List Of Integrals Of Exponential Functions
The following is a list of integrals of exponential functions . For a complete list of integral functions, please see the list of integrals .
Indefinite integral [ edit ] Indefinite integrals are antiderivative functions. A constant (the constant of integration ) may be added to the right hand side of any of these formulas, but has been suppressed here in the interest of brevity.
Integrals of polynomials [ edit ]
∫
x
e
c
x
d
x
=
e
c
x
(
c
x
−
1
c
2
)
for
c
≠
0
;
{\displaystyle \int xe^{cx}\,dx=e^{cx}\left({\frac {cx-1}{c^{2}}}\right)\qquad {\text{ for }}c\neq 0;}
∫
x
2
e
c
x
d
x
=
e
c
x
(
x
2
c
−
2
x
c
2
+
2
c
3
)
{\displaystyle \int x^{2}e^{cx}\,dx=e^{cx}\left({\frac {x^{2}}{c}}-{\frac {2x}{c^{2}}}+{\frac {2}{c^{3}}}\right)}
∫
x
n
e
c
x
d
x
=
1
c
x
n
e
c
x
−
n
c
∫
x
n
−
1
e
c
x
d
x
=
(
∂
∂
c
)
n
e
c
x
c
=
e
c
x
∑
i
=
0
n
(
−
1
)
i
n
!
(
n
−
i
)
!
c
i
+
1
x
n
−
i
=
e
c
x
∑
i
=
0
n
(
−
1
)
n
−
i
n
!
i
!
c
n
−
i
+
1
x
i
{\displaystyle {\begin{aligned}\int x^{n}e^{cx}\,dx&={\frac {1}{c}}x^{n}e^{cx}-{\frac {n}{c}}\int x^{n-1}e^{cx}\,dx\\&=\left({\frac {\partial }{\partial c}}\right)^{n}{\frac {e^{cx}}{c}}\\&=e^{cx}\sum _{i=0}^{n}(-1)^{i}{\frac {n!}{(n-i)!c^{i+1}}}x^{n-i}\\&=e^{cx}\sum _{i=0}^{n}(-1)^{n-i}{\frac {n!}{i!c^{n-i+1}}}x^{i}\end{aligned}}}
∫
e
c
x
x
d
x
=
ln
|
x
|
+
∑
n
=
1
∞
(
c
x
)
n
n
⋅
n
!
{\displaystyle \int {\frac {e^{cx}}{x}}\,dx=\ln |x|+\sum _{n=1}^{\infty }{\frac {(cx)^{n}}{n\cdot n!}}}
∫
e
c
x
x
n
d
x
=
1
n
−
1
(
−
e
c
x
x
n
−
1
+
c
∫
e
c
x
x
n
−
1
d
x
)
(for
n
≠
1
)
{\displaystyle \int {\frac {e^{cx}}{x^{n}}}\,dx={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\text{(for }}n\neq 1{\text{)}}}
Integrals involving only exponential functions [ edit ]
∫
f
′
(
x
)
e
f
(
x
)
d
x
=
e
f
(
x
)
{\displaystyle \int f'(x)e^{f(x)}\,dx=e^{f(x)}}
∫
e
c
x
d
x
=
1
c
e
c
x
{\displaystyle \int e^{cx}\,dx={\frac {1}{c}}e^{cx}}
∫
a
x
d
x
=
a
x
ln
a
for
a
>
0
,
a
≠
1
{\displaystyle \int a^{x}\,dx={\frac {a^{x}}{\ln a}}\qquad {\text{ for }}a>0,\ a\neq 1}
Integrals involving the error function [ edit ] In the following formulas, erf is the error function and Ei is the exponential integral .
