Integrazioa kalkulu integralaren oinarrizko bi eragiketako bat da. Deribazioak arau errazak dituen bitartean funtzio konplexu bat aurkitzeko hura osatzen duten funtzio bakunen diferentziazioa egiten, integrazioak ez, horregatik oso baliagarriak dira ezagututako integralen taulak. Orrialde honek jatorrizko ezagunenak zerrendatzen ditu.
Integralen zerrenden historia-bilakaera
Aldatu
Integralen zerrenda baten bilduma (Integraltafeln) eta kalkulu integralaren teknikak Meyer Hirsch Alemaniako matematikariak argitaratu zituen 1810ean. Taula horiek berrargitaratu zituzten Erresuma Batuan 1823an. 1858an, David de Bierens de Haan Herbehereetako matematikariak taula luzeagoak bildu zituen. Edizio berri bat 1862an argitaratu zuten. Taula horiek, zeinetan nagusiki oinarrizko funtzioen integralak dauden, 20. mendearen erdira arte jarraitu zituzten erabiltzen. Gero, taula horien ordez Gradshteyn en eta Ryzhik en taula handiagoak erabiltzen hasi ziren. Gradshteyn en eta Ryzhik en tauletan Bierens en liburutik hartutako integralak BI letrekin adierazten dituzte.
Integralen zerrenda
Aldatu
K erabiltzen da integrazio-konstante gisa. Konstante hori zehaztu daiteke soilik integralaren balioa ezaguna baldin bada puntu batean. Horrela, funtzio bakoitzak jatorrizkoen kopuru infinitua dauka.
Funtzio arrazionalak
Aldatu
∫
d
x
=
x
+
K
{\displaystyle \int \,dx=x+K}
∫
x
n
d
x
=
x
n
+
1
n
+
1
+
K
baldin
n
≠
−
1
bada
{\displaystyle \int x^{n}\,dx={\frac {x^{n+1}}{n+1}}+K\qquad {\mbox{ baldin }}n\neq -1{\mbox{ bada }}}
∫
d
x
x
=
ln
|
x
|
+
K
{\displaystyle \int {dx \over x}=\ln {\left|x\right|}+K}
∫
d
x
a
2
+
b
2
x
2
=
1
a
b
arctan
b
x
a
+
K
{\displaystyle \int {dx \over {a^{2}+b^{2}x^{2}}}={1 \over ab}\arctan {bx \over a}+K}
Funtzio irrazionalak
Aldatu
∫
d
x
a
2
−
x
2
=
sin
−
1
x
a
+
K
{\displaystyle \int {dx \over {\sqrt {a^{2}-x^{2}}}}=\sin ^{-1}{x \over a}+K}
∫
−
d
x
a
2
−
x
2
=
cos
−
1
x
a
+
K
{\displaystyle \int {-dx \over {\sqrt {a^{2}-x^{2}}}}=\cos ^{-1}{x \over a}+K}
∫
d
x
x
x
2
−
a
2
=
1
a
sec
−
1
|
x
|
a
+
K
{\displaystyle \int {dx \over x{\sqrt {x^{2}-a^{2}}}}={1 \over a}\sec ^{-1}{|x| \over a}+K}
Funtzio logaritmikoak
Aldatu
∫
ln
x
d
x
=
x
ln
x
−
x
+
K
{\displaystyle \int \ln {x}\,dx=x\ln {x}-x+K}
∫
log
b
x
d
x
=
x
log
b
x
−
x
log
b
e
+
K
{\displaystyle \int \log _{b}{x}\,dx=x\log _{b}{x}-x\log _{b}{e}+K}
Funtzio esponentzialak
Aldatu
∫
e
x
d
x
=
e
x
+
K
{\displaystyle \int e^{x}\,dx=e^{x}+K}
∫
a
x
d
x
=
a
x
ln
a
+
K
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln a}}+K}
Funtzio trigonometrikoak
Aldatu
∫
sin
x
d
x
=
−
cos
x
+
K
{\displaystyle \int \sin {x}\,dx=-\cos {x}+K}
