Llista d'integrals definides

En matemàtiques, la integral definida:

∫ a b f ( x ) d x {\displaystyle \int _{a}^{b}f(x)\,dx}

és l'àrea de la regió del pla xy delimitada per la gràfica de f, l'eix x i les rectes x = a i x = b, tal qu el l'àrea sobre l'eix x suma en el total i l'àrea sota l'eix x resta del total.

El teorema fonamental del càlcul estableix una relació entre les integrals definides i les primitives i introdueix una tècnica per avaluar les integrals indefinides.

Si l'interval és infinit, la integral definida s'anomena integral impròpia i és definida usant els procediments de límits adquats. Per exemple:

∫ a ∞ f ( x ) d x = lim b → ∞ ∫ a b f ( x ) d x {\displaystyle \int _{a}^{\infty }f(x)\,dx=\lim _{b\to \infty }\int _{a}^{b}f(x)\,dx}

La següent és una llista de les integral definides més comunes. Per una llista d'integrals indefinides, veure taula d'integrals.

Amb expressions racionals i irracionals

∫ 0 ∞ d x x 2 + a 2 = π 2 a {\displaystyle \int _{0}^{\infty }{\frac {dx}{x^{2}+a^{2}}}={\frac {\pi }{2a}}}
∫ 0 ∞ x m d x x n + a n = π a m − n + 1 n sin ⁡ [ ( m + 1 ) π / n ) ] , 0 < m + 1 < n {\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin[(m+1)\pi /n)]}},0<m+1<n}
∫ 0 ∞ x p − 1 d x 1 + x = π sin ⁡ p π   0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x}}={\frac {\pi }{\sin p\pi }}\ \,0<p<1}
∫ 0 ∞ x m d x 1 + 2 x cos ⁡ β + x 2 = π sin ⁡ ( m π ) sin ⁡ ( m β ) sin ⁡ β {\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{1+2x\cos \beta +x^{2}}}={\frac {\pi }{\sin(m\pi )}}{\frac {\sin(m\beta )}{\sin \beta }}}
∫ 0 ∞ d x a 2 − x 2 = π 2 {\displaystyle \int _{0}^{\infty }{\frac {dx}{\sqrt {a^{2}-x^{2}}}}={\frac {\pi }{2}}}
∫ 0 a a 2 − x 2 d x = π a 2 4 {\displaystyle \int _{0}^{a}{\sqrt {a^{2}-x^{2}}}\,dx={\frac {\pi a^{2}}{4}}}
∫ 0 a x m ( a n − x n ) p d x = a m + 1 + n p Γ [ ( m + 1 ) / n ] Γ ( p + 1 ) n Γ [ ( ( m + 1 ) / n ) + p + 1 ] {\displaystyle \int _{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]}}}
∫ 0 ∞ x m d x ( x n + a n ) r = ( − 1 ) r − 1 π a m + 1 − n r Γ [ ( m + 1 ) / n ] n sin ⁡ [ ( m + 1 ) π / n ] ( r − 1 ) ! Γ [ ( m + 1 ) / n − r + 1 ]   n ( r − 2 ) < m + 1 < n r {\displaystyle \int _{0}^{\infty }{\frac {x^{m}\,dx}{({x^{n}+a^{n})}^{r}}}={\frac {(-1)^{r-1}\pi a^{m+1-nr}\Gamma [(m+1)/n]}{n\sin[(m+1)\pi /n](r-1)!\Gamma [(m+1)/n-r+1]}}\ \,n(r-2)<m+1<nr}

