Αυτή η σελίδα περιέχει έναν κατάλογο ορισμένων ολοκληρωμάτων [ 1] [ 2] . Για τα λοιπά ολοκληρώματα δείτε τους αντίστοιχους καταλόγους ολοκληρωμάτων.
Στα μαθηματικά , το ορισμένο ολοκλήρωμα
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\,dx}
είναι το εμβαδόν της περιοχής στο επίπεδο xy που οριοθετείται από τη γραφική παράσταση της f , τον άξονα x και τις ευθείες x = a και x = b , έτσι ώστε το εμβαδόν πάνω από τον άξονα x να προστίθεται στο σύνολο και το εμβαδόν κάτω από τον άξονα x να αφαιρείται από το σύνολο.[ 3]
Το θεμελιώδες θεώρημα του λογισμού καθορίζει τη σχέση μεταξύ αόριστων και ορισμένων ολοκληρωμάτων και εισάγει μια τεχνική για την αξιολόγηση ορισμένων ολοκληρωμάτων.
Εάν το διάστημα είναι άπειρο, το ορισμένο ολοκλήρωμα ονομάζεται ακατάλληλο ολοκλήρωμα και ορίζεται με τη χρήση κατάλληλων διαδικασιών περιορισμού. για παράδειγμα:
∫
a
∞
f
(
x
)
d
x
=
lim
b
→
∞
[
∫
a
b
f
(
x
)
d
x
]
{\displaystyle \int _{a}^{\infty }f(x)\,dx=\lim _{b\to \infty }\left[\int _{a}^{b}f(x)\,dx\right]}
Μια σταθερά, όπως το π, που μπορεί να οριστεί από το ολοκλήρωμα μιας αλγεβρικής συνάρτησης σε ένα αλγεβρικό πεδίο, είναι γνωστή ως περίοδος.
Ακολουθεί ένας κατάλογος μερικών από τα πιο συνηθισμένα ή ενδιαφέροντα ορισμένα ολοκληρώματα[ 4] . Για έναν κατάλογο των αόριστων ολοκληρωμάτων βλέπε Κατάλογος αόριστων ολοκληρωμάτων.
∫
0
∞
d
x
1
+
x
p
=
π
/
p
sin
(
π
/
p
)
for
ℜ
(
p
)
>
1
{\displaystyle \int _{0}^{\infty }{\frac {dx}{1+x^{p}}}={\frac {\pi /p}{\sin(\pi /p)}}\quad {\text{for }}\Re (p)>1}
∫
0
∞
x
p
−
1
d
x
1
+
x
=
π
sin
(
p
π
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {x^{p-1}dx}{1+x}}={\frac {\pi }{\sin(p\pi )}}\quad {\text{for }}0<p<1}
∫
0
∞
x
m
d
x
x
n
+
a
n
=
π
a
m
−
n
+
1
n
sin
(
m
+
1
n
π
)
for
0
<
m
+
1
<
n
{\displaystyle \int _{0}^{\infty }{\frac {x^{m}dx}{x^{n}+a^{n}}}={\frac {\pi a^{m-n+1}}{n\sin \left({\dfrac {m+1}{n}}\pi \right)}}\quad {\text{for }}0<m+1<n}
∫
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 )}}\cdot {\frac {\sin(m\beta )}{\sin(\beta )}}}
∫
0
a
d
x
a
2
−
x
2
=
π
2
{\displaystyle \int _{0}^{a}{\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 \left({\dfrac {m+1}{n}}\right)\Gamma (p+1)}{n\Gamma \left({\dfrac {m+1}{n}}+p+1\right)}}}
∫
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
n
π
)
(
r
−
1
)
!
Γ
(
m
+
1
n
−
r
+
1
)
for
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 \left({\dfrac {m+1}{n}}\right)}{n\sin \left({\dfrac {m+1}{n}}\pi \right)(r-1)!\,\Gamma \left({\dfrac {m+1}{n}}-r+1\right)}}\quad {\text{for }}n(r-2)<m+1<nr}
∫
0
π
sin
(
m
x
)
sin
(
n
x
)
d
x
=
{
0
if
m
≠
n
π
2
if
m
=
n
for
m
,
n
positive integers
{\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}}\quad {\text{for }}m,n{\text{ positive integers}}}
∫
0
π
cos
(
m
x
)
cos
(
n
x
)
d
x
=
{
0
if
m
≠
n
π
2
if
m
=
n
for
m
,
n
positive integers
{\displaystyle \int _{0}^{\pi }\cos(mx)\cos(nx)dx={\begin{cases}0&{\text{if }}m\neq n\\\\{\dfrac {\pi }{2}}&{\text{if }}m=n\end{cases}}\quad {\text{for }}m,n{\text{ positive integers}}}
∫
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
for
m
,
n
integers
.