∫
cos
x
d
x
=
sin
x
+
K
{\displaystyle \int \cos {x}\,dx=\sin {x}+K}
∫
tan
x
d
x
=
−
ln
|
cos
x
|
+
K
=
ln
|
sec
x
|
+
K
{\displaystyle \int \tan {x}\,dx=-\ln {\left|\cos {x}\right|}+K=\ln {\left|\sec {x}\right|}+K}
∫
cot
x
d
x
=
ln
|
sin
x
|
+
K
{\displaystyle \int \cot {x}\,dx=\ln {\left|\sin {x}\right|}+K}
∫
sec
x
d
x
=
ln
|
sec
x
+
tan
x
|
+
K
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+K}
∫
csc
x
d
x
=
−
ln
|
csc
x
+
cot
x
|
+
K
{\displaystyle \int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+K}
∫
sec
2
x
d
x
=
tan
x
+
K
{\displaystyle \int \sec ^{2}x\,dx=\tan x+K}
∫
csc
2
x
d
x
=
−
cot
x
+
K
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+K}
∫
sec
x
tan
x
d
x
=
sec
x
+
K
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+K}
∫
csc
x
cot
x
d
x
=
−
csc
x
+
K
{\displaystyle \int \csc {x}\,\cot {x}\,dx=-\csc {x}+K}
∫
sin
2
x
d
x
=
1
2
(
x
−
sin
2
x
2
)
+
K
=
1
2
(
x
−
sin
x
cos
x
)
+
K
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}\left(x-{\frac {\sin 2x}{2}}\right)+K={\frac {1}{2}}(x-\sin x\cos x)+K}
∫
cos
2
x
d
x
=
1
2
(
x
+
sin
2
x
2
)
+
K
=
1
2
(
x
+
sin
x
cos
x
)
+
K
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}\left(x+{\frac {\sin 2x}{2}}\right)+K={\frac {1}{2}}(x+\sin x\cos x)+K}
∫
sec
3
x
d
x
=
1
2
sec
x
tan
x
+
1
2
ln
|
sec
x
+
tan
x
|
+
K
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+K}
(ikusi sekantearen kuboaren integrala )
∫
sin
n
x
d
x
=
−
sin
n
−
1
x
cos
x
n
+
n
−
1
n
∫
sin
n
−
2
x
d
x
{\displaystyle \int \sin ^{n}x\,dx=-{\frac {\sin ^{n-1}{x}\cos {x}}{n}}+{\frac {n-1}{n}}\int \sin ^{n-2}{x}\,dx}
∫
cos
n
x
d
x
=
cos
n
−
1
x
sin
x
n
+
n
−
1
n
∫
cos
n
−
2
x
d
x
{\displaystyle \int \cos ^{n}x\,dx={\frac {\cos ^{n-1}{x}\sin {x}}{n}}+{\frac {n-1}{n}}\int \cos ^{n-2}{x}\,dx}
Alderantzizko funtzio trigonometrikoak
Aldatu
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
K
{\displaystyle \int \arcsin {x}\,dx=x\,\arcsin {x}+{\sqrt {1-x^{2}}}+K}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
K
{\displaystyle \int \arccos {x}\,dx=x\,\arccos {x}-{\sqrt {1-x^{2}}}+K}
∫
arctan
x
d
x
=
x
arctan
x
−
1
2
ln
|
1
+
x
2
|
+
K
{\displaystyle \int \arctan {x}\,dx=x\,\arctan {x}-{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+K}
∫
arccot
x
d
x
=
x
arccot
x
+
1
2
ln
|
1
+
x
2
|
+
K
{\displaystyle \int \operatorname {arccot} {x}\,dx=x\,\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\left|1+x^{2}\right|}+K}
∫
arcsec
x
d
x
=
x
arcsec
x
−
artanh
1
−
1
x
2
+
K
{\displaystyle \int \operatorname {arcsec} {x}\,dx=x\,\operatorname {arcsec} {x}-\operatorname {artanh} \,{\sqrt {1-{\frac {1}{x^{2}}}}}+K}