Amb funcions trigonomètriques

∫ 0 π sin ⁡ m x sin ⁡ n x d x = { 0 if  m ≠ n π 2 if  m = n     m , n  enters {\displaystyle \int _{0}^{\pi }\sin mx\sin nx\,dx={\begin{cases}0&{\text{if }}m\neq n\\{\dfrac {\pi }{2}}&{\text{if }}m=n\end{cases}}\ \ m,n{\text{ enters}}}
∫ 0 π cos ⁡ m x cos ⁡ n x d x = { 0 if  m ≠ n π 2 if  m = n     m , n  enters {\displaystyle \int _{0}^{\pi }\cos mx\cos nxdx={\begin{cases}0&{\text{if }}m\neq n\\{\dfrac {\pi }{2}}&{\text{if }}m=n\end{cases}}\ \ m,n{\text{ enters}}}
∫ 0 π sin ⁡ m x cos ⁡ n x d x = { 0 if  m + n  even 2 m m 2 − n 2 if  m + n  odd     m , n  enters . {\displaystyle \int _{0}^{\pi }\sin mx\cos nx\,dx={\begin{cases}0&{\text{if }}m+n{\text{ even}}\\{\dfrac {2m}{m^{2}-n^{2}}}&{\text{if }}m+n{\text{ odd}}\end{cases}}\ \ m,n{\text{ enters}}.}
∫ 0 π 2 sin 2 ⁡ x d x = ∫ 0 π 2 cos 2 ⁡ x d x = π 4 {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2}x\,dx={\frac {\pi }{4}}}
∫ 0 π 2 sin 2 m ⁡ x d x = ∫ 0 π 2 cos 2 m ⁡ x d x = 1 × 3 × 5 × ⋯ × ( 2 m − 1 ) 2 × 4 × 6 × ⋯ × 2 m π 2     m = 1 , 2 , 3 , … {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m}x\,dx={\frac {1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}}{\frac {\pi }{2}}\ \ m=1,2,3,\ldots }
∫ 0 π 2 sin 2 m + 1 ⁡ x d x = ∫ 0 π 2 cos 2 m + 1 ⁡ x d x = 2 × 4 × 6 × ⋯ × 2 m 1 × 3 × 5 × ⋯ × ( 2 m + 1 )     m = 1 , 2 , 3 , … {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2m+1}x\,dx=\int _{0}^{\frac {\pi }{2}}\cos ^{2m+1}x\,dx={\frac {2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m+1)}}\ \ m=1,2,3,\ldots }
∫ 0 π 2 sin 2 p − 1 ⁡ x cos 2 q − 1 ⁡ x d x = Γ ( p ) Γ ( q ) 2 Γ ( p + q ) = 1 2 B ( p , q ) {\displaystyle \int _{0}^{\frac {\pi }{2}}\sin ^{2p-1}x\cos ^{2q-1}x\,dx={\frac {\Gamma (p)\Gamma (q)}{2\Gamma (p+q)}}={\frac {1}{2}}{\text{B}}(p,q)}
∫ 0 ∞ sin ⁡ p x x d x = { π 2 if  p > 0 0 if  p = 0 − π 2 if  p < 0 {\displaystyle \int _{0}^{\infty }{\frac {\sin px}{x}}\,dx={\begin{cases}{\dfrac {\pi }{2}}&{\text{if }}p>0\\\\0&{\text{if }}p=0\\\\-{\dfrac {\pi }{2}}&{\text{if }}p<0\end{cases}}}
∫ 0 ∞ sin ⁡ p x cos ⁡ q x x   d x = { 0  if  p > q > 0 π 2  if  0 < p < q π 4  if  p = q > 0 {\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}}\ dx={\begin{cases}0&{\text{ if }}p>q>0\\\\{\dfrac {\pi }{2}}&{\text{ if }}0<p<q\\\\{\dfrac {\pi }{4}}&{\text{ if }}p=q>0\end{cases}}}