{\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}}\quad {\text{for }}m,n{\text{ integers}}.}
∫
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
for
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}}\cdot {\frac {\pi }{2}}\quad {\text{for }}m=1,2,3\ldots }
∫
0
x
sin
2
m
(
t
)
d
t
=
(
2
m
−
1
)
!
!
(
2
m
)
!
!
(
x
−
sin
(
x
)
cos
(
x
)
(
1
+
∑
k
=
1
∞
sin
2
k
(
x
)
(
2
k
)
!
!
(
2
k
+
1
)
!
!
)
)
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{x}\sin ^{2m}(t)dt={\frac {(2m-1)!!}{(2m)!!}}{\Biggl (}x-\sin(x)\cos(x){\Biggl (}1+\sum _{k=1}^{\infty }{\frac {\sin ^{2k}(x)(2k)!!}{(2k+1)!!}}{\Biggr )}{\Biggr )}\quad {\text{for }}m=1,2,3\ldots }
∫
0
x
cos
2
m
(
t
)
d
t
=
(
2
m
−
1
)
!
!
(
2
m
)
!
!
(
x
−
sin
(
x
)
cos
(
x
)
(
1
+
∑
k
=
1
∞
cos
2
k
(
x
)
(
2
k
)
!
!
(
2
k
+
1
)
!
!
)
)
for
m
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{x}\cos ^{2m}(t)dt={\frac {(2m-1)!!}{(2m)!!}}{\Biggl (}x-\sin(x)\cos(x){\Biggl (}1+\sum _{k=1}^{\infty }{\frac {\cos ^{2k}(x)(2k)!!}{(2k+1)!!}}{\Biggr )}{\Biggr )}\quad {\text{for }}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
)
for
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)}}\quad {\text{for }}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}}}
(βλ. ολοκλήρωμα Ντίριχλετ)
∫
0
∞
sin
p
x
cos
q
x
x
d
x
=
{
0
if
q
>
p
>
0
π
2
if
0
<
q
<
p
π
4
if
p
=
q
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\sin px\cos qx}{x}}\ dx={\begin{cases}0&{\text{ if }}q>p>0\\\\{\dfrac {\pi }{2}}&{\text{ if }}0<q<p\\\\{\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}}}\left(1-e^{-ma}\right)}
∫
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}\left({\dfrac {b}{a}}\right)}{\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
for
0
<
a
<
1
{\displaystyle \int _{0}^{2\pi }{\frac {dx}{1-2a\cos x+a^{2}}}={\frac {2\pi }{1-a^{2}}}\quad {\text{for }}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}{\dfrac {\pi }{a}}\ln \left|1+a\right|&{\text{if }}|a|<1\\\\{\dfrac {\pi }{a}}\ln \left|1+{\dfrac {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
for
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 {\text{for }}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
for
n
>
1
{\displaystyle \int _{0}^{\infty }\sin ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\sin {\frac {\pi }{2n}}\quad {\text{for }}n>1}
∫
0
∞
cos
a
x
n
=
1
n
a
1
/
n
Γ
(
1
n
)
cos
π
2
n
for
n
>
1
{\displaystyle \int _{0}^{\infty }\cos ax^{n}={\frac {1}{na^{1/n}}}\Gamma \left({\frac {1}{n}}\right)\cos {\frac {\pi }{2n}}\quad {\text{for }}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
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {\sin x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\sin \left({\dfrac {p\pi }{2}}\right)}}\quad {\text{for }}0<p<1}
∫
0
∞
cos
x
x
p
d
x
=
π
2
Γ
(
p
)
cos
(
p
π
2
)
for
0
<
p
<
1
