積分問題集

3804

関数についての補足は青いところをタップしてもらえばwikipediaに飛べます.

初級

\begin{array}{r} (1) \int_{- π / 2}^{π / 2} \frac{\cos^{2} x}{1 + e^{\sin x}} d x = \frac{π}{4} \end{array}

\begin{array}{r} (2) \int_{0}^{π} \arccos \cos^{2 n + 1} x d x = \frac{π^{2}}{2} \end{array}

\begin{array}{r} (3) \int \prod_{k = 1}^{n} a_{k} (x) d x = a_{n} (x) + C (a_{n + 1} (x) = e^{a_{n} (x)}, a_{1} (x) = e^{x}) \end{array}

\begin{array}{rcl} (4) \int_{0}^{π} \frac{x}{a + \cos^{2} x} d x & = & \frac{π^{2}}{2 \sqrt{a (1 + a)}} (a > 0) \end{array}

\begin{array}{rcl} (5) \int_{0}^{\infty} \frac{1}{(1 + x^{2}) (9 + x^{2}) (25 + x^{2})} d x & = & \frac{π}{640} \end{array}

\begin{array}{rcl} (6) \int_{0}^{2 π} \frac{d θ}{\cosh φ + \cos θ} & = & \frac{2 π}{\sinh φ} (φ > 0) \end{array}

\begin{array}{rcl} (7) \int_{0}^{1} \frac{x}{x^{5} + x^{4} + x^{3} + x^{2} + x + 1} d x & = & \frac{π}{12 \sqrt{3}} + \frac{1}{4} \ln 3 - \frac{1}{3} \ln 2 \end{array}

\begin{array}{rcl} (8) \int_{0}^{1} \frac{\ln^{s - 1} \frac{1}{x}}{1 - x} d x & = & Γ (s) ζ (s) (Re s > 1) \end{array}

\begin{array}{rcl} (9) \int_{0}^{1} \frac{x}{\sqrt{1 - x^{2}}} \ln \frac{1 + z x}{1 - z x} d x & = & π \frac{1 - \sqrt{1 - z^{2}}}{z} (| z | < 1) \end{array}

\begin{array}{rcl} (10) \int_{0}^{\infty} \frac{\sin x}{x} e^{- a x} d x & = & \arctan \frac{1}{a} (a \geq 0) \end{array}

${a}^{n} = {a, a, \dots, a}$ (n個)とする

\begin{array}{r} (1) \int_{0}^{π / 2} \tan x \arctan \frac{a \sin 2 x}{1 - a \cos 2 x} d x = \frac{π}{2} \ln (1 + a) (| a | < 1) \end{array}

\begin{array}{r} (2) \int_{0}^{\infty} \frac{\ln (a^{2} + x^{2})}{1 + x^{2}} d x = π \ln (1 + a) (a \geq 0) \end{array}

\begin{array}{r} (3) \int_{0}^{\infty} \frac{\sin a x \sinh x}{\cosh^{2} x} d x = \frac{π a}{2 \cosh \frac{π a}{2}} \end{array}

\begin{array}{r} (4) \int_{0}^{1} \sin π x \ln Γ (x) d x = \frac{1}{π} (\ln \frac{π}{2} + 1) \end{array}

\begin{array}{r} (5) \int_{0}^{1} \frac{1 + x^{2}}{\sqrt{1 + x^{4}} + x^{2}} \ln x d x = - \frac{Γ (\frac{1}{4})^{2}}{18 π} \end{array}

\begin{array}{r} (6) \int_{0}^{\infty} \frac{\tanh x}{x} e^{- 4 a x} d x = \ln \frac{2^{4 a - 2} Γ (a + 1) Γ (a)^{3}}{Γ (2 a)^{2} π} (a > 0) \end{array}

\begin{array}{r} (7) \int_{0}^{\infty} \frac{d x}{(1 + z \cosh^{2} x)^{s}} = \frac{2^{2 s - 2}}{(1 + z)^{s}} \frac{Γ (s)^{2}}{Γ (2 s)}_{2} F_{1} [\begin{matrix} s, \frac{1}{2} \\ s + \frac{1}{2} \end{matrix}; \frac{1}{1 + z}] \end{array}

\begin{array}{r} (8) \int_{- \infty}^{\infty} \frac{\sin x}{\prod_{k = 0}^{n} (x + k)} d x = \frac{2^{n}}{n!} π \end{array}

\begin{array}{r} (9) \int_{0}^{\infty} \frac{\sin x}{x} \prod_{k = 1}^{n} \frac{\sin a_{k} x}{a_{k} x} d x = \frac{π}{2} (a_{m} > 0, \sum_{k = 1}^{n} a_{k} < 1) \end{array}

\begin{array}{r} (10) \int_{0}^{1} \frac{\ln^{n} x \ln^{m} (1 - x)}{x (1 - x)} d x = n! m! (- 1)^{n + m} (ζ ({1}^{m}, n + 1) + ζ ({1}^{m - 1}, n + 2)) \end{array}

\begin{array}{r} (1) \int_{0}^{1} \frac{\ln x \ln (1 + x)}{x (1 + x)} d x = - \frac{5}{8} ζ (3) \end{array}

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

\begin{array}{r} (3) \int_{0}^{1} \frac{\arctan^{3} x}{x^{3}} d x = \frac{3}{2} β (2) + \frac{3}{8} π \ln 2 - \frac{3}{32} π^{2} - \frac{π^{3}}{64} \end{array}

\begin{array}{r} (4) \int_{0}^{π / 2} \frac{x^{4}}{\tan x} d x = \frac{π^{2}}{16} \ln 2 + \frac{93}{32} ζ (5) - \frac{9}{16} π^{2} ζ (3) \end{array}

\begin{array}{r} (5) \int_{0}^{\infty} \frac{\cos α x}{\cosh x + \cos β π} d x = \frac{π \sinh α β x}{\sinh α π \sin β π} \end{array}

\begin{array}{r} (6) \int_{0}^{π} \sin^{2 n + 1} x e^{- t x} d x = \frac{(2 n + 1)!}{\prod_{k = 0}^{n} (t^{2} + (2 k + 1)^{2})} (1 + e^{- π t}) \end{array}

\begin{array}{r} (7) \int_{0}^{\infty} \frac{x^{s + 1}}{x^{2} + 1} \frac{\sin (x - s \arctan \frac{1}{a x})}{(1 + a^{2} x^{2})^{s / 2}} d x = \frac{π}{2 e (a + 1)^{s}} (a, s > 0) \end{array}

\begin{array}{r} (8) \int_{0}^{\infty} {(\frac{\sin x}{x} \arcsin a \sin x)}^{2} d x = \frac{π}{4} {Li}_{2} (a^{2}) (| a | < 1) \end{array}

\begin{array}{r} (9) \int_{0}^{\infty} x \ln (1 + \frac{1}{\cosh x}) d x = \frac{21}{16} ζ (3) \end{array}

\begin{array}{r} (10) \int_{0}^{1} \frac{\ln (1 - x)}{1 + s^{2} x^{2}} d x = \frac{\arctan s \ln (1 + s^{2})}{2 s} - \sum_{n \geq 0} \frac{(- 1)^{n}}{(2 n + 1)^{2}} s^{2 n} (| s | < 1) \end{array}

$\begin{array}{r} ζ_{a} (k) := \sum_{n \geq 0} \frac{1}{(n + a)^{k}} \end{array}$
のようにフルヴィッツのゼータ関数とする.
wikiとは少し違うが多重ゼータ値と見分けるためにこのように定義した.

$\begin{array}{r} K (k) := \int_{0}^{π / 2} \frac{d x}{\sqrt{1 - k^{2} \sin^{2} x}} \end{array}$
楕円積分とする.

\begin{array}{r} (1) \int_{0}^{1} \frac{K (t x)}{\sqrt{1 - x^{2}}} d x = K {(\sqrt{\frac{1 - \sqrt{1 - t^{2}}}{2}})}^{2} \end{array}

\begin{array}{r} (2) \int_{0}^{π / 2} \frac{\ln (t \cos x)}{\ln^{2} (t \cos x) + x^{2}} d x = \frac{π}{2 \ln \frac{t}{2}} (0 < t < 1) \end{array}

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

\begin{array}{r} (4) \int_{0}^{\infty} \frac{\cos (s \tan^{- 1} x)}{(1 + x^{2})^{s / 2}} \frac{d x}{\cosh π x + \cos 2 π t} = \frac{1}{2^{s} \sin 2 π t} (ζ_{1 - t} (s) - ζ_{1 + t} (s)) \end{array}

(5)✕

\begin{array}{r} (6) \int_{0}^{π} \frac{x^{2}}{\ln^{2} (t \sin x) + x^{2}} d x = \frac{2 π^{2}}{4 \ln^{2} \frac{t}{2} + π^{2}} (0 < t < 1) \end{array}

\begin{array}{r} (7) \int_{- \infty}^{\infty} \frac{1 + s \cosh π x}{1 + 2 s \cosh π x + s^{2}} \frac{d x}{x^{2} + 1} = π (\frac{1}{1 - s} + \frac{1}{\ln s}) \end{array}

\begin{array}{r} (8) \int_{0}^{\infty} \arctan^{2} \frac{2 a x}{1 - a + (1 + a) x^{2}} \frac{d x}{1 + x^{2}} = \frac{π}{4} {Li}_{2} (a) \end{array}

\begin{array}{r} (9) \int_{- \infty}^{\infty} \frac{\sin (s \tan^{- 1} \frac{x}{z})}{(z^{2} + x^{2})^{s / 2}} \frac{d x}{e^{2 π x} - 1} d x = - \frac{1}{2 z^{s}} + \sum_{n \geq 0} \frac{1}{(n + z)^{s}} \end{array}

\begin{array}{r} (10) \int_{0}^{1} x K (x)^{2} d x = \frac{7}{4} ζ (3) \end{array}

投稿日：2023年12月23日

更新日：2024年1月28日

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