∫ 0 ∞ sin ⁡ p x sin ⁡ q x x 2   d x = { π p 2  if  0 < p ≤ q π q 2  if  0 < q ≤ p {\displaystyle \int _{0}^{\infty }{\frac {\sin px\sin qx}{x^{2}}}\ dx={\begin{cases}{\dfrac {\pi p}{2}}&{\text{ if }}0<p\leq q\\\\{\dfrac {\pi q}{2}}&{\text{ if }}0<q\leq p\end{cases}}}
∫ 0 ∞ sin 2 ⁡ p x x 2   d x = π p 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}px}{x^{2}}}\ dx={\frac {\pi p}{2}}}
∫ 0 ∞ 1 − cos ⁡ p x x 2   d x = π p 2 {\displaystyle \int _{0}^{\infty }{\frac {1-\cos px}{x^{2}}}\ dx={\frac {\pi p}{2}}}
∫ 0 ∞ cos ⁡ p x − cos ⁡ q x x   d x = ln ⁡ q p {\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x}}\ dx=\ln {\frac {q}{p}}}
∫ 0 ∞ cos ⁡ p x − cos ⁡ q x x 2   d x = π ( q − p ) 2 {\displaystyle \int _{0}^{\infty }{\frac {\cos px-\cos qx}{x^{2}}}\ dx={\frac {\pi (q-p)}{2}}}
∫ 0 ∞ cos ⁡ m x x 2 + a 2   d x = π 2 a e − m a {\displaystyle \int _{0}^{\infty }{\frac {\cos mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2a}}e^{-ma}}
∫ 0 ∞ x sin ⁡ m x x 2 + a 2   d x = π 2 e − m a {\displaystyle \int _{0}^{\infty }{\frac {x\sin mx}{x^{2}+a^{2}}}\ dx={\frac {\pi }{2}}e^{-ma}}
∫ 0 ∞ sin ⁡ m x x ( x 2 + a 2 )   d x = π 2 a 2 ( 1 − e − m a ) {\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{x(x^{2}+a^{2})}}\ dx={\frac {\pi }{2a^{2}}}(1-e^{-ma})}
∫ 0 2 π d x a + b sin ⁡ x = 2 π a 2 − b 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\sin x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}
∫ 0 2 π d x a + b cos ⁡ x = 2 π a 2 − b 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{a+b\cos x}}={\frac {2\pi }{\sqrt {a^{2}-b^{2}}}}}
∫ 0 π 2 d x a + b cos ⁡ x = cos − 1 ⁡ ( b / a ) a 2 − b 2 {\displaystyle \int _{0}^{\frac {\pi }{2}}{\frac {dx}{a+b\cos x}}={\frac {\cos ^{-1}(b/a)}{\sqrt {a^{2}-b^{2}}}}}
∫ 0 2 π d x ( a + b sin ⁡ x ) 2 = ∫ 0 2 π d x ( a + b cos ⁡ x ) 2 = 2 π a ( a 2 − b 2 ) 3 / 2 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{(a+b\sin x)^{2}}}=\int _{0}^{2\pi }{\frac {dx}{(a+b\cos x)^{2}}}={\frac {2\pi a}{(a^{2}-b^{2})^{3/2}}}}
