فهرست‌های انتگرال‌ها
انتگرال‌گیری یکی از دو عمل اصلی در حساب دیفرانسیل و انتگرال است. برخلاف دیفرانسیل که قواعد ساده‌ای دارد که با استفاده از دیفرانسیل تابع‌های سادهٔ مشابه یک تابع پیچیده، می‌توان دیفرانسیل آن را یافت، انتگرال‌ها این‌گونه نیستند. از این‌رو جدول‌های انتگرال بسیار کاربردی هستند. این صفحه فهرست برخی از پرکاربردترین انتگرال‌ها را دربردارد.
برای جزئیات بیشتر صفحه‌های زیر را ببینید:
∫ ∫ -->
1
x
d
x
=
ln
-->
|
x
|
+
C
{\displaystyle \int {1 \over x}\,dx=\ln \left|x\right|+C}
∫ ∫ -->
1
x
d
x
=
ln
-->
|
x
|
+
{
A
if
x
>
0
;
B
if
x
<
0.
{\displaystyle \int {1 \over x}\,dx=\ln |x|+{\begin{cases}A&{\text{if }}x>0;\\B&{\text{if }}x<0.\end{cases}}}
انتگرال‌های بیشتر: فهرست انتگرال توابع گویا این تابع‌ها در نقطهٰ صفر برای a <-۱ یک تکینگی دارند.
∫ ∫ -->
k
d
x
=
k
x
+
C
{\displaystyle \int k\,dx=kx+C}
∫ ∫ -->
x
a
d
x
=
x
a
+
1
a
+
1
+
C
{\displaystyle \int x^{a}\,dx={\frac {x^{a+1}}{a+1}}+C}
Cavalieri's quadrature formula )
∫ ∫ -->
(
a
x
+
b
)
n
d
x
=
(
a
x
+
b
)
n
+
1
a
(
n
+
1
)
+
C
(for
n
≠ ≠ -->
− − -->
1
)
{\displaystyle \int (ax+b)^{n}dx={\frac {(ax+b)^{n+1}}{a(n+1)}}+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
∫ ∫ -->
1
x
d
x
=
ln
-->
|
x
|
+
C
{\displaystyle \int {1 \over x}\,dx=\ln \left|x\right|+C}
∫ ∫ -->
c
a
x
+
b
d
x
=
c
a
ln
-->
|
a
x
+
b
|
+
C
{\displaystyle \int {\frac {c}{ax+b}}dx={\frac {c}{a}}\ln \left|ax+b\right|+C}
انتگرال‌های بیشتر: فهرست انتگرال تابع‌های نمایی
∫ ∫ -->
e
x
d
x
=
e
x
+
C
{\displaystyle \int e^{x}\,dx=e^{x}+C}
∫ ∫ -->
a
x
d
x
=
a
x
ln
-->
a
− − -->
C
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln a}}-C}
انتگرال‌های بیشتر: فهرست انتگرال توابع لگاریتمی
∫ ∫ -->
ln
-->
x
d
x
=
x
ln
-->
x
− − -->
x
+
C
{\displaystyle \int \ln x\,dx=x\ln x-x+C}
∫ ∫ -->
log
a
-->
x
d
x
=
x
log
a
-->
x
− − -->
x
ln
-->
a
+
C
{\displaystyle \int \log _{a}x\,dx=x\log _{a}x-{\frac {x}{\ln a}}+C}
انتگرال‌های بیشتر: فهرست انتگرال توابع مثلثاتی
∫ ∫ -->
sin
-->
x
d
x
=
− − -->
cos
-->
x
+
C
{\displaystyle \int \sin {x}\,dx=-\cos {x}+C}
∫ ∫ -->
cos
-->
x
d
x
=
sin
-->
x
+
C
{\displaystyle \int \cos {x}\,dx=\sin {x}+C}
∫ ∫ -->
tan
-->
x
d
x
=
− − -->
ln
-->
|
cos
-->
x
|
+
C
=
ln
-->
|
sec
-->
x
|
+
C
{\displaystyle \int \tan {x}\,dx=-\ln {\left|\cos {x}\right|}+C=\ln {\left|\sec {x}\right|}+C}
∫ ∫ -->
cot
-->
x
d
x
=
ln
-->
|
sin
-->
x
|
+
C
{\displaystyle \int \cot {x}\,dx=\ln {\left|\sin {x}\right|}+C}
∫ ∫ -->
sec
-->
x
d
x
=
ln
-->
|
sec
-->
x
+
tan
-->
x
|
+
C
{\displaystyle \int \sec {x}\,dx=\ln {\left|\sec {x}+\tan {x}\right|}+C}
∫ ∫ -->
csc
-->
x
d
x
=
− − -->
ln
-->
|
csc
-->
x
+
