t^{2}} e r f (t)^{3} d t \\ = & {(\frac{\sqrt{π}}{2})}^{4} \int_{0}^{\infty} t e^{- t^{2}} e r f (t)^{4} d t \\ = & \frac{π^{2}}{16} \int_{0}^{\infty} t e^{- t^{2}} {(1 - \frac{4}{π} \int_{0}^{1} \frac{e^{- t^{2} (1 + x^{2})}}{1 + x^{2}} d x)}^{2} d t \\ = & \frac{π^{2}}{16} \int_{0}^{\infty} t e^{- t^{2}} d t - \frac{π}{2} \int_{0}^{\infty} t \int_{0}^{1} \frac{e^{- t^{2} (2 + x^{2})}}{1 + x^{2}} d x d t + \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})} d x d y d t \\ = & \frac{π^{2}}{32} - \frac{π}{4} \int_{0}^{1} \frac{d x}{(1 + x^{2}) (2 + x^{2})} + \frac{1}{2} \int_{0}^{1} \int_{0}^{1} \frac{d x d y}{(1 + x^{2}) (1 + y^{2}) (3 + x^{2} + y^{2})} \\ = & \frac{π^{2}}{32} - \frac{π}{4} \int_{0}^{1} \frac{d x}{(1 + x^{2}) (2 + x^{2})} + \frac{1}{2} \int_{0}^{1} \frac{1}{(1 + x^{2}) (2 + x^{2})} (\frac{π}{4} - \frac{\tan^{- 1} \frac{1}{\sqrt{3 + x^{2}}}}{\sqrt{3 + x^{2}}}) d x \\ = & \frac{π^{2}}{32} - \frac{π}{8} \int_{0}^{1} \frac{d x}{(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}}} d x \\ = & \frac{π^{2}}{32} - \frac{π}{8} (\frac{π}{4} - \frac{1}{\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}}} d x \\ = & \frac{π}{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}}} d x \end{aligned}

$a \geq 0, R > 0$ に対して

\begin{array}{r} \int_{0}^{1} \dots \int_{0}^{1} \frac{d x_{1} \dots d x_{2 m - 1}}{{(R + 1 + x_{1}^{2} + \dots + x_{2 m - 1}^{2})}^{m + a}} = \frac{Γ (1 + a)}{m Γ (m + a)} \sum_{k = 1}^{m} (- 1)^{k - 1} (\binom{m}{k}) {(\frac{π}{4})}^{m - k} \int_{0}^{1} \dots \int_{0}^{1} (1 - \frac{R}{R + k + x_{1}^{2} + \dots + x_{k}^{2}}) \frac{d x_{1} \dots d x_{k}}{(1 + x_{1}^{2}) \dots (1 + x_{k}^{2}) {(R + k + x_{1}^{2} + \dots + x_{k}^{2})}^{a}} \end{array}