∫
arccsc
x
d
x
=
x
arccsc
x
+
artanh
1
−
1
x
2
+
K
{\displaystyle \int \operatorname {arccsc} {x}\,dx=x\,\operatorname {arccsc} {x}+\operatorname {artanh} \,{\sqrt {1-{\frac {1}{x^{2}}}}}+K}
Funtzio hiperbolikoak
Aldatu
∫
sinh
x
d
x
=
cosh
x
+
K
{\displaystyle \int \sinh x\,dx=\cosh x+K}
∫
cosh
x
d
x
=
sinh
x
+
K
{\displaystyle \int \cosh x\,dx=\sinh x+K}
∫
tanh
x
d
x
=
ln
|
cosh
x
|
+
K
{\displaystyle \int \tanh x\,dx=\ln |\cosh x|+K}
∫
csch
x
d
x
=
ln
|
tanh
x
2
|
+
K
{\displaystyle \int {\mbox{csch}}\,x\,dx=\ln \left|\tanh {x \over 2}\right|+K}
∫
sech
x
d
x
=
arcsin
(
tanh
x
)
+
K
{\displaystyle \int {\mbox{sech}}\,x\,dx=\arcsin \,(\tanh x)+K}
∫
coth
x
d
x
=
ln
|
sinh
x
|
+
K
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+K}
Alderantzizko funtzio hiperbolikoak
Aldatu
∫
arsinh
x
d
x
=
x
arsinh
x
−
x
2
+
1
+
K
{\displaystyle \int \operatorname {arsinh} \,x\,dx=x\,\operatorname {arsinh} \,x-{\sqrt {x^{2}+1}}+K}
∫
arcosh
x
d
x
=
x
arcosh
x
−
x
+
1
x
−
1
+
K
{\displaystyle \int \operatorname {arcosh} \,x\,dx=x\,\operatorname {arcosh} \,x-{\sqrt {x+1}}\,{\sqrt {x-1}}+K}
∫
artanh
x
d
x
=
x
artanh
x
+
ln
(
1
−
x
2
)
2
+
K
{\displaystyle \int \operatorname {artanh} \,x\,dx=x\,\operatorname {artanh} \,x+{\frac {\ln \left(1-x^{2}\right)}{2}}+K}
∫
arcoth
x
d
x
=
x
arcoth
x
+
ln
(
1
−
x
2
)
2
+
K
{\displaystyle \int \operatorname {arcoth} \,x\,dx=x\,\operatorname {arcoth} \,x+{\frac {\ln \left(1-x^{2}\right)}{2}}+K}
∫
arsech
x
d
x
=
x
arsech
x
−
2
arctan
1
−
x
1
+
x
+
K
{\displaystyle \int \operatorname {arsech} \,x\,dx=x\,\operatorname {arsech} \,x-2\,\arctan {\sqrt {\frac {1-x}{1+x}}}+K}
∫
arcsch
x
d
x
=
x
arcsch
x
+
artanh
1
x
2
+
1
+
K
{\displaystyle \int \operatorname {arcsch} \,x\,dx=x\,\operatorname {arcsch} \,x+\operatorname {artanh} {\sqrt {{\frac {1}{x^{2}}}+1}}+K}
Xehetasun gehiagorako ondorengo orrietara jo:
Jatorrizko itxia ez duten integral mugatuak
Aldatu
Badaude zenbait funtzio zeinen jatorrizkoak ezin diren adierazi forma itxian, hau da, ezin dira adierazi funtzio arrazional, irrazional, esponentzial, logaritmiko, trigonometriko edo alderantzizko funtzio trigonometrikoen konposizio, batuketa edo biderketa gisa. Ostera, zenbait tarte komunetan, funtzio horien integral mugatuen balioak era sinbolikoan kalkula daitezke eta balio zehatza ere lortu. Kasu horietan, baliabidetariko batzuk hauek ditugu:
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
Gamma funtzioa ere)
∫
0
∞
e
−
x
2
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{e^{-x^{2}}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