cot
-->
x
|
+
C
{\displaystyle \int \csc {x}\,dx=-\ln {\left|\csc {x}+\cot {x}\right|}+C}
∫ ∫ -->
sec
2
-->
x
d
x
=
tan
-->
x
+
C
{\displaystyle \int \sec ^{2}x\,dx=\tan x+C}
∫ ∫ -->
csc
2
-->
x
d
x
=
− − -->
cot
-->
x
+
C
{\displaystyle \int \csc ^{2}x\,dx=-\cot x+C}
∫ ∫ -->
sec
-->
x
tan
-->
x
d
x
=
sec
-->
x
+
C
{\displaystyle \int \sec {x}\,\tan {x}\,dx=\sec {x}+C}
∫ ∫ -->
csc
-->
x
cot
-->
x
d
x
=
− − -->
csc
-->
x
+
C
{\displaystyle \int \csc {x}\,\cot {x}\,dx=-\csc {x}+C}
∫ ∫ -->
sin
2
-->
x
d
x
=
1
2
(
x
− − -->
sin
-->
2
x
2
)
+
C
=
1
2
(
x
− − -->
sin
-->
x
cos
-->
x
)
+
C
{\displaystyle \int \sin ^{2}x\,dx={\frac {1}{2}}\left(x-{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x-\sin x\cos x)+C}
∫ ∫ -->
cos
2
-->
x
d
x
=
1
2
(
x
+
sin
-->
2
x
2
)
+
C
=
1
2
(
x
+
sin
-->
x
cos
-->
x
)
+
C
{\displaystyle \int \cos ^{2}x\,dx={\frac {1}{2}}\left(x+{\frac {\sin 2x}{2}}\right)+C={\frac {1}{2}}(x+\sin x\cos x)+C}
∫ ∫ -->
sec
3
-->
x
d
x
=
1
2
sec
-->
x
tan
-->
x
+
1
2
ln
-->
|
sec
-->
x
+
tan
-->
x
|
+
C
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln |\sec x+\tan x|+C}
(ببینید انتگرال مکعب سکانت )
∫ ∫ -->
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}
انتگرال‌های بیشتر: فهرست انتگرال توابع وارون مثلثانی
∫ ∫ -->
arcsin
-->
x
d
x
=
x
arcsin
-->
x
+
1
− − -->
x
2
+
C
,
for
|
x
|
≤ ≤ -->
+
1
{\displaystyle \int \arcsin {x}\,dx=x\arcsin {x}+{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq +1}
∫ ∫ -->
arccos
-->
x
d
x
=
x
arccos
-->
x
− − -->
1
− − -->
x
2
+
C
,
for
|
x
|
≤ ≤ -->
+
1
{\displaystyle \int \arccos {x}\,dx=x\arccos {x}-{\sqrt {1-x^{2}}}+C,{\text{ for }}\vert x\vert \leq +1}
∫ ∫ -->
arctan
-->
x
d
x
=
x
arctan
-->
x
− − -->
1
2
ln
-->
|
1
+
x
2
|
+
C
,
for all real
x
{\displaystyle \int \arctan {x}\,dx=x\arctan {x}-{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x}
∫ ∫ -->
arccot
-->
x
d
x
=
x
arccot
-->
x
+
1
2
ln
-->
|
1
+
x
2
|
+
C
,
for all real
x
{\displaystyle \int \operatorname {arccot} {x}\,dx=x\operatorname {arccot} {x}+{\frac {1}{2}}\ln {\vert 1+x^{2}\vert }+C,{\text{ for all real }}x}
∫ ∫ -->
arcsec
-->
x
d
x
=
x
arcsec
-->
x
− − -->
ln
-->
|
x
(
1
+
1
− − -->
x
− − -->
2
)
|
+
C
,
for
|
x
|
≥ ≥ -->
1
{\displaystyle \int \operatorname {arcsec} {x}\,dx=x\operatorname {arcsec} {x}-\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1}
∫ ∫ -->
arccsc
-->
x
d
x
=
x
arccsc
-->
x
+
ln
-->
|
x
(
1
+
1
− − -->
x
− − -->
2
)
|
+
C
,
for
|
x
|
≥ ≥ -->
1
{\displaystyle \int \operatorname {arccsc} {x}\,dx=x\operatorname {arccsc} {x}+\ln \left\vert x\,\left(1+{\sqrt {1-x^{-2}}}\,\right)\right\vert +C,{\text{ for }}\vert x\vert \geq 1}