tips

\begin{aligned} \frac{Γ (m + a)}{2} \int_{0}^{1} \dots \int_{0}^{1} \frac{d x_{1} \dots d x_{2 m - 1}}{{(R + 1 + x_{1}^{2} + \dots + x_{2 m - 1}^{2})}^{m + a}} \\ = & \int_{0}^{1} \dots \int_{0}^{1} \int_{0}^{\infty} t^{2 a + 2 m - 1} e^{- t^{2} (R + 1 + x_{1}^{2} + \dots + x_{2 m - 1}^{2})} d t d x_{1} \dots d x_{2 m - 1} \\ = & \int_{0}^{\infty} t^{2 a + 2 m - 1} e^{- t^{2} (R + 1)} {(\int_{0}^{1} e^{- t^{2} x^{2}} d x)}^{2 m - 1} d t \\ = & \int_{0}^{\infty} t^{2 a + 2 m - 1} e^{- t^{2} (R + 1)} {(\frac{\sqrt{π}}{2 t} e r f (t))}^{2 m - 1} d t \\ = & {(\frac{\sqrt{π}}{2})}^{2 m - 1} \int_{0}^{\infty} t^{2 a} e^{- t^{2} (R + 1)} e r f (t)^{2 m - 1} d t \\ = & \frac{1}{m} {(\frac{\sqrt{π}}{2})}^{2 m} \int_{0}^{\infty} (R t^{2} - a) t^{2 a - 1} e^{- R t^{2}} e r f (t)^{2 m} d t \\ = & \frac{1}{m} {(\frac{π}{4})}^{m} \int_{0}^{\infty} (R t^{2} - a) t^{2 a - 1} e^{- R t^{2}} {(1 - \frac{4}{π} \int_{0}^{1} \frac{e^{- t^{2} (1 + x^{2})}}{1 + x^{2}} d x)}^{m} d t \\ = & \frac{1}{m} {(\frac{π}{4})}^{m} \sum_{k = 0}^{m} (- 1)^{k} (\binom{m}{k}) {(\frac{4}{π})}^{k} \int_{0}^{\infty} (R t^{2} - a) t^{2 a - 1} e^{- R t^{2}} {(\int_{0}^{1} \frac{e^{- t^{2} (1 + x^{2})}}{1 + x^{2}} d x)}^{k} d t \\ = & \frac{1}{m} \sum_{k = 0}^{m} (- 1)^{k} (\binom{m}{k}) {(\frac{π}{4})}^{m - k} \int_{0}^{\infty} (R t^{2} - a) t^{2 a - 1} e^{- R t^{2}} (\int_{0}^{1} \dots \int_{0}^{1} \frac{e^{- t^{2} (k + x_{1}^{2} + \dots + x_{k}^{2})}}{(1 + x_{1}^{2}) \dots (1 + x_{k}^{2})} d x_{1} \dots d x_{k}) d t \\ = & \frac{1}{m} \sum_{k = 0}^{m} (- 1)^{k} (\binom{m}{k}) {(\frac{π}{4})}^{m - k} \int_{0}^{1} \dots \int_{0}^{1} \int_{0}^{\infty} (R t^{2} - a) t^{2 a - 1} e^{- t^{2} (R + k + x_{1}^{2} + \dots + x_{k}^{2})} d t \frac{d x_{1} \dots d x_{k}}{(1 + x_{1}^{2}) \dots (1 + x_{k}^{2})} \\ = & \frac{Γ (1 + a)}{2 m} \sum_{k = 0}^{m} (- 1)^{k} (\binom{m}{k}) {(\frac{π}{4})}^{m - k} \int_{0}^{1} \dots \int_{0}^{1} (\frac{R}{R + k + x_{1}^{2} + \dots + x_{k}^{2}} - 1) \frac{d x_{1} \dots d x_{k}}{(1 + x_{1}^{2}) \dots (1 + x_{k}^{2}) {(R + k + x_{1}^{2} + \dots + x_{k}^{2})}^{a}} \\ = & \frac{Γ (1 + a)}{2 m} \sum_{k = 1}^{m} (- 1)^{k - 1} (\binom{m}{k}) {(\frac{π}{4})}^{m - k} \int_{0}^{1} \dots \int_{0}^{1} (1 - \frac{R}{R + k + x_{1}^{2} + \dots + x_{k}^{2}}) \frac{d x_{1} \dots d x_{k}}{(1 + x_{1}^{2}) \dots (1 + x_{k}^{2}) {(R + k + x_{1}^{2} + \dots + x_{k}^{2})}^{a}} \end{aligned}

結果的には実数の範囲では

a \notin - N, R > - 1

で成り立つか。

\begin{array}{r} \int_{0}^{1} (\frac{1}{1 + x^{2}} + \frac{1}{2 + x^{2}}) \frac{\tan^{- 1} \frac{1}{\sqrt{3 + x^{2}}}}{\sqrt{3 + x^{2}}} d x = \frac{π \tan^{- 1} \frac{1}{\sqrt{2}}}{4 \sqrt{2}} \end{array}

tips

\begin{array}{r} \int_{0}^{\infty} e^{- 2 t^{2}} e r f (t) (1 - e r f (t)^{2}) d t \end{array}