{\displaystyle \int _{0}^{\infty }{\frac {\cos x}{x^{p}}}\ dx={\frac {\pi }{2\Gamma (p)\cos \left({\dfrac {p\pi }{2}}\right)}}\quad {\text{for }}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)}
∫
0
∞
x
e
−
x
d
x
=
1
2
π
{\displaystyle \int _{0}^{\infty }{\sqrt {x}}\,e^{-x}\,dx={\frac {1}{2}}{\sqrt {\pi }}}
(βλ. επίσης Συνάρτηση Γάμμα )
∫
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
−
cos
(
b
x
)
x
d
x
=
ln
b
a
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-\cos(bx)}{x}}\,dx=\ln {\frac {b}{a}}}
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π
a
for
a
>
0
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}\quad {\text{for }}a>0}
(το γκαουσιανό ολοκλήρωμα)
∫
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^{\left({\frac {-b^{2}}{4a}}\right)}}
∫
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
,
where
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^{\left({\frac {b^{2}-4ac}{4a}}\right)}\cdot \operatorname {erfc} {\frac {b}{2{\sqrt {a}}}},{\text{ where }}\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^{\left({\frac {b^{2}-4ac}{4a}}\right)}}
∫
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
2
e
−
a
x
2
d
x
=
1
4
π
a
3
for
a
>
0
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}}\quad {\text{for }}a>0}
∫
0
∞
x
2
n
e
−
a
x
2
d
x
=
2
n
−
1
2
a
∫
0
∞
x
2
(
n
−
1
)
e
−
a
x
2
d
x
=
(
2
n
−
1
)
!
!
2
n
+
1
π
a
2
n
+
1
=
(
2
n
)
!
n
!
2
2
n
+
1
π
a
2
n
+
1
for
a
>
0
,
n
=
1
,
2
,
3
…
{\displaystyle \int _{0}^{\infty }x^{2n}e^{-ax^{2}}\,dx={\frac {2n-1}{2a}}\int _{0}^{\infty }x^{2(n-1)}e^{-ax^{2}}\,dx={\frac {(2n-1)!!}{2^{n+1}}}{\sqrt {\frac {\pi }{a^{2n+1}}}}={\frac {(2n)!}{n!2^{2n+1}}}{\sqrt {\frac {\pi }{a^{2n+1}}}}\quad {\text{for }}a>0\ ,\ n=1,2,3\ldots }
(όπου !! είναι το διπλό παραγοντικό)
∫
0
∞
x
3
e
−
a
x
2
d
x
=
1
2
a
2
for
a
>
0
{\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2}}\,dx}={\frac {1}{2a^{2}}}\quad {\text{for }}a>0}
∫
0
∞
x
2
n
+
1
e
−
a
x
2
d
x
=
n
a
∫
0
∞
x
2
n
−
1
e
−
a
x
2
d
x
=
n
!
2
a
n
+
1
for
a
>
0
,
n
=
0
,
1
,
2
…
{\displaystyle \int _{0}^{\infty }x^{2n+1}e^{-ax^{2}}\,dx={\frac {n}{a}}\int _{0}^{\infty }x^{2n-1}e^{-ax^{2}}\,dx={\frac {n!}{2a^{n+1}}}\quad {\text{for }}a>0\ ,\ n=0,1,2\ldots }
∫
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 \left({\dfrac {m+1}{2}}\right)}{2a^{\left({\frac {m+1}{2}}\right)}}}}
∫
0
∞
e
(
−
a
x
2
−
b
x
2
)
d
x
=
1
2
π
a
e
−
2
a
b
{\displaystyle \int _{0}^{\infty }e^{\left(-ax^{2}-{\frac {b}{x^{2}}}\right)}\ 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
∞
x
n
e
x
+
1
d
x
=
n
!
⋅
(
2
n
−
1
2
n
)
ζ
(
n
+
1
)
{\displaystyle \int _{0}^{\infty }{\frac {x^{n}}{e^{x}+1}}\ dx=n!\cdot \left({\frac {2^{n}-1}{2^{n}}}\right)\zeta (n+1)}
∫
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 }
(όπου
γ
{\displaystyle \gamma }
είναι η σταθερά Όιλερ-Μαστσερόνι)
∫
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
−
1
2
x
2
d
x
=
(
2
n
+
1
)
!