انتگرال‌های بیشتر: فهرست انتگرال تابع‌های هیپربولیک
∫ ∫ -->
sinh
-->
x
d
x
=
cosh
-->
x
+
C
{\displaystyle \int \sinh x\,dx=\cosh x+C}
∫ ∫ -->
cosh
-->
x
d
x
=
sinh
-->
x
+
C
{\displaystyle \int \cosh x\,dx=\sinh x+C}
∫ ∫ -->
tanh
-->
x
d
x
=
ln
(
cosh
-->
x
)
+
C
{\displaystyle \int \tanh x\,dx=\ln \,(\cosh x)+C}
∫ ∫ -->
coth
-->
x
d
x
=
ln
-->
|
sinh
-->
x
|
+
C
,
for
x
≠ ≠ -->
0
{\displaystyle \int \coth x\,dx=\ln |\sinh x|+C,{\text{ for }}x\neq 0}
∫ ∫ -->
sech
x
d
x
=
arctan
(
sinh
-->
x
)
+
C
{\displaystyle \int \operatorname {sech} \,x\,dx=\arctan \,(\sinh x)+C}
∫ ∫ -->
csch
x
d
x
=
ln
-->
|
tanh
-->
x
2
|
+
C
,
for
x
≠ ≠ -->
0
{\displaystyle \int \operatorname {csch} \,x\,dx=\ln \left|\tanh {x \over 2}\right|+C,{\text{ for }}x\neq 0}
انتگرال‌های بیشتر: فهرست انتگرال تابع‌های وارون هیپربولیک
∫ ∫ -->
arsinh
x
d
x
=
x
arsinh
x
− − -->
x
2
+
1
+
C
,
for all real
x
{\displaystyle \int \operatorname {arsinh} \,x\,dx=x\,\operatorname {arsinh} \,x-{\sqrt {x^{2}+1}}+C,{\text{ for all real }}x}
∫ ∫ -->
arcosh
x
d
x
=
x
arcosh
x
− − -->
x
2
− − -->
1
+
C
,
for
x
≥ ≥ -->
1
{\displaystyle \int \operatorname {arcosh} \,x\,dx=x\,\operatorname {arcosh} \,x-{\sqrt {x^{2}-1}}+C,{\text{ for }}x\geq 1}
∫ ∫ -->
artanh
x
d
x
=
x
artanh
x
+
ln
-->
(
1
− − -->
x
2
)
2
+
C
,
for
|
x
|
<
1
{\displaystyle \int \operatorname {artanh} \,x\,dx=x\,\operatorname {artanh} \,x+{\frac {\ln \left(\,1-x^{2}\right)}{2}}+C,{\text{ for }}\vert x\vert <1}
∫ ∫ -->
arcoth
x
d
x
=
x
arcoth
x
+
ln
-->
(
x
2
− − -->
1
)
2
+
C
,
for
|
x
|
>
1
{\displaystyle \int \operatorname {arcoth} \,x\,dx=x\,\operatorname {arcoth} \,x+{\frac {\ln \left(x^{2}-1\right)}{2}}+C,{\text{ for }}\vert x\vert >1}
∫ ∫ -->
arsech
x
d
x
=
x
arsech
x
+
arcsin
-->
x
+
C
,
for
0
<
x
≤ ≤ -->
1
{\displaystyle \int \operatorname {arsech} \,x\,dx=x\,\operatorname {arsech} \,x+\arcsin x+C,{\text{ for }}0<x\leq 1}
∫ ∫ -->
arcsch
x
d
x
=
x
arcsch
x
+
|
arsinh
x
|
+
C
,
for
x
≠ ≠ -->
0
{\displaystyle \int \operatorname {arcsch} \,x\,dx=x\,\operatorname {arcsch} \,x+\vert \operatorname {arsinh} \,x\vert +C,{\text{ for }}x\neq 0}
∫ ∫ -->
cos
-->
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
a
sin
-->
a
x
+
b
cos
-->
a
x
)
+
C
{\displaystyle \int \cos ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(a\sin ax+b\cos ax\right)+C}
∫ ∫ -->
sin
-->
a
x
e
b
x
d
x
=
e
b
x
a
2
+
b
2
(
b
sin
-->
a
x
− − -->
a
cos
-->
a
x
)
+
C
{\displaystyle \int \sin ax\,e^{bx}\,dx={\frac {e^{bx}}{a^{2}+b^{2}}}\left(b\sin ax-a\cos ax\right)+C}