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

\begin{array}{r} \int_{0}^{1} \frac{e^{- x^{2} (1 + t^{2})}}{1 + t^{2}} d t = \frac{π}{4} (1 - e r f (x)^{2}) \dots A \end{array}

\begin{array}{r} \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}}} d x = \frac{2 (2 - \sqrt{3})}{3} β (2) \end{array}

tips

\begin{aligned} \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}}} d x & = \frac{3 (1 + \sqrt{3})}{2} \int_{0}^{\frac{π}{24}} \frac{\tan 2 θ \ln \frac{1}{\tan θ}}{\sqrt{\cos 4 θ - \frac{\sqrt{3}}{2}}} d θ \\ = \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}}} d x \\ = \frac{3 (1 + \sqrt{3})}{8} \int_{0}^{\frac{1}{2} (1 - \frac{\sqrt{3}}{2})} \frac{\tanh^{- 1} \sqrt{1 - x}}{(1 - x) \sqrt{1 - \frac{\sqrt{3}}{2} - 2 x}} d x \\ = \frac{3 (1 + \sqrt{3})}{4} \int_{0}^{\frac{- 1 + \sqrt{3}}{2}} \frac{\tanh^{- 1} \sqrt{\frac{1}{2} ({(\frac{1 + \sqrt{3}}{2})}^{2} + x^{2})}}{{(\frac{1 + \sqrt{3}}{2})}^{2} + x^{2}} d x \\ = \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}} d x \\ = \frac{3}{2} \int_{0}^{\frac{π}{12}} \tanh^{- 1} \frac{\cos \frac{π}{12}}{\cos θ} d θ \\ = \frac{π}{8} \int_{0}^{1} \tanh^{- 1} \frac{\cos \frac{π}{12}}{\cos \frac{π v}{12}} d v \end{aligned}

f (t) = \tanh^{- 1} \frac{\cos t}{\cos t v}

とすれば，

f^{'} (t) = - \frac{1}{2} (\frac{1 + v}{\sin (1 + v) t} + \frac{1 - v}{\sin (1 - v) t})

となるので

\begin{aligned} \int_{0}^{1} f^{'} (t) d v & = - \frac{1}{2} \int_{0}^{1} (\frac{1 + v}{\sin (1 + v) t} + \frac{1 - v}{\sin (1 - v) t}) d v \\ = - \frac{1}{2} \int_{0}^{2} \frac{v}{\sin t v} d v \end{aligned}

また，

f (\frac{π}{2}) = 0

なので

\begin{aligned} \frac{π}{8} \int_{0}^{1} \tanh^{- 1} \frac{\cos \frac{π}{12}}{\cos \frac{π v}{12}} d v & = \frac{π}{16} \int_{\frac{π}{12}}^{\frac{π}{2}} \int_{0}^{2} \frac{v}{\sin t v} d v d t \\ = \frac{π}{16} \int_{0}^{2} \int_{\frac{π}{12}}^{\frac{π}{2}} \frac{v}{\sin t v} d t d v \\ = \frac{π}{16} \int_{0}^{2} \ln \frac{\tan \frac{π v}{4}}{\tan \frac{π v}{24}} d v \\ = \frac{π}{8} \int_{0}^{1} \ln \frac{\tan \frac{π v}{2}}{\tan \frac{π v}{12}} d v \\ = \frac{π}{8} \sum_{n = 0}^{\infty} \frac{2}{2 n + 1} \int_{0}^{1} (\cos \frac{(2 n + 1) π v}{6} - \cos (2 n + 1) π v) d v \\ = \frac{3}{2} \sum_{n = 0}^{\infty} \frac{\sin \frac{(2 n + 1) π}{6}}{(2 n + 1)^{2}} \\ = β (2) \end{aligned}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

\begin{array}{r}  \end{array}

投稿日：2023年3月30日

OptHub AI Competition

この記事を高評価した人

高評価したユーザはいません

この記事に送られたバッジ

バッジはありません。

バッチを贈って投稿者を応援しよう

バッチを贈ると投稿者に現金やAmazonのギフトカードが還元されます。

投稿者

＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿

291

18792

コメント

他の人のコメント

コメントはありません。

読み込み中

＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿＿