جدول تكاملات محددة

analysis

Definite Integrals' Table

تكاملات تحوي الدوال المثلثية :

$\int\limits_0^{\pi /2} {\sin ^2 } xdx = \int\limits_0^{\pi /2} {\cos ^2 } dx = \frac{\pi }{4}$

$\beta (p,q) = 2\int\limits_0^{\pi /2} {\sin ^{2p - 1} } x\cos ^{2q - 1} xdx = \frac{{\Gamma (p)\Gamma (q)}}{{\Gamma (p + q)}}$

$\int\limits_0^\infty {} \frac{{1 - \cos px}}{{x^2 }}dx = \frac{{\pi p}}{2}$

$\int\limits_0^\infty {} \frac{{\cos px - \cos qx}}{x}dx = \ln \frac{q}{p}$

$\int\limits_0^\infty {} \frac{{\cos px - \cos qx}}{{x^2 }}dx = \frac{{\pi (q - p)}}{2}$

$\int\limits_0^\infty {} \frac{{\cos mx}}{{x^2 + a^2 }}dx = \frac{\pi }{{2a}}e^{ - ma}$

$\int\limits_0^\infty {} \frac{{x\sin mx}}{{x^2 + a^2 }}dx = \frac{\pi }{2}e^{ - ma}$

$\int\limits_0^\infty {} \frac{{\sin mx}}{{x(x^2 + a^2 )}}dx = \frac{\pi }{{2a^2 }}(1 - e^{ - ma} )$

$\int\limits_0^{2\pi } {} \frac{{dx}}{{a + b\sin x}} = \frac{{2\pi }}{{\sqrt {a^2 - b^2 } }}$

$\int\limits_0^{2\pi } {} \frac{{dx}}{{a + b\cos x}} = \frac{{2\pi }}{{\sqrt {a^2 - b^2 } }}$

$\int\limits_0^{\pi /2} {} \frac{{dx}}{{a + b\cos x}} = \frac{{cos^{ - 1} (b/a)}}{{\sqrt {a^2 - b^2 } }}$

$\int\limits_0^{2\pi } {\frac{{dx}}{{(a + b\sin x)^2 }}} = \int\limits_0^{2\pi } {\frac{{dx}}{{(a + b\cos x)^2 }}} = \frac{{2\pi a}}{{(a^2 - b^2 )^{3/2} }}$

$\int\limits_0^{2\pi } {} \frac{{dx}}{{1 - 2a\cos x + a^2 }} = \frac{{2\pi }}{{1 - a^2 }},0 < a < 1$

$\int\limits_0^\pi {} \frac{{\cos mxdx}}{{1 - 2a\cos x + a^2 }} = \frac{{\pi a^m }}{{1 - a^2 }},a^2 < 1$

$\int\limits_0^\infty {\sin } \,ax^2 dx = \int\limits_0^\infty {\cos } \,ax^2 dx = \frac{1}{2}\sqrt {\frac{\pi }{{2a}}}$

$\int_0^\infty\frac{\sin(x)}{x}\,dx=\frac{\pi}{2}$

$\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot n}\frac{\pi}{2}$

إذا كان n عدداً زوجياً أكبر من الصفر

$\int_0^\frac{\pi}{2}\sin^n{x}\,dx=\int_0^\frac{\pi}{2}\cos^n{x}\,dx=\frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (n-1)}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot n}$

إذا كان n عدداً فردياً أكبر من الواحد

$\int_0^\infty\frac{\sin^2{x}}{x^2}\,dx=\frac{\pi}{2}$

$\int\limits_0^\infty {\sin } \,ax^n dx = \frac{1}{{na^{1/n} }}\Gamma (1/n)\sin \frac{\pi }{{2n}},n > 1$

$\int\limits_0^\infty {\cos \,} ax^n dx = \frac{1}{{na^{1/n} }}\Gamma (1/n)\cos \frac{\pi }{{2n}},n > 1$

$\int\limits_0^\infty {\frac{{\sin x}}{{\sqrt x }}} dx = \int\limits_0^\infty {\frac{{\cos x}}{{\sqrt x }}} dx = \sqrt {\frac{\pi }{2}}$

$\int\limits_0^\infty {} \frac{{\sin x}}{{x^p }}dx = \frac{\pi }{{2\Gamma (p)\sin (p\pi /2)}},0 < p < 1$

$\int\limits_0^\infty {} \frac{{\cos x}}{{x^p }}dx = \frac{\pi }{{2\Gamma (p)\cos (p\pi /2)}},0 < p < 1$