2
n
n
!
2
π
for
n
=
0
,
1
,
2
,
…
{\displaystyle \int _{-\infty }^{\infty }x^{2(n+1)}e^{-{\frac {1}{2}}x^{2}}\,dx={\frac {(2n+1)!}{2^{n}n!}}{\sqrt {2\pi }}\quad {\text{for }}n=0,1,2,\ldots }
∫
0
1
x
m
(
ln
x
)
n
d
x
=
(
−
1
)
n
n
!
(
m
+
1
)
n
+
1
for
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 {\text{for }}m>-1,n=0,1,2,\ldots }
∫
1
∞
x
m
(
ln
x
)
n
d
x
=
(
−
1
)
n
+
1
n
!
(
m
+
1
)
n
+
1
for
m
<
−
1
,
n
=
0
,
1
,
2
,
…
{\displaystyle \int _{1}^{\infty }x^{m}(\ln x)^{n}\,dx={\frac {(-1)^{n+1}n!}{(m+1)^{n+1}}}\quad {\text{for }}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
)
for
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 {\text{for }}a,b>0}
∫
0
∞
ln
x
x
2
+
a
2
d
x
=
π
ln
a
2
a
for
a
>
0
{\displaystyle \int _{0}^{\infty }{\frac {\ln x}{x^{2}+a^{2}}}\ dx={\frac {\pi \ln a}{2a}}\quad {\text{for }}a>0}
∫
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}}\cdot {\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}}}}
∫
0
∞
x
2
n
+
1
sinh
a
x
d
x
=
c
2
n
+
1
(
π
a
)
2
(
n
+
1
)
,
c
2
n
+
1
=
(
−
1
)
n
2
(
1
2
−
∑
k
=
0
n
−
1
(
−
1
)
k
(
2
n
+
1
2
k
+
1
)
c
2
k
+
1
)
,
c
1
=
1
4
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n+1}}{\sinh ax}}\ dx=c_{2n+1}\left({\frac {\pi }{a}}\right)^{2(n+1)},\quad c_{2n+1}={\frac {(-1)^{n}}{2}}\left({\frac {1}{2}}-\sum _{k=0}^{n-1}(-1)^{k}{2n+1 \choose 2k+1}c_{2k+1}\right),\quad c_{1}={\frac {1}{4}}}
∫
0
∞
1
cosh
a
x
d
x
=
π
2
a
{\displaystyle \int _{0}^{\infty }{\frac {1}{\cosh ax}}\ dx={\frac {\pi }{2a}}}
∫
0
∞
x
2
n
cosh
a
x
d
x
=
d
2
n
(
π
a
)
2
n
+
1
,
d
2
n
=
(
−
1
)
n
2
(
1
4
n
−
∑
k
=
0
n
−
1
(
−
1
)
k
(
2
n
2
k
)
d
2
k
)
,
d
0
=
1
2
{\displaystyle \int _{0}^{\infty }{\frac {x^{2n}}{\cosh ax}}\ dx=d_{2n}\left({\frac {\pi }{a}}\right)^{2n+1},\quad d_{2n}={\frac {(-1)^{n}}{2}}\left({\frac {1}{4^{n}}}-\sum _{k=0}^{n-1}(-1)^{k}{2n \choose 2k}d_{2k}\right),\quad d_{0}={\frac {1}{2}}}
∫
0
∞
f
(
a
x
)
−
f
(
b
x
)
x
d
x
=
(
lim
x
→
0
f
(
x
)
−
lim
x
→
∞
f
(
x
)
)
ln
(
b
a
)
{\displaystyle \int _{0}^{\infty }{\frac {f(ax)-f(bx)}{x}}\ dx=\left(\lim _{x\to 0}f(x)-\lim _{x\to \infty }f(x)\right)\ln \left({\frac {b}{a}}\right)}
ισχύει εάν το ολοκλήρωμα υπάρχει και το
f
′
(
x
)
{\displaystyle f'(x)}
είναι συνεχές.
Zwillinger, Daniel· Jeffrey, Alan (23 Φεβρουαρίου 2007). Table of Integrals, Series, and Products . Elsevier. ISBN 978-0-08-047111-2 .
