Визначені інтеграли без явних первісних
Деякі функції, чиї первісні не можуть бути представлені явно, тим не менш їхні деякі визначені інтеграли можуть бути обчислені.
∫ 0 ∞ x n − 1 e − x d x = ∫ 0 1 ( ln 1 x ) n − 1 d x = Γ ( n ) = ( n − 1 ) ! {\displaystyle \int _{0}^{\infty }{x^{n-1}\,e^{-x}\,dx}=\int _{0}^{1}{\left(\ln {\frac {1}{x}}\right)^{n-1}\,dx}=\Gamma (n)=(n-1)!} Γ ( x ) {\displaystyle \Gamma (x)\!} Гамма-функція )∫ 0 ∞ x e − x d x = Γ ( 3 2 ) = π 2 {\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}=\Gamma \left({\frac {3}{2}}\right)={\frac {\sqrt {\pi }}{2}}} ∫ 0 ∞ x n e − a 2 x 2 d x = Γ ( n + 1 2 ) 2 a n + 1 {\displaystyle \int _{0}^{\infty }{x^{n}e^{-a^{2}x^{2}}\,dx}={\frac {\Gamma ({\frac {n+1}{2}})}{2a^{n+1}}}} a > 0 {\displaystyle a>0\!} Гамма-функція )
∫ − ∞ ∞ ( 1 + x 2 / ν ) − ( ν + 1 ) / 2 d x = ν π Γ ( ν / 2 ) Γ ( ( ν + 1 ) / 2 ) {\displaystyle \int _{-\infty }^{\infty }{(1+x^{2}/\nu )^{-(\nu +1)/2}dx}={\frac {{\sqrt {\nu \pi }}\ \Gamma (\nu /2)}{\Gamma ((\nu +1)/2)}}\,} ν > 0 {\displaystyle \nu >0\,} густини ймовірності для T-розподілу Стьюдента
∫ 0 1 x m − 1 ( 1 − x ) n − 1 d x = Γ ( m ) Γ ( n ) Γ ( m + n ) = B ( m , n ) {\displaystyle \int _{0}^{1}{x^{m-1}\,(1-x)^{n-1}\,dx}={\frac {\Gamma (m)\,\Gamma (n)}{\Gamma (m+n)}}=B(m,n)} m , n > 0 {\displaystyle m,n\,>\,0\!} B ( m , n ) {\displaystyle B(m,n)\!} Бета-функція ∫ 0 a x m − 1 ( a − x ) n − 1 d x = a m + n − 1 Γ ( m ) Γ ( n ) Γ ( m + n ) = a m + n − 1 B ( m , n ) {\displaystyle \int _{0}^{a}{x^{m-1}\,(a-x)^{n-1}\,dx}=a^{m+n-1}\,{\frac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)}}=a^{m+n-1}\,B(m,n)} a , m , n > 0 {\displaystyle a,m,n\,>\,0\!} B ( m , n ) {\displaystyle B(m,n)\!} ∫ 0 1 x m − 1 + x n − 1 ( 1 + x ) m + n d x = Γ ( m ) Γ ( n ) Γ ( m + n ) = B ( m , n ) {\displaystyle \int _{0}^{1}{{\frac {x^{m-1}+x^{n-1}}{(1+x)^{m+n}}}\,dx}={\frac {\Gamma (m)\,\Gamma (n)}{\Gamma (m+n)}}=B(m,n)} m , n > 0 {\displaystyle m,n\,>\,0\!}
∫ 0 1 x m 1 − x 2 d x {\displaystyle \int _{0}^{1}{{\frac {x^{m}}{\sqrt {1-x^{2}}}}\,dx}} = π 2 Γ ( m + 1 2 ) Γ ( m + 2 2 ) {\displaystyle ={\frac {\sqrt {\pi }}{2}}\,{\frac {\Gamma \left({\frac {m+1}{2}}\right)}{\Gamma \left({\frac {m+2}{2}}\right)}}} m > − 1 {\displaystyle m\,>\,-1\!} m {\displaystyle m\!} = 2 ⋅ 4 ⋅ 6 ⋅ . . . ⋅ ( m − 1 ) 3 ⋅ 5 ⋅ 7 ⋅ . . . ⋅ m {\displaystyle ={\frac {2\cdot 4\cdot 6\cdot ...\cdot (m-1)}{3\cdot 5\cdot 7\cdot ...\cdot m}}} m > 1 {\displaystyle m\,>\,1\!} m {\displaystyle m\!} непарне число ;
= 1 ⋅ 3 ⋅ 5 ⋅ . . . ⋅ ( m − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ . . . ⋅ m π 2 {\displaystyle ={\frac {1\cdot 3\cdot 5\cdot ...\cdot (m-1)}{2\cdot 4\cdot 6\cdot ...\cdot m}}\,{\frac {\pi }{2}}} m > 0 {\displaystyle m\,>\,0\!} m {\displaystyle m\!} парне число ;