∫ ∫ -->
cos
-->
a
x
cosh
-->
b
x
d
x
=
1
a
2
+
b
2
(
a
sin
-->
a
x
cosh
-->
b
x
+
b
cos
-->
a
x
sinh
-->
b
x
)
+
C
{\displaystyle \int \cos ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(a\sin ax\,\cosh bx+b\cos ax\,\sinh bx\right)+C}
∫ ∫ -->
sin
-->
a
x
cosh
-->
b
x
d
x
=
1
a
2
+
b
2
(
b
sin
-->
a
x
sinh
-->
b
x
− − -->
a
cos
-->
a
x
cosh
-->
b
x
)
+
C
{\displaystyle \int \sin ax\,\cosh bx\,dx={\frac {1}{a^{2}+b^{2}}}\left(b\sin ax\,\sinh bx-a\cos ax\,\cosh bx\right)+C}
∫ ∫ -->
|
(
a
x
+
b
)
n
|
d
x
=
(
a
x
+
b
)
n
+
2
a
(
n
+
1
)
|
a
x
+
b
|
+
C
[
n
is odd, and
n
≠ ≠ -->
− − -->
1
]
{\displaystyle \int \left|(ax+b)^{n}\right|\,dx={(ax+b)^{n+2} \over a(n+1)\left|ax+b\right|}+C\,\,[\,n{\text{ is odd, and }}n\neq -1\,]}
∫ ∫ -->
|
sin
-->
a
x
|
d
x
=
− − -->
1
a
|
sin
-->
a
x
|
cot
-->
a
x
+
C
{\displaystyle \int \left|\sin {ax}\right|\,dx={-1 \over a}\left|\sin {ax}\right|\cot {ax}+C}
∫ ∫ -->
|
cos
-->
a
x
|
d
x
=
1
a
|
cos
-->
a
x
|
tan
-->
a
x
+
C
{\displaystyle \int \left|\cos {ax}\right|\,dx={1 \over a}\left|\cos {ax}\right|\tan {ax}+C}
∫ ∫ -->
|
tan
-->
a
x
|
d
x
=
tan
-->
(
a
x
)
[
− − -->
ln
-->
|
cos
-->
a
x
|
]
a
|
tan
-->
a
x
|
+
C
{\displaystyle \int \left|\tan {ax}\right|\,dx={\tan(ax)[-\ln \left|\cos {ax}\right|] \over a\left|\tan {ax}\right|}+C}
∫ ∫ -->
|
csc
-->
a
x
|
d
x
=
− − -->
ln
-->
|
csc
-->
a
x
+
cot
-->
a
x
|
sin
-->
a
x
a
|
sin
-->
a
x
|
+
C
{\displaystyle \int \left|\csc {ax}\right|\,dx={-\ln \left|\csc {ax}+\cot {ax}\right|\sin {ax} \over a\left|\sin {ax}\right|}+C}
∫ ∫ -->
|
sec
-->
a
x
|
d
x
=
ln
-->
|
sec
-->
a
x
+
tan
-->
a
x
|
cos
-->
a
x
a
|
cos
-->
a
x
|
+
C
{\displaystyle \int \left|\sec {ax}\right|\,dx={\ln \left|\sec {ax}+\tan {ax}\right|\cos {ax} \over a\left|\cos {ax}\right|}+C}
∫ ∫ -->
|
cot
-->
a
x
|
d
x
=
tan
-->
(
a
x
)
[
ln
-->
|
sin
-->
a
x
|
]
a
|
tan
-->
a
x
|
+
C
{\displaystyle \int \left|\cot {ax}\right|\,dx={\tan(ax)[\ln \left|\sin {ax}\right|] \over a\left|\tan {ax}\right|}+C}
∫ ∫ -->
0
∞ ∞ -->
x
e
− − -->
x
d
x
=
1
2
π π -->
{\displaystyle \int _{0}^{\infty }{{\sqrt {x}}\,e^{-x}\,dx}={\frac {1}{2}}{\sqrt {\pi }}}
تابع گاما )
∫ ∫ -->
0
∞ ∞ -->
e
− − -->
a
x
2
d
x
=
1
2
π π -->
a
{\displaystyle \int _{0}^{\infty }{e^{-ax^{2}}\,dx}={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}}
انتگرال گاوسی )
∫ ∫ -->
0
∞ ∞ -->
x
2
e
− − -->
a
x
2
d
x
=
1
4
π π -->
a
3
{\displaystyle \int _{0}^{\infty }{x^{2}e^{-ax^{2}}\,dx}={\frac {1}{4}}{\sqrt {\frac {\pi }{a^{3}}}}}
∫ ∫ -->
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 π -->
a
2
n
+
1
=
(
2
n
)
!