$\int\limits_0^\infty {\sin } \,ax^2 \cos 2bxdx = \frac{1}{2}\sqrt {\frac{\pi }{{2a}}} \left( {\cos \frac{{b^2 }}{a} - \sin \frac{{b^2 }}{a}} \right)$

$\int\limits_0^\infty {\cos \,} ax^2 \cos 2bxdx = \frac{1}{2}\sqrt {\frac{\pi }{{2a}}} \left( {\cos \frac{{b^2 }}{a} + \sin \frac{{b^2 }}{a}} \right)$

$\int\limits_0^\infty {} \frac{{\sin ^3 x}}{{x^3 }}dx = \frac{{3\pi }}{8}$

$\int\limits_0^\infty {} \frac{{\sin ^4 x}}{{x^4 }}dx = \frac{\pi }{3}$

$\int\limits_0^\infty {} \frac{{\tan x}}{x}dx = \frac{\pi }{2}$

$\int\limits_0^{\pi /2} {} \frac{{dx}}{{1 + \tan ^r x}} = \frac{\pi }{4}$

$\int\limits_0^1 {} \frac{{\sin ^{ - 1} x}}{x}dx = \frac{\pi }{2}\ln 2$

$\int\limits_0^1 {} \frac{{1 - \cos x}}{x}dx - \int\limits_1^\infty {\frac{{\cos x}}{x}dx} = \gamma \approx 0.5772156649\,\,({\rm{Euler's Constant)}}$

$\int\limits_0^\infty {\left( {\frac{1}{{1 + x^2 }} - \cos x} \right)} \frac{{dx}}{x} = \gamma \approx 0.5772156649\,\,({\rm{Euler's Constant)}}$

$\int\limits_0^\infty {} \frac{{\tan ^{ - 1} px - \tan ^{ - 1} qx}}{x}dx = \frac{\pi }{2}\ln \frac{p}{q}$

تكاملات تحوي الدالة الأسية :

$\int_0^\infty{\sqrt{x}\,e^{-x}\,dx} = \frac{1}{2}\sqrt \pi$

$\int\limits_0^\infty {e^{ - ax^2 } } dx = \frac{1}{2}\sqrt {\frac{\pi }{a}}$

$\int_0^\infty{\frac{x^3}{e^x-1}\,dx} = \frac{\pi^4}{15}$

$\int_0^\infty{\frac{x}{e^x-1}\,dx} = \frac{\pi^2}{6}$

$\int\limits_0^\infty {} \frac{{x^{n - 1} }}{{e^x - 1}}dx = \Gamma (n + 1)\left( {1 + \frac{1}{{2^n }} + \frac{1}{{3^n }} + \cdot\cdot\cdot} \right)$

$\int\limits_0^\infty {} \frac{{xdx}}{{e^x + 1}} = 1 - \frac{1}{{2^2 }} + \frac{1}{{3^2 }} - \frac{1}{{4^2 }} + \cdot\cdot\cdot = \frac{{\pi ^2 }}{{12}}$

$\int\limits_0^\infty {} \frac{{x^{n - 1} }}{{e^x + 1}}dx = \Gamma (n + 1)\left( {1 - \frac{1}{{2^n }} + \frac{1}{{3^n }} - \cdot\cdot\cdot} \right)$

$\int\limits_0^\infty {e^{ - ax} } \sin bxdx = \frac{b}{{a^2 + b^2 }}$

$\int\limits_0^\infty {e^{ - ax} } \cos bxdx = \frac{a}{{a^2 + b^2 }}$

$\int\limits_0^\infty {} \frac{{e^{ - ax} \sin bx}}{x}dx = \tan ^{ - 1} \frac{b}{a}$

$\int\limits_0^\infty {} \frac{{e^{ - ax} - e^{ - bx} }}{x}dx = \ln \frac{b}{a}$

$\int\limits_0^\infty {e^{ - ax^2 } } \cos bxdx = \frac{1}{2}\sqrt {\frac{\pi }{a}} e^{ - b^2 /4a}$

$\int\limits_0^\infty {e^{ - (ax^2 + bx + c)} } dx = \frac{1}{2}\sqrt {\frac{\pi }{a}} e^{(b^2 - 4ac)/4a}$

$\int\limits_{ - \infty }^\infty {e^{ - (ax^2 + bx + c)} } dx = \sqrt {\frac{\pi }{a}} e^{(b^2 - 4ac)/4a}$

$\int\limits_0^\infty {e^{ - (ax^2 + b/x^2 )} } dx = \frac{1}{2}\sqrt {\frac{\pi }{a}} e^{ - 2\sqrt {ab} }$

$\int\limits_0^\infty {} \frac{{\sin mx}}{{e^{2\pi x} - 1}}dx = \frac{1}{4}\coth \frac{m}{2} - \frac{1}{{2m}}$