Gauss en integrala
∫
0
∞
x
e
x
−
1
d
x
=
π
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
Bernoulli ren zenbakia
∫
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
∞
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 \dots \cdot (n-1)}{2\cdot 4\cdot 6\cdot \dots \cdot n}}{\frac {\pi }{2}}}
n bikoiti osoa eta
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 \dots \cdot (n-1)}{3\cdot 5\cdot 7\cdot \dots \cdot n}}}
n
{\displaystyle \scriptstyle {n}}
n
≥
3
{\displaystyle \scriptstyle {n\geq 3}}
∫
0
∞
sin
2
x
x
2
d
x
=
π
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫
0
∞
x
z
−
1
e
−
x
d
x
=
Γ
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
Γ
(
z
)
{\displaystyle \Gamma (z)}
Gamma funtzioa den)
∫
−
∞
∞
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]}
e
u
{\displaystyle e^{u}}
funtzio esponentziala den, eta
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)}
Bessel en funtzio
∫
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)}
∫
−
∞
∞
(
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\,}
Student en t banaketarenprobabilitatearen dentsitate-funtzioari lotuta dago)Exhauzio-metodoak formula bat ematen du kasu orokorrerako jatorrizkorik ez dagoenean:
∫
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
e
x
⋅
ln
a
+
(
1
−
x
)
⋅
ln
b
d
x
=
∫
0
1
(
a
b
)
x
⋅
b
d
x
=
∫
0
1
a
x
⋅
b
1
−
x
d
x
=
a
−
b
ln
a
−
ln
b
{\displaystyle \int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\;\mathrm {d} x=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\;\mathrm {d} x=\int _{0}^{1}a^{x}\cdot b^{1-x}\;\mathrm {d} x={\frac {a-b}{\ln a-\ln b}}}
a
>
0
,
b
>
0
,
a
≠
b
{\displaystyle a>0,\ b>0,\ a\neq b}
batez besteko logaritmikoa dena
∫
0
∞
e
−
a
x
d
x
=
1
a
{\displaystyle \int _{0}^{\infty }e^{-ax}\,\mathrm {d} x={\frac {1}{a}}}
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π
a
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)}
Gauss en integrala)
∫
−
∞
∞
e
−
a
x
2
d
x
=
π
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,\mathrm {d} x={\sqrt {\pi \over a}}\quad (a>0)}
∫
−
∞
∞
e
−
a
x
2
e
2
b
x
d
x
=
π
a
e
b
2
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)}
∫
−
∞
∞
x
e
−
a
(
x
−
b
)
2
d
x
=
b
π
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\pi \over a}}\quad (a>0)}
∫
−
∞
∞
x
2
e
−
a
x
2
d
x
=
1
2
π
a
3
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,\mathrm {d} x={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)}
∫
0
∞
x
n
e
−
a
x
2
d
x
=
{
1
2
Γ
(
n
+
1
2
)
/
a
n
+
1
2
(
n
>
−
1
,
a
>
0
)
(
2
k
−
1
)
!
!
2
k
+
1
a
k
π
a
(
n
=
2
k
,
k
zenbaki osoa
,
a
>
0
)
k
!