n
!
2
2
n
+
1
π π -->
a
2
n
+
1
{\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}}}}}
فاکتوریل .
∫ ∫ -->
0
∞ ∞ -->
x
3
e
− − -->
a
x
2
d
x
=
1
2
a
2
{\displaystyle \int _{0}^{\infty }{x^{3}e^{-ax^{2}}\,dx}={\frac {1}{2a^{2}}}}
∫ ∫ -->
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
{\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}}}}
∫ ∫ -->
0
∞ ∞ -->
x
e
x
− − -->
1
d
x
=
π π -->
2
6
{\displaystyle \int _{0}^{\infty }{{\frac {x}{e^{x}-1}}\,dx}={\frac {\pi ^{2}}{6}}}
Bernoulli number )
∫ ∫ -->
0
∞ ∞ -->
x
3
e
x
− − -->
1
d
x
=
π 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 π -->
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin {x}}{x}}\,dx={\frac {\pi }{2}}}
تابع سینک و انتگرال سینوسی )
∫ ∫ -->
0
∞ ∞ -->
sin
2
-->
x
x
2
d
x
=
π π -->
2
{\displaystyle \int _{0}^{\infty }{\frac {\sin ^{2}{x}}{x^{2}}}\,dx={\frac {\pi }{2}}}
∫ ∫ -->
0
π π -->
2
sin
n
-->
x
d
x
=
∫ ∫ -->
0
π π -->
2
cos
n
-->
x
d
x
=
1
⋅ ⋅ -->
3
⋅ ⋅ -->
5
⋅ ⋅ -->
⋯ ⋯ -->
⋅ ⋅ -->
(
n
− − -->
1
)
2
⋅ ⋅ -->
4
⋅ ⋅ -->
6
⋅ ⋅ -->
⋯ ⋯ -->
⋅ ⋅ -->
n
π 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 is an even integer and
n
≥ ≥ -->
2
{\displaystyle \scriptstyle {n\geq 2}}
∫ ∫ -->
0
π π -->
2
sin
n
-->
x
d
x
=
∫ ∫ -->
0
π 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 -->
π π -->
cos
-->
(
α α -->
x
)
cos
n
-->
(
β β -->
x
)
d
x
=
{
2
π π -->
2
n
(
n
m
)
|
α α -->
|
=
|
β 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 α -->
,
β β -->
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β β -->
≠ ≠ -->
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥ ≥ -->
0
{\displaystyle \scriptstyle m,n\geq 0}
∫ ∫ -->
− − -->
π π -->
π π -->
sin
-->
(
α α -->
x
)
cos
n
-->
(
β β -->
x
)
d
x
=
0
{\displaystyle \int _{-\pi }^{\pi }\sin(\alpha x)\cos ^{n}(\beta x)dx=0}
α α -->
,
β β -->
{\displaystyle \scriptstyle \alpha ,\beta }
n
{\displaystyle \scriptstyle n}
تقارن )
∫ ∫ -->
− − -->
π π -->
π π -->
sin
-->
(
α α -->
x
)
sin
n
-->
(
β β -->
x
)
d
x
=
{
(
− − -->
1
)
(
n
+
1
)
/
2
(
− − -->
1
)
m
2
π π -->
2
n
(
n
m
)
n
odd
,
α α -->
=
β 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 α -->
,
β β -->
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β β -->
≠ ≠ -->
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥ ≥ -->
0
{\displaystyle \scriptstyle m,n\geq 0}
Binomial coefficient )
∫ ∫ -->
− − -->
π π -->
π π -->
cos
-->
(
α α -->
x
)
sin
n
-->
(
β β -->
x
)
d
x
=
{
(
− − -->
1
)
n
/
2
(
− − -->
1
)
m
2
π π -->
2
n
(
n
m
)
n