∫ 0 2 π d x 1 − 2 a cos ⁡ x + a 2 = 2 π 1 − a 2       0 < a < 1 {\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2}}}={\frac {2\pi }{1-a^{2}}}\ \ \,\ 0<a<1}
∫ 0 π x sin ⁡ x   d x 1 − 2 a cos ⁡ x + a 2 = { π a ln ⁡ | 1 + a | if  | a | < 1 π a ln ⁡ | 1 + 1 a | if  | a | > 1 {\displaystyle \int _{0}^{\pi }{\frac {x\sin x\ dx}{1-2a\cos x+a^{2}}}={\begin{cases}{\frac {\pi }{a}}\ln \left|1+a\right|&{\text{if }}|a|<1\\{\frac {\pi }{a}}\ln \left|1+{\frac {1}{a}}\right|&{\text{if }}|a|>1\end{cases}}}
∫ 0 π cos ⁡ m x   d x 1 − 2 a cos ⁡ x + a 2 = π a m 1 − a 2 , a 2 < 1 ,   m = 0 , 1 , 2 , … {\displaystyle \int _{0}^{\pi }{\frac {\cos mx\ dx}{1-2a\cos x+a^{2}}}={\frac {\pi a^{m}}{1-a^{2}}}\quad ,a^{2}<1,\ m=0,1,2,\dots }
∫ 0 ∞ sin ⁡ a x 2   d x = ∫ 0 ∞ cos ⁡ a x 2 = 1 2 π 2 a {\displaystyle \int _{0}^{\infty }\sin ax^{2}\ dx=\int _{0}^{\infty }\cos ax^{2}={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}}
∫ 0 ∞ sin ⁡ a x n = 1 n a 1 / n Γ ( 1 / n ) sin ⁡ π 2 n , n > 1 {\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n}}}\Gamma (1/n)\sin {\frac {\pi }{2n}}\quad ,n>1}
∫ 0 ∞ cos ⁡ a x n = 1 n a 1 / n Γ ( 1 / n ) cos ⁡ π 2 n , n > 1 {\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n}}}\Gamma (1/n)\cos {\frac {\pi }{2n}}\quad ,n>1}
∫ 0 ∞ sin ⁡ x x   d x = ∫ 0 ∞ cos ⁡ x x   d x = π 2 {\displaystyle \int _{0}^{\infty }{\frac {\sin x}{\sqrt {x}}}\ dx=\int _{0}^{\infty }{\frac {\cos x}{\sqrt {x}}}\ dx={\sqrt {\frac {\pi }{2}}}}
∫ 0 ∞ sin ⁡ x x p   d x = π 2 Γ ( p ) sin ⁡ ( p π / 2 ) , 0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\sin(p\pi /2)}},\quad 0<p<1}
∫ 0 ∞ cos ⁡ x x p   d x = π 2 Γ ( p ) cos ⁡ ( p π / 2 ) , 0 < p < 1 {\displaystyle \int _{0}^{\infty }{\frac {\cos x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\cos(p\pi /2)}},\quad 0<p<1}
∫ 0 ∞ sin ⁡ a x 2 cos ⁡ 2 b x   d x = 1 2 π 2 a ( cos ⁡ b 2 a − sin ⁡ b 2 a ) {\displaystyle \int _{0}^{\infty }\sin ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}-\sin {\frac {b^{2}}{a}}\right)}
∫ 0 ∞ cos ⁡ a x 2 cos ⁡ 2 b x   d x = 1 2 π 2 a ( cos ⁡ b 2 a + sin ⁡ b 2 a ) {\displaystyle \int _{0}^{\infty }\cos ax^{2}\cos 2bx\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{2a}}}\left(\cos {\frac {b^{2}}{a}}+\sin {\frac {b^{2}}{a}}\right)}