2
a
k
+
1
(
n
=
2
k
+
1
,
k
zenbaki osoa
,
a
>
0
)
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,\mathrm {d} x={\begin{cases}{\frac {1}{2}}\Gamma \left({\frac {n+1}{2}}\right)/a^{\frac {n+1}{2}}&(n>-1,a>0)\\{\frac {(2k-1)!!}{2^{k+1}a^{k}}}{\sqrt {\frac {\pi }{a}}}&(n=2k,k\;{\text{zenbaki osoa}},a>0)\\{\frac {k!}{2a^{k+1}}}&(n=2k+1,k\;{\text{zenbaki osoa}},a>0)\end{cases}}}
Faktorial bikoitza da)
∫
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∞
x
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a
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d
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{
Γ
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{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\,\mathrm {d} x={\begin{cases}{\frac {\Gamma (n+1)}{a^{n+1}}}&(n>-1,a>0)\\{\frac {n!}{a^{n+1}}}&(n=0,1,2,\ldots ,a>0)\\\end{cases}}}
∫
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sin
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{\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,\mathrm {d} x={\frac {b}{a^{2}+b^{2}}}\quad (a>0)}
∫
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−
a
x
cos
b
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a
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{\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,\mathrm {d} x={\frac {a}{a^{2}+b^{2}}}\quad (a>0)}
∫
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∞
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2
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2
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{\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,\mathrm {d} x={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
∫
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∞
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x
cos
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d
x
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2
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{\displaystyle \int _{0}^{\infty }xe^{-ax}\cos bx\,\mathrm {d} x={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
∫
0
1
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−
x
d
x
=
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1
∞
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n
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=
1.291285997
…
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∫
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x
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∑
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1
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n
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n
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=
0.783430510712
…
)
{\displaystyle {\begin{aligned}\int _{0}^{1}x^{-x}\,dx&=\sum _{n=1}^{\infty }n^{-n}&&(=1.291285997\dots )\\\int _{0}^{1}x^{x}\,dx&=\sum _{n=1}^{\infty }-(-1)^{n}n^{-n}&&(=0.783430510712\dots )\end{aligned}}}
Johann Bernoulli da ustezko egilea.
I.S. Gradshteyn (И.С. Градштейн), I.M. Ryzhik (И.М. Рыжик); Alan Jeffrey, Daniel Zwillinger, argitaratzaileak. Table of Integrals, Series, and Products , zazpigarren edizioa. Academic Press, 2007. ISBN 978-0-12-373637-6 . Akatsa. (Aurreko edizio asko ondo daude.) A.P. Prudnikov (А.П. Прудников), Yu.A. Brychkov (Ю.А. Брычков), O.I. Marichev (О.И. Маричев). Integrals and Series . Lehenengo edizioa (errusieraz), 1–5 liburukiak, Nauka , 1981−1986. Lehenengo edizioa (ingelesez, N.M. Queen-ek errusieratik itzulita), 1–5 liburukiak, Gordon & Breach Science Publishers/CRC Press , 1988–1992, ISBN 2-88124-097-6 . Bigarren edizio berrikusia (errusieraz), 1–3 liburukiak, Fiziko-Matematicheskaya Literatura, 2003. Yu.A. Brychkov (Ю.А. Брычков), Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas . Errusierazko edizioa, Fiziko-Matematicheskaya Literatura, 2006. Ingelesezko edizioa, Chapman & Hall/CRC Press, 2008, ISBN 1-58488-956-X . Daniel Zwillinger. CRC Standard Mathematical Tables and Formulae , 31. edizioa. Chapman & Hall/CRC Press, 2002. ISBN 1-58488-291-3 . (Aurreko edizio asko ondo daude.) Meyer Hirsch, Integraltafeln, oder, Sammlung von Integralformeln (Duncker und Humblot, Berlin, 1810)
Meyer Hirsch, Integral Tables, Or, A Collection of Integral Formulae (Baynes and son, London, 1823) [Integraltafeln -en ingelesezko itzulpena]
David Bierens de Haan, Nouvelles Tables d'Intégrales définies (Engels, Leiden, 1862)
Benjamin O. Pierce A short table of integrals - edizio berrikusia (Ginn & co., Boston, 1899) Kanpo loturak
Aldatu
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