even
,
|
α α -->
|
=
|
β 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 α -->
,
β β -->
,
m
,
n
{\displaystyle \scriptstyle \alpha ,\beta ,m,n}
β β -->
≠ ≠ -->
0
{\displaystyle \scriptstyle \beta \neq 0}
m
,
n
≥ ≥ -->
0
{\displaystyle \scriptstyle m,n\geq 0}
Binomial coefficient )
∫ ∫ -->
− − -->
∞ ∞ -->
∞ ∞ -->
e
− − -->
(
a
x
2
+
b
x
+
c
)
d
x
=
π 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]}
تابع نمایی
e
u
{\displaystyle e^{u}}
a
>
0
{\displaystyle a>0}
∫ ∫ -->
0
∞ ∞ -->
x
z
− − -->
1
e
− − -->
x
d
x
=
Γ Γ -->
(
z
)
{\displaystyle \int _{0}^{\infty }x^{z-1}\,e^{-x}\,dx=\Gamma (z)}
Γ Γ -->
(
z
)
{\displaystyle \Gamma (z)}
تابع گاما )
∫ ∫ -->
0
1
x
m
− − -->
1
(
1
− − -->
x
)
n
− − -->
1
d
x
=
Γ Γ -->
(
m
)
Γ Γ -->
(
n
)
Γ Γ -->
(
m
+
n
)
{\displaystyle \int _{0}^{1}x^{m-1}(1-x)^{n-1}dx={\frac {\Gamma (m)\Gamma (n)}{\Gamma (m+n)}}}
تابع بتا )
∫ ∫ -->
0
2
π π -->
e
x
cos
-->
θ θ -->
d
θ θ -->
=
2
π 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)}
تابع بسل of the first kind)
∫ ∫ -->
0
2
π π -->
e
x
cos
-->
θ θ -->
+
y
sin
-->
θ θ -->
d
θ θ -->
=
2
π 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 -->
)
− − -->
(
ν ν -->
+
1
)
/
2
d
x
=
ν ν -->
π π -->
Γ Γ -->
(
ν ν -->
/
2
)
Γ Γ -->
(
(
ν 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 ν -->
>
0
{\displaystyle \nu >0\,}
تابع چگالی احتمال of the توزیع تی-استیودنت )
The method of exhaustion provides a formula for the general case when no antiderivative exists:
∫ ∫ -->
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!}
Start by using the substitution
x
=
artanh
t
{\displaystyle x=\operatorname {artanh} \,t}
I
p
=
∫ ∫ -->
0
1
[
ln
-->
(
1
/
x
)
]
p
d
x
=
∫ ∫ -->
0
∞ ∞ -->
[
ln
-->
(
1
/
artanh
t
)
]
p
d
t
1
− − -->
t
2
{\displaystyle I_{p}=\int _{0}^{1}[\ln(1/x)]^{p}\;\mathrm {d} x=\int _{0}^{\infty }\left[\ln(1/\operatorname {artanh} \,t)\right]^{p}\;{\frac {\mathrm {d} t}{1-t^{2}}}}
This brings the integral to the general form
I
n
=
∫ ∫ -->
a
b
(
ln
-->
f
)
n
f
′
d
t
{\displaystyle I_{n}=\int _{a}^{b}(\ln f)^{n}f^{'}\;\mathrm {d} t}
which after integration by parts yields
[
f
(
ln
-->
f
)
n
]
a
b
− − -->
n
∫ ∫ -->
a
b
(
ln
-->
f
)
n
− − -->
1
f
′
d
t
{\displaystyle \left[f(\ln f)^{n}\right]_{a}^{b}-n\int _{a}^{b}(\ln f)^{n-1}f^{'}\;\mathrm {d} t}
and provided the first term vanishes at the end points, we get the recurrence relation
I
n
=
− − -->
n
I
n
− − -->
1
{\displaystyle I_{n}=-n\,I_{n-1}}
which upon computation gives
I
n
=
(
− − -->
1
)
n
n
!