∫ 0 1 x m + 1 − x 2 d x = 1 m + 2 ∫ 0 1 x m 1 − x 2 d x {\displaystyle \int _{0}^{1}{x^{m}+{\sqrt {1-x^{2}}}\,dx}={\frac {1}{m+2}}\,\int _{0}^{1}{{\frac {x^{m}}{\sqrt {1-x^{2}}}}\,dx}} m {\displaystyle m\!} m > − 1 {\displaystyle m\,>\,-1\!}
∫ 0 1 d x 1 − x n = π n Γ ( 1 n ) Γ ( 1 n + 1 2 ) {\displaystyle \int _{0}^{1}{\frac {dx}{\sqrt {1-x^{n}}}}={\frac {\sqrt {\pi }}{n}}\,{\frac {\Gamma \left({\frac {1}{n}}\right)}{\Gamma \left({\frac {1}{n}}+{\frac {1}{2}}\right)}}} n > 0 {\displaystyle n\,>\,0\!} ∫ 0 1 x m d x 1 − x n = π n Γ ( m + 1 n ) Γ ( m + 1 n + 1 2 ) {\displaystyle \int _{0}^{1}{\frac {x^{m}\,dx}{\sqrt {1-x^{n}}}}={\frac {\sqrt {\pi }}{n}}\,{\frac {\Gamma \left({\frac {m+1}{n}}\right)}{\Gamma \left({\frac {m+1}{n}}+{\frac {1}{2}}\right)}}} m + 1 , n > 0 {\displaystyle m+1,n\,>\,0\!}
∫ 0 1 x m ( 1 − x 2 ) p d x = Γ ( p + 1 ) Γ ( m + 1 2 ) 2 Γ ( p + m + 3 2 ) {\displaystyle \int _{0}^{1}{x^{m}\,(1-x^{2})^{p}\,dx}={\frac {\Gamma (p+1)\Gamma \left({\frac {m+1}{2}}\right)}{2\Gamma \left(p+{\frac {m+3}{2}}\right)}}} p + 1 , m + 1 > 0 {\displaystyle p+1,m+1\,>\,0\!} ∫ 0 1 x m ( 1 − x n ) p d x = Γ ( p + 1 ) Γ ( m + 1 n ) n Γ ( p + 1 + m + 1 n ) {\displaystyle \int _{0}^{1}{x^{m}\,(1-x^{n})^{p}\,dx}={\frac {\Gamma (p+1)\Gamma \left({\frac {m+1}{n}}\right)}{n\Gamma \left(p+1+{\frac {m+1}{n}}\right)}}} p + 1 , m + 1 , n > 0 {\displaystyle p+1,m+1,n\,>\,0\!} Інші інтеграли
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∫ 0 1 x n d x 1 + x = ( − 1 ) n [ ln 2 − 1 + 1 2 − 1 3 + . . . + ( − 1 ) n n ] {\displaystyle \int _{0}^{1}{\frac {x^{n}\,dx}{1+x}}=(-1)^{n}\left[\ln {2}-1+{\frac {1}{2}}-{\frac {1}{3}}+...+{\frac {(-1)^{n}}{n}}\right]} n = 1 , 2 , 3 , . . . {\displaystyle n\,=\,1,2,3,...\!}
∫ 0 1 x n − 1 d x 1 + x m = 1 n − 1 n + m + 1 n + 2 m − 1 n + 3 m + . . . {\displaystyle \int _{0}^{1}{\frac {x^{n-1}\,dx}{1+x^{m}}}={\frac {1}{n}}-{\frac {1}{n+m}}+{\frac {1}{n+2m}}-{\frac {1}{n+3m}}+...} m , n > 0 {\displaystyle m,n\,>\,0\!}
∫ 0 1 d x 1 + x + x 2 = π 3 3 {\displaystyle \int _{0}^{1}{\frac {dx}{1+x+x^{2}}}={\frac {\pi }{3{\sqrt {3}}}}}
∫ 0 1 d x 1 − x + x 2 = 2 π 3 3 {\displaystyle \int _{0}^{1}{\frac {dx}{1-x+x^{2}}}={\frac {2\pi }{3{\sqrt {3}}}}}
∫ 0 1 x p − 1 d x ( 1 − x ) p = π sin p π {\displaystyle \int _{0}^{1}{\frac {x^{p-1}\,dx}{(1-x)^{p}}}={\frac {\pi }{\sin {p\pi }}}} 0 < p < 1 {\displaystyle 0\,<\,p\,<\,1\!}
∫ 0 1 x p + x − p 1 + x 2 d x = π 2 cos p π 2 {\displaystyle \int _{0}^{1}{{\frac {x^{p}+x^{-p}}{1+x^{2}}}\,dx}={\frac {\pi }{2\cos {\frac {p\pi }{2}}}}} − 1 < p < 1 {\displaystyle -1\,<\,p\,<\,1\!}