Baeza, Ricardo (2004). Algebraic and Arithmetic Theory of Quadratic Forms: Proceedings of the International Conference on the Algebraic and Arithmetic Theory of Quadratic Forms, December 11-18, 2002, Universidad de Talca, Chile . American Mathematical Soc. ISBN 978-0-8218-3441-1 .
Board, Oswaal Editorial (9 Σεπτεμβρίου 2024). Oswaal ISC 10 Sample Question Papers Class 12 (Set of 5 Books) Physics, Chemistry, Maths, English Paper 1 & 2 For 2025 Board Exam (Based On The Latest CISCE/ICSE Specimen Paper) . Oswaal Books. ISBN 978-93-6239-374-6 .
Álvarez-Cónsul, Luis· Burgos-Gil, José Ignacio (24 Σεπτεμβρίου 2015). Feynman Amplitudes, Periods and Motives . American Mathematical Soc. ISBN 978-1-4704-2247-9 .
Bond, Thomas· Hughes, Chris (4 Νοεμβρίου 2013). A-level Mathematics Effective Guide (Concise) (Yellowreef) . Yellowreef Limited.
Board, Oswaal Editorial (9 Σεπτεμβρίου 2024). Oswaal ISC 10 Sample Question Papers Class 12 (Set of 5 Books) Physics, Chemistry, Maths, English Paper 1 & 2 For 2025 Board Exam (Based On The Latest CISCE/ICSE Specimen Paper) . Oswaal Books. ISBN 978-93-6239-374-6 .
Experts, Disha (27 Οκτωβρίου 2021). Guide to Indian Navy Senior Secondary Recruit (SSR) & Artificer Apprentice (AA) Exam 2021-22 . Disha Publications. ISBN 978-93-91551-67-4 .
Gerdt, Vladimir P.· Koepf, Wolfram (10 Σεπτεμβρίου 2015). Computer Algebra in Scientific Computing: 17th International Workshop, CASC 2015, Aachen, Germany, September 14-18, 2015, Proceedings . Springer. ISBN 978-3-319-24021-3 .
Wilson, R. L. (9 Μαρτίου 2013). Much Ado About Calculus: A Modern Treatment with Applications Prepared for Use with the Computer . Springer Science & Business Media. ISBN 978-1-4615-9644-8 .
Dua, Amita· Singh, Chayannika (2 Αυγούστου 2024). Quantum Chemistry: Classical to Computational . Taylor & Francis. ISBN 978-1-040-10058-5 .
«Derivation of Logarithmic and Logarithmic Hyperbolic Tangent Integrals Expressed in Terms of Special Functions». Mathematics 8 (687): 687. 2020. doi :10.3390/math8050687 .
«A Definite Integral Involving the Logarithmic Function in Terms of the Lerch Function». Mathematics 7 (1148): 1148. 2019. doi :10.3390/math7121148 .
«Definite Integral of Arctangent and Polylogarithmic Functions Expressed as a Series». Mathematics 7 (1099): 1099. 2019. doi :10.3390/math7111099 .
«Eigenschaften Einiger Bestimmten Integrale». Hof, K.K., Ed. . 1861.
Spiegel, Murray R.· Lipschutz, Seymour· Liu, John (2009). Mathematical handbook of formulas and tables (3rd έκδοση). McGraw-Hill . ISBN 978-0071548557 .
Zwillinger, Daniel (2003). CRC standard mathematical tables and formulae (32nd έκδοση). CRC Press . ISBN 978-143983548-7 .
Moll, Victor Hugo (27 Οκτωβρίου 2015). Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume II . Series: Monographs and Research Notes in Mathematics . II (1 έκδοση). Chapman and Hall /CRC Press . ISBN 978-1-48225-653-6 . Ανακτήθηκε στις 12 Φεβρουαρίου 2016 .
Toyesh Prakash Sharma , https://www.isroset.org/pdf_paper_view.php?paper_id=2214&7-ISROSET-IJSRMSS-05130.pdf
Segal, I. E.· Kunze, R. A. (6 Δεκεμβρίου 2012). Integrals and Operators . Springer Science & Business Media. ISBN 978-3-642-66693-3 .
«Integrals of Particular Functions: Proofs with Solved Examples» . allen.in . Ανακτήθηκε στις 8 Μαρτίου 2025 .