{\displaystyle I_{n}=(-1)^{n}\,n!}
Applying to our integral, we notice that
[
ln
-->
(
1
/
x
)
]
p
=
(
− − -->
1
)
p
[
ln
-->
(
x
)
]
p
{\displaystyle [\ln(1/x)]^{p}=(-1)^{p}\;[\ln(x)]^{p}}
Hence the final answer is:
I
p
=
(
− − -->
1
)
p
(
− − -->
1
)
p
p
!
=
p
!
{\displaystyle I_{p}=(-1)^{p}\,(-1)^{p}\,p!=p!}
M. Abramowitz and I.A. Stegun , editors. Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables I.S. Gradshteyn (И. С. Градштейн), I.M. Ryzhik (И. М. Рыжик); Alan Jeffrey, Daniel Zwillinger, editors. Table of Integrals, Series, and Products , seventh edition. Academic Press, 2007. ISBN 978-0-12-373637-6 . Errata. (Several previous editions as well.)
A.P. Prudnikov (А. П. Прудников), Yu.A. Brychkov (Ю. А. Брычков), O.I. Marichev (О. И. Маричев). Integrals and Series . First edition (Russian), volume 1–5, Nauka , 1981−1986. First edition (English, translated from the Russian by N.M. Queen), volume 1–5, Gordon & Breach Science Publishers/انتشارات سی‌آرسی ، 1988–1992, شابک ۲−۸۸۱۲۴−۰۹۷−۶ . Second revised edition (Russian), volume 1–3, Fiziko-Matematicheskaya Literatura, 2003.
Yu.A. Brychkov (Ю. А. Брычков), Handbook of Special Functions: Derivatives, Integrals, Series and Other Formulas . Russian edition, Fiziko-Matematicheskaya Literatura, 2006. English edition, Chapman & Hall/CRC Press, 2008, ISBN 1-58488-956-X .
Daniel Zwillinger. CRC Standard Mathematical Tables and Formulae , 31st edition. Chapman & Hall/CRC Press, 2002. ISBN 1-58488-291-3 . (Many earlier editions as well.)
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) [English translation of Integraltafeln ]
David Bierens de Haan, Nouvelles Tables d'Intégrales définies (Engels, Leiden, 1862)
Benjamin O. Pierce A short table of integrals - revised edition (Ginn & co. , Boston, 1899)
![{\displaystyle \int \left|(ax+b)^{n}\right|\,dx={(ax+b)^{n+2} \over a(n+1)\left|ax+b\right|}+C\,\,[\,n{\text{ is odd, and }}n\neq -1\,]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9db551fb843fee76fb6a9325999ab8c0515a8d1f)
![{\displaystyle \int \left|\tan {ax}\right|\,dx={\tan(ax)[-\ln \left|\cos {ax}\right|] \over a\left|\tan {ax}\right|}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/138aa84438bbb1353aa6d046f7987681f7223958)
![{\displaystyle \int \left|\cot {ax}\right|\,dx={\tan(ax)[\ln \left|\sin {ax}\right|] \over a\left|\tan {ax}\right|}+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b8bc94eb6f71b1b8b2f126a0dc1182e080ee756)
![{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\frac {\pi }{a}}}\exp \left[{\frac {b^{2}-4ac}{4a}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9aac3de5d20ec70a35c8d52bafaa6a82c148b566)
![{\displaystyle \exp[u]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e3cee4965602d51126500dce76d424788844c7d)
![{\displaystyle \int _{0}^{1}[\ln(1/x)]^{p}\,dx=p!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e5028566f81ea22ae449458f859edfc08a532cd)
![{\displaystyle I_{p}=\int _{0}^{1}[\ln(1/x)]^{p}\;\mathrm {d} x=\int _{0}^{\infty }\left[\ln(1/\operatorname {artanh} \,t)\right]^{p}\;{\frac {\mathrm {d} t}{1-t^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f5a721b0b712e4ab3daebe9614ab3e50c371e5f1)
![{\displaystyle \left[f(\ln f)^{n}\right]_{a}^{b}-n\int _{a}^{b}(\ln f)^{n-1}f^{'}\;\mathrm {d} t}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e047b074c3f80b8471b2708f80e1acc6a45d66e)
![{\displaystyle [\ln(1/x)]^{p}=(-1)^{p}\;[\ln(x)]^{p}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f4125eb9e7d0801e23b2dab9b75de6df3c8f5f3)