∫ 0 ∞ e − x 2 d x = 1 2 π {\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}} Гаусовий інтеграл )∫ 0 ∞ x e − a 2 x 2 d x = 1 2 a 2 {\displaystyle \int _{0}^{\infty }{xe^{-a^{2}x^{2}}\,dx}={\frac {1}{2a^{2}}}} ∫ 0 ∞ x 2 e − a 2 x 2 d x = π 4 a 3 {\displaystyle \int _{0}^{\infty }{x^{2}e^{-a^{2}x^{2}}\,dx}={\frac {\sqrt {\pi }}{4a^{3}}}} a > 0 {\displaystyle a>0\!} ∫ 0 ∞ x 2 i + 1 e − a 2 x 2 d x = a ! 2 a 2 i + 2 {\displaystyle \int _{0}^{\infty }{x^{2i+1}e^{-a^{2}x^{2}}\,dx}={\frac {a!}{2a^{2i+2}}}} a > 0 ; i = 1 , 2 , 3... {\displaystyle a>0;\;\;\!i=1,2,3...} ∫ 0 ∞ x 2 i + 1 e − a 2 x 2 d x = 1 ⋅ 3 ⋅ 5 ⋅ . . . ⋅ ( 2 i − 1 ) 2 i + 1 a 2 i + 1 π {\displaystyle \int _{0}^{\infty }{x^{2i+1}e^{-a^{2}x^{2}}\,dx}={\frac {1\cdot 3\cdot 5\cdot ...\cdot (2i-1)}{2^{i+1}a^{2i+1}}}{\sqrt {\pi }}} a > 0 ; i = 1 , 2 , 3... {\displaystyle a>0;\;\;\!i=1,2,3...} ∫ 0 ∞ x e x − 1 d x = π 2 6 {\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}} числа Бернуллі )∫ 0 ∞ x 3 e x − 1 d x = π 4 15 {\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}={\frac {\pi ^{4}}{15}}} ∫ 0 ∞ 1 e a x + 1 d x = ln 2 a {\displaystyle \int _{0}^{\infty }{{\frac {1}{e^{ax}+1}}\,dx}={\frac {\ln {2}}{a}}} a > 0 {\displaystyle a>0\!} ∫ 0 ∞ x e a x + 1 d x = π 2 12 a 2 {\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{ax}+1}}\,dx}={\frac {\pi ^{2}}{12a^{2}}}} a > 0 {\displaystyle a>0\!} ∫ 0 ∞ x 3 e a x + 1 d x = 7 120 π 4 a 4 {\displaystyle \int _{0}^{\infty }{{\frac {x^{3}}{e^{ax}+1}}\,dx}={\frac {7}{120}}{\frac {\pi ^{4}}{a^{4}}}} a > 0 {\displaystyle a>0\!} ∫ 0 ∞ sin ( x ) x d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}} ∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ ( n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ n π 2 {\displaystyle \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}}} n парне число і n ≥ 2 {\displaystyle \scriptstyle {n\geq 2}} ∫ 0 π 2 sin n x d x = ∫ 0 π 2 cos n x d x = 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ ( n − 1 ) 3 ⋅ 5 ⋅ 7 ⋅ ⋯ ⋅ n {\displaystyle \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}}} n {\displaystyle \scriptstyle {n}} n ≥ 3 {\displaystyle \scriptstyle {n\geq 3}} ∫ − π π cos ( α x ) cos n ( β x ) d x = { 2 π 2 n ( n m ) | α | = | β ( 2 m − n ) | 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\cos ^{n}(\beta x)dx=\left\{{\begin{array}{cc}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&|\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.