Amb funcions exponencials

∫ 0 ∞ e − a x cos ⁡ b x d x = a a 2 + b 2 {\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}}
∫ 0 ∞ e − a x sin ⁡ b x d x = b a 2 + b 2 {\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}}
∫ 0 ∞ e − a x sin ⁡ b x x d x = tan − 1 ⁡ b a {\displaystyle \int _{0}^{\infty }{\frac {{}e^{-ax}\sin bx}{x}}\,dx=\tan ^{-1}{\frac {b}{a}}}
∫ 0 ∞ e − a x − e − b x x d x = ln ⁡ b a {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\,dx=\ln {\frac {b}{a}}}
∫ 0 ∞ e − a x 2 d x = 1 2 π a {\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}}
∫ 0 ∞ e − a x 2 cos ⁡ b x d x = 1 2 π a e − b 2 / 4 a {\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}}\cos bx\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{-b^{2}/4a}}
∫ 0 ∞ e − ( a x 2 + b x + c ) d x = 1 2 π a e ( b 2 − 4 a c ) / 4 a   erfc ⁡ b 2 a ,  on  erfc ⁡ ( p ) = 2 π ∫ p ∞ e − x 2 d x {\displaystyle \int _{0}^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{(b^{2}-4ac)/4a}\ \operatorname {erfc} {\frac {b}{2{\sqrt {a}}}},{\text{ on }}\operatorname {erfc} (p)={\frac {2}{\sqrt {\pi }}}\int _{p}^{\infty }e^{-x^{2}}\,dx}
∫ − ∞ + ∞ e − ( a x 2 + b x + c )   d x = π a e ( b 2 − 4 a c ) / 4 a {\displaystyle \int _{-\infty }^{+\infty }e^{-(ax^{2}+bx+c)}\ dx={\sqrt {\frac {\pi }{a}}}e^{(b^{2}-4ac)/4a}}
∫ 0 ∞ x n e − a x   d x = Γ ( n + 1 ) a n + 1 {\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\ dx={\frac {\Gamma (n+1)}{a^{n+1}}}}
∫ 0 ∞ x m e − a x 2   d x = Γ [ ( m + 1 ) / 2 ] 2 a ( m + 1 ) / 2 {\displaystyle \int _{0}^{\infty }x^{m}e^{-ax^{2}}\ dx={\frac {\Gamma [(m+1)/2]}{2a^{(m+1)/2}}}}
∫ 0 ∞ e − a x 2 − b / x 2   d x = 1 2 π a e − 2 a b {\displaystyle \int _{0}^{\infty }e^{-ax^{2}-b/x^{2}}\ dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}}
∫ 0 ∞ x e x − 1   d x = ζ ( 2 ) = π 2 6 {\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}-1}}\ dx=\zeta (2)={\frac {\pi ^{2}}{6}}}
∫ 0 ∞ x n − 1 e x − 1   d x = Γ ( n ) ζ ( n ) {\displaystyle \int _{0}^{\infty }{\frac {x^{n-1}}{e^{x}-1}}\ dx=\Gamma (n)\zeta (n)}
∫ 0 ∞ x e x + 1   d x = 1 1 2 − 1 2 2 + 1 3 2 − 1 4 2 + ⋯ = π 2 12 {\displaystyle \int _{0}^{\infty }{\frac {x}{e^{x}+1}}\ dx={\frac {1}{1^{2}}}-{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}-{\frac {1}{4^{2}}}+\dots ={\frac {\pi ^{2}}{12}}}
∫ 0 ∞ sin ⁡ m x e 2 π x − 1   d x = 1 4 coth ⁡ m 2 − 1 2 m {\displaystyle \int _{0}^{\infty }{\frac {\sin mx}{e^{2\pi x}-1}}\ dx={\frac {1}{4}}\coth {\frac {m}{2}}-{\frac {1}{2m}}}
∫ 0 ∞ ( 1 1 + x − e − x )   d x x = γ {\displaystyle \int _{0}^{\infty }\left({\frac {1}{1+x}}-e^{-x}\right)\ {\frac {dx}{x}}=\gamma }
∫ 0 ∞ e − x 2 − e − x x   d x = γ 2 {\displaystyle \int _{0}^{\infty }{\frac {e^{-x^{2}}-e^{-x}}{x}}\ dx={\frac {\gamma }{2}}}
∫ 0 ∞ ( 1 e x − 1 − e − x x )   d x = γ {\displaystyle \int _{0}^{\infty }\left({\frac {1}{e^{x}-1}}-{\frac {e^{-x}}{x}}\right)\ dx=\gamma }
∫ 0 ∞ e − a x − e − b x x sec ⁡ p x   d x = 1 2 ln ⁡ b 2 + p 2 a 2 + p 2 {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\sec px}}\ dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}}
∫ 0 ∞ e − a x − e − b x x csc ⁡ p x   d x = tan − 1 ⁡ b p − tan − 1 ⁡ a p {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x\csc px}}\ dx=\tan ^{-1}{\frac {b}{p}}-\tan ^{-1}{\frac {a}{p}}}
∫ 0 ∞ e − a x ( 1 − cos ⁡ x ) x 2   d x = cot − 1 ⁡ a − a 2 ln ⁡ | a 2 + 1 a 2 | {\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\ dx=\cot ^{-1}a-{\frac {a}{2}}\ln \left|{\frac {a^{2}+1}{a^{2}}}\right|}
∫ − ∞ ∞ e − x 2 d x = π {\displaystyle \int _{-\infty }^{\infty }e^{-x^{2}}\,dx={\sqrt {\pi }}}
∫ − ∞ ∞ x 2 ( n + 1 ) e − x 2 / 2 d x = ( 2 n + 1 ) ! 2 n n ! 2 π n = 0 , 1 , 2 , … {\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-x^{2}/2}\,dx={\frac {(2n+1)!}{2^{n}n!}}{\sqrt {2\pi }}\quad n=0,1,2,\ldots }

