Taula da integraal

Quest articol chi l'è scrivuu in Koiné occidentala.

Integraal definiit

Sa la cjama integrala definida da la funziun f in l'intervall [a,b] la grandezza

$\left[F(x)\right]_{a}^{b}=\int _{a}^{b}f(x)\,dx=F(b)-F(a)$

intúe $F\,$ l'è una primitiva qual-sa-vöör da $f\,$ e $a\,$, $b\,$ a inn i cunfin da l'integrala.

I esiist di funziun che a inn integràbil, ma da che vargüna primitiva la pöö mía vess esprimüda in tèrmin da funziun elementaar. Da tüta manera, la valuur da vargüna da sti integrall a pöö vess calcülada e a l'è mustrada chí-da-sota:

$\int _{0}^{+\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}$

$\int _{0}^{+\infty }{e^{-{\frac {x^{2}}{2}}}\,dx}={\sqrt {\frac {\pi }{2}}}$ (integrala da Gauss)

$\int _{0}^{+\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}$

$\int _{0}^{+\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}$

$\int _{0}^{+\infty }{{\frac {x^{3}}{e^{x}-1}}\,dx}={\frac {\pi ^{4}}{15}}$

$\int _{0}^{+\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}$

$\int _{0}^{+\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)$ ($\Gamma \,$ a l'è la funziun Gamma, definida pour z > 0)

$\int _{0}^{1}{\frac {1}{\sqrt {1-t^{3}}}}\,dt={\frac {1}{3}}\mathrm {B} \left({\frac {1}{3}},{\frac {1}{2}}\right)$ (integrala elítica)

$\int _{0}^{\frac {\pi }{2}}\ln(\cos(x))\,dx=\int _{0}^{\frac {\pi }{2}}\ln(\sin(x))\,dx=-{\frac {\pi }{2}}\ln(2)$

$\int _{-\infty }^{+\infty }\cos(x^{2})\,dx=\int _{-\infty }^{+\infty }\sin(x^{2})\,dx={\sqrt {\frac {\pi }{2}}}$ (integrall da Fresnel)

$\int _{0}^{\pi }\ln(1-2\alpha \cos \,x+\alpha ^{2})\,dx=2\pi \ln |\alpha |$ si $|\alpha |>1\,$ e $0\,$ si $|\alpha |\leq 1$.

$\int _{0}^{+\infty }{xe^{-x^{3}}\,dx}={\frac {1}{3}}\Gamma \left({\frac {2}{3}}\right)$

$\int _{0}^{\frac {\pi }{2}}\sin ^{n}(x)\,dx=I_{n}$ (integrall da Wallis)

Videe igualameent

• Taula da primitiif
• Integrala
• Integrala indefinida