$\int\limits_0^\infty {\left( {\frac{1}{{1 + x}} - e^{ - x} } \right)} \frac{{dx}}{x} = \gamma$

$\int\limits_0^\infty {} \frac{{e^{ - x^2 } - e^{ - x} }}{x}dx = \frac{1}{2}\gamma$

$\int\limits_0^\infty {\left( {\frac{1}{{e^x - 1}} - \frac{{e^{ - x} }}{x}} \right)} dx = \gamma$

$\int\limits_0^\infty {} \frac{{e^{ - ax} - e^{ - bx} }}{{x\sec px}}dx = \frac{1}{2}\ln \left( {\frac{{b^2 + p^2 }}{{a^2 + p^2 }}} \right)$

$\int\limits_0^\infty {} \frac{{e^{ - ax} - e^{ - bx} }}{{x\csc px}}dx = \tan ^{ - 1} \frac{b}{p} - \tan ^{ - 1} \frac{a}{p}$

$\int\limits_0^\infty {} \frac{{e^{ - ax} (1 - \cos x)}}{{x^2 }}dx = \cot ^{ - 1} a - \frac{a}{2}\ln (a^2 + 1)$

$\int\limits_0^\infty {x^n } e^{ - ax} dx = \frac{{\Gamma (n + 1)}}{{a^n + 1}}$

$\int\limits_0^\infty {x^m } e^{ - ax^2 } dx = \frac{{\Gamma [(m + 1)/2]}}{{2a^{(m + 1)/2} }}$

$\int_{-\infty}^\infty e^{-(ax^2+bx+c)}\,dx=\sqrt{\frac{\pi}{a}}\exp\left[\frac{b^2-4ac}{4a}\right ]$

$\int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x)$

(حيث $I_{0}(x)$ هي دالة بسل المعدلة من النوع الأول )

$\int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left(\sqrt{x^2 + y^2}\right)$

تكاملات تحوي لوغاريثمات :

$\int\limits_0^1 {x^m } (\ln x)^n dx = \frac{{( - 1)^n n!}}{{(m + 1)^{n + 1} }}\,,\,m > - 1,n = 0,1,2,...$

$\int\limits_0^1 {} \frac{{\ln x}}{{1 + x}}dx = - \frac{{\pi ^2 }}{{12}}$

$\int\limits_0^1 {} \frac{{\ln x}}{{1 - x}}dx = - \frac{{\pi ^2 }}{6}$

$\int\limits_0^1 {} \frac{{\ln (1 + x)}}{x}dx = \frac{{\pi ^2 }}{{12}}$

$\int\limits_0^1 {} \frac{{\ln (1 - x)}}{x}dx = - \frac{{\pi ^2 }}{6}$

$\int\limits_0^\infty {} \frac{{\ln (1 + x^p )}}{{x^q }}dx = \frac{\pi }{{q - 1}}\csc \left[ {(q - 1)\frac{\pi }{p}} \right]\,\,,\,p > 0\,\,\,q > 1$

$\int\limits_0^1 {\ln \,} x\ln (1 + x)dx = 2 - 2\ln 2 - \frac{{\pi ^2 }}{{12}}$

$\int\limits_0^1 {\ln \,} x\ln (1 - x)dx = 2 - \frac{{\pi ^2 }}{6}$

$\int\limits_0^\infty {} \frac{{x^{p - 1} \ln x}}{{1 + x}}dx = - \pi ^2 \csc p\pi \cot p\pi \,\,,\,\,0 < p < 1$

$\int\limits_0^1 {} \frac{{x^m - x^n }}{{\ln x}}dx = \ln \frac{{m + 1}}{{n + 1}}$

$\int\limits_0^\infty {e^{ - x} } \ln xdx = - \gamma$

$\int\limits_0^\infty {e^{ - x^2 } } \ln x\,dx = - \frac{{\sqrt \pi }}{4}(\gamma + 2\ln 2)$

$\int\limits_0^\infty {\ln } \left( {\frac{{e^x + 1}}{{e^x - 1}}} \right)dx = \frac{{\pi ^2 }}{4}$

$\int\limits_0^{\pi /2} {\ln } \sin xdx = \int\limits_0^{\pi /2} {\ln } \cos xdx = - \frac{\pi }{2}\ln 2$

$\int\limits_0^{\pi /2} ( \ln \sin x)^2 dx = \int\limits_0^{\pi /2} ( \ln \cos x)^2 dx = \frac{\pi }{2}(\ln 2)^2 + \frac{{\pi ^3 }}{{24}}$