} α , β , m , n {\displaystyle \scriptstyle \alpha ,\beta ,m,n} β ≠ 0 {\displaystyle \scriptstyle \beta \neq 0} m , n ≥ 0 {\displaystyle \scriptstyle m,n\geq 0} Біноміальний коефіцієнт )∫ − π π sin ( α x ) cos n ( β x ) d x = 0 {\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\cos ^{n}(\beta x)dx=0} α , β {\displaystyle \scriptstyle \alpha ,\beta } n {\displaystyle \scriptstyle n} Симетрія )∫ − π π sin ( α x ) sin n ( β x ) d x = { ( − 1 ) ( n + 1 ) / 2 ( − 1 ) m 2 π 2 n ( n m ) n odd , α = β ( 2 m − n ) 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{(n+1)/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ odd}},\ \alpha =\beta (2m-n)\\0&{\mbox{otherwise}}\\\end{array}}\right.} α , β , m , n {\displaystyle \scriptstyle \alpha ,\beta ,m,n} β ≠ 0 {\displaystyle \scriptstyle \beta \neq 0} m , n ≥ 0 {\displaystyle \scriptstyle m,n\geq 0} Біноміальний коефіцієнт )∫ − π π cos ( α x ) sin n ( β x ) d x = { ( − 1 ) n / 2 ( − 1 ) m 2 π 2 n ( n m ) n even , | α | = | β ( 2 m − n ) | 0 otherwise {\displaystyle \int _{-\pi }^{\pi }\cos(\alpha x)\sin ^{n}(\beta x)dx=\left\{{\begin{array}{cc}(-1)^{n/2}(-1)^{m}{\frac {2\pi }{2^{n}}}{\binom {n}{m}}&n{\mbox{ even}},\ |\alpha |=|\beta (2m-n)|\\0&{\mbox{otherwise}}\\\end{array}}\right.} α , β , m , n {\displaystyle \scriptstyle \alpha ,\beta ,m,n} β ≠ 0 {\displaystyle \scriptstyle \beta \neq 0\!} m , n ≥ 0 {\displaystyle \scriptstyle m,n\geq 0\!} Біноміальний коефіцієнт )∫ 0 ∞ sin 2 x x 2 d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}} ∫ − ∞ ∞ e − ( a x 2 + b x + c ) d x = π a exp [ b 2 − 4 a c 4 a ] {\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]} exp [ u ] {\displaystyle \exp[u]\!} експонента e u {\displaystyle e^{u}} a > 0 {\displaystyle a>0\!} ∫ 0 2 π e x cos θ d θ = 2 π I 0 ( x ) {\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)} I 0 ( x ) {\displaystyle I_{0}(x)\!} Функція Бесселя першого роду)∫ 0 2 π e x cos θ + y sin θ d θ = 2 π I 0 ( x 2 + y 2 ) {\displaystyle \int _{0}^{2\pi }e^{x\cos \theta +y\sin \theta }d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}}}\right)} ∫ 0 π 2 ln sin x d x = − π 2 ln 2 {\displaystyle \int _{0}^{\frac {\pi }{2}}\ln \sin x\,dx=-\,{\frac {\pi }{2}}\ln 2} Для загального випадку, якщо первісної не існує, застосовується метод вичерпання :
∫ a b f ( x ) d x = ( b − a ) ∑ n = 1 ∞ ∑ m = 1 2 n − 1 ( − 1 ) m + 1 2 − n f ( a + m ( b − a ) 2 − n ) . {\displaystyle \int _{a}^{b}{f(x)\,dx}=(b-a)\sum \limits _{n=1}^{\infty }{\sum \limits _{m=1}^{2^{n}-1}{\left({-1}\right)^{m+1}}}2^{-n}f(a+m\left({b-a}\right)2^{-n}).} ∫ 0 1 [ ln ( 1 / x ) ] p d x = p ! {\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!} 
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