Amb funcions logarítmiques

∫ 0 1 x m ( ln ⁡ x ) n d x = ( − 1 ) n n ! ( m + 1 ) n + 1 m > − 1 , n = 0 , 1 , 2 , … {\displaystyle \int _{0}^{1}x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n}n!}{(m+1)^{n+1}}}\quad m>-1,n=0,1,2,\ldots }
∫ 0 1 ln ⁡ x 1 + x d x = − π 2 12 {\displaystyle \int _{0}^{1}{\frac {\ln x}{1+x}}\,dx=-{\frac {\pi ^{2}}{12}}}
∫ 0 1 ln ⁡ x 1 − x d x = − π 2 6 {\displaystyle \int _{0}^{1}{\frac {\ln x}{1-x}}\,dx=-{\frac {\pi ^{2}}{6}}}
∫ 0 1 ln ⁡ ( 1 + x ) x d x = π 2 12 {\displaystyle \int _{0}^{1}{\frac {\ln(1+x)}{x}}\,dx={\frac {\pi ^{2}}{12}}}
∫ 0 1 ln ⁡ ( 1 − x ) x d x = − π 2 6 {\displaystyle \int _{0}^{1}{\frac {\ln(1-x)}{x}}\,dx=-{\frac {\pi ^{2}}{6}}}
∫ 0 ∞ ln ⁡ ( a 2 + x 2 ) b 2 + x 2   d x = π b ln ⁡ ( a + b ) a , b > 0 {\displaystyle \int _{0}^{\infty }{\frac {\ln(a^{2}+x^{2})}{b^{2}+x^{2}}}\ dx={\frac {\pi }{b}}\ln(a+b)\quad a,b>0}
∫ 0 ∞ ln ⁡ x x 2 + a 2   d x = π ln ⁡ a 2 a a > 0 {\displaystyle \int _{0}^{\infty }{\frac {\ln x}{x^{2}+a^{2}}}\ dx={\frac {\pi \ln a}{2a}}\quad a>0}

Amb funcions hiperbòliques

∫ 0 ∞ sin ⁡ a x sinh ⁡ b x   d x = π 2 b tanh ⁡ a π 2 b {\displaystyle \int _{0}^{\infty }{\frac {\sin ax}{\sinh bx}}\ dx={\frac {\pi }{2b}}\tanh {\frac {a\pi }{2b}}}

∫ 0 ∞ cos ⁡ a x cosh ⁡ b x   d x = π 2 b 1 cosh ⁡ a π 2 b {\displaystyle \int _{0}^{\infty }{\frac {\cos ax}{\cosh bx}}\ dx={\frac {\pi }{2b}}{\frac {1}{\cosh {\frac {a\pi }{2b}}}}}

∫ 0 ∞ x sinh ⁡ a x   d x = π 2 4 a 2 {\displaystyle \int _{0}^{\infty }{\frac {x}{\sinh ax}}\ dx={\frac {\pi ^{2}}{4a^{2}}}}

∫ − ∞ ∞ 1 cosh ⁡ x   d x = π {\displaystyle \int _{-\infty }^{\infty }{\frac {1}{\cosh x}}\ dx=\pi }

Altres

∫ 0 ∞ f ( a x ) − f ( b x ) x   d x = [ f ( 0 ) − f ( ∞ ) ] ln ⁡ b a {\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\ dx=[{f(0)-f(\infty )}]\ln {\frac {b}{a}}}

Vegeu també

Referències