$\int\limits_0^\pi x \ln \sin xdx = - \frac{{\pi ^2 }}{2}\ln 2$

$\int\limits_0^{\pi /2} {\sin } \,x\ln \sin xdx = \ln 2 - 1$

$\int\limits_0^{2\pi } {\ln } (a + b\sin x)dx = \int\limits_0^{2\pi } {\ln } (a + b\cos x)dx = 2\pi \ln (a + \sqrt {a^2 - b^2 } )$

$\int\limits_0^\pi {\ln } (a + b\cos x)dx = \pi \ln \left( {\frac{{a + \sqrt {a^2 - b^2 } }}{2}} \right)$

$\int\limits_0^\pi {\ln } (a^2 - 2ab\cos x + b^2 )dx = \left\{ {\begin{array}{*{20}c} {2\pi \ln a} & {a \ge b > 0} \\ {2\pi \ln b,} & {b \ge a > 0} \\ \end{array}} \right.$

$\int\limits_0^{\pi /4} {\ln } (1 + \tan x)dx = \frac{\pi }{8}\ln 2$

$\int\limits_0^{\pi /2} {\sec } x\ln \left( {\frac{{1 + b\cos x}}{{1 + a\cos x}}} \right)dx = \frac{1}{2}[(\cos ^{ - 1} a)^2 - (\cos ^{ - 1} b)^2 ]$

$\int\limits_0^a {\ln } \left( {2\sin \frac{x}{2}} \right)dx = - \sum\limits_{n = 1}^\infty {\frac{{\sin na}}{{n^2 }}}$

تكاملات تحوي الدوال الزائدية

$\int\limits_0^\infty {} \frac{{\sin ax}}{{\sinh bx}}dx = \frac{\pi }{{2b}}\tanh \frac{{a\pi }}{{2b}}$

$\int\limits_0^\infty {} \frac{{\cos ax}}{{\cosh bx}}dx = \frac{\pi }{{2b}}\frac{1}{{\cosh (a\pi /2b)}}$

$\int\limits_0^\infty {} \frac{{xdx}}{{\sinh ax}} = \frac{{\pi ^2 }}{{4a^2 }}$

$\int\limits_0^\infty {} \frac{{x^n dx}}{{\sinh ax}} = \frac{{2^{n + 1} - 1}}{{2^n a^{n + 1} }}\Gamma (n + 1)\left( {1 + \frac{1}{{2^{n + 1} }} + \frac{1}{{3^{n + 1} }} + \cdot\cdot\cdot} \right)$

$\int\limits_0^\infty {} \frac{{\sinh ax}}{{e^{bx} + 1}}dx = \frac{\pi }{{2b}}\csc \frac{{a\pi }}{b} - \frac{1}{{2a}}$

$\int\limits_0^\infty {} \frac{{\sinh ax}}{{e^{bx} - 1}}dx = \frac{1}{{2a}} - \frac{\pi }{{2b}}\cot \frac{{a\pi }}{b}$

تكاملات تحوي صيغاً نسبية ولانسبية

$\int\limits_0^\infty {\frac{{dx}}{{x^2 + a^2 }}} = \frac{\pi }{{2a}}$

$\int\limits_0^\infty {\frac{{x^{p - 1} dx}}{{1 + x}}} = \frac{\pi }{{\sin (p\pi )}}\,\,\,,\,\,\,0 < p < 1$

$\int\limits_0^\infty {\frac{{x^m }}{{x^n + a^n }}} = \frac{{\pi a^{m + 1 - n} }}{{n\sin [(m + 1)\pi /n]}}\,\,,\,\,0 < m + 1 < n$

$\int\limits_0^\infty {\frac{{x^m }}{{1 + 2x\cos (b) + x^2 }}} = \frac{\pi }{{\sin (m\pi )}}\frac{{\sin (mb)}}{{\sin (b)}}$

$\int\limits_0^a {\frac{{dx}}{{\sqrt {a^2 - x^2 } }}} = \frac{\pi }{2}$

$\int\limits_0^a {\sqrt {a^2 - x^2 } } dx = \frac{{\pi a^2 }}{4}$

$\int\limits_0^a {x^m } (a^n - x^n )^p dx = \frac{{a^{m + 1 + np} \Gamma [(m + 1)/n]\Gamma (p + 1)}}{{n\Gamma [(m + 1)/(n + p + 1)]}}$

$\int\limits_0^\infty {\frac{{x^m }}{{(a^n + x^n )^p }}} dx = \frac{{( - 1)^{p - 1} \pi a^{m + 1 - np} \Gamma \left[ {(m + 1)/n} \right]}}{{n\sin [(m + 1)\pi /n](p - 1)!\Gamma [(m + 1)/(n - p + 1)]}}\,\,,\,\,0 < m + 1 < np$