∫
e
c
x
ln
x
d
x
=
1
c
(
e
c
x
ln
|
x
|
−
Ei
(
c
x
)
)
{\displaystyle \int e^{cx}\ln x\,dx={\frac {1}{c}}\left(e^{cx}\ln |x|-\operatorname {Ei} (cx)\right)}
∫
x
e
c
x
2
d
x
=
1
2
c
e
c
x
2
{\displaystyle \int xe^{cx^{2}}\,dx={\frac {1}{2c}}e^{cx^{2}}}
∫
e
−
c
x
2
d
x
=
π ぱい
4
c
erf
(
c
x
)
{\displaystyle \int e^{-cx^{2}}\,dx={\sqrt {\frac {\pi }{4c}}}\operatorname {erf} ({\sqrt {c}}x)}
∫
x
e
−
c
x
2
d
x
=
−
1
2
c
e
−
c
x
2
{\displaystyle \int xe^{-cx^{2}}\,dx=-{\frac {1}{2c}}e^{-cx^{2}}}
∫
e
−
x
2
x
2
d
x
=
−
e
−
x
2
x
−
π ぱい
erf
(
x
)
{\displaystyle \int {\frac {e^{-x^{2}}}{x^{2}}}\,dx=-{\frac {e^{-x^{2}}}{x}}-{\sqrt {\pi }}\operatorname {erf} (x)}
∫
1
σ しぐま
2
π ぱい
e
−
1
2
(
x
−
μ みゅー
σ しぐま
)
2
d
x
=
1
2
erf
(
x
−
μ みゅー
σ しぐま
2
)
{\displaystyle \int {{\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}}\,dx={\frac {1}{2}}\operatorname {erf} \left({\frac {x-\mu }{\sigma {\sqrt {2}}}}\right)}
(Note that the value of the expression is independent of the value of n , which is why it does not appear in the integral.)
∫
x
x
⋅
⋅
x
⏟
m
d
x
=
∑
n
=
0
m
(
−
1
)
n
(
n
+
1
)
n
−
1
n
!
Γ がんま (
n
+
1
,
−
ln
x
)
+
∑
n
=
m
+
1
∞
(
−
1
)
n
a
m
n
Γ がんま (
n
+
1
,
−
ln
x
)
(for
x
>
0
)
{\displaystyle {\int \underbrace {x^{x^{\cdot ^{\cdot ^{x}}}}} _{m}dx=\sum _{n=0}^{m}{\frac {(-1)^{n}(n+1)^{n-1}}{n!}}\Gamma (n+1,-\ln x)+\sum _{n=m+1}^{\infty }(-1)^{n}a_{mn}\Gamma (n+1,-\ln x)\qquad {\text{(for }}x>0{\text{)}}}}
where
a
m
n
=
{
1
if
n
=
0
,
1
n
!
if
m
=
1
,
1
n
∑
j
=
1
n
j
a
m
,
n
−
j
a
m
−
1
,
j
−
1
otherwise
{\displaystyle a_{mn}={\begin{cases}1&{\text{if }}n=0,\\\\{\dfrac {1}{n!}}&{\text{if }}m=1,\\\\{\dfrac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}}
and Γ がんま x ,y )upper incomplete gamma function .
∫
1
a
e
λ らむだ x
+
b
d
x
=
x
b
−
1
b
λ らむだ
ln
(
a
e
λ らむだ x
+
b
)
{\displaystyle \int {\frac {1}{ae^{\lambda x}+b}}\,dx={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)}
b
≠
0
{\displaystyle b\neq 0}
λ らむだ ≠
0
{\displaystyle \lambda \neq 0}
a
e
λ らむだ x
+
b
>
0.
{\displaystyle ae^{\lambda x}+b>0.}
∫
e
2
λ らむだ x
a
e
λ らむだ x
+
b
d
x
=
1
a
2
λ らむだ
[
a
e
λ らむだ x
+
b
−
b
ln
(
a
e
λ らむだ x
+
b
)
]
{\displaystyle \int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\,dx={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]}
a
≠
0
{\displaystyle a\neq 0}
λ らむだ ≠
0
{\displaystyle \lambda \neq 0}
a
e
λ らむだ x
+
b
>
0.
{\displaystyle ae^{\lambda x}+b>0.}
∫
a
e
c
x
−
1
b
e
c
x
−
1
d
x
=
(
a
−
b
)
log
(
1
−
b
e
c
x
)
b
c
+
x
.
{\displaystyle \int {\frac {ae^{cx}-1}{be^{cx}-1}}\,dx={\frac {(a-b)\log(1-be^{cx})}{bc}}+x.}
∫
e
x
(
f
(
x
)
+
f
′
(
x
)
)
dx
=
e
x
f
(
x
)
+
C
{\displaystyle \int {e^{x}\left(f\left(x\right)+f'\left(x\right)\right){\text{dx}}}=e^{x}f\left(x\right)+C}
∫
e
x
(
f
(
x
)
−
(
−
1
)
n
d
n
f
(
x
)
d
x
n
)
d
x
=
e
x
∑
k
=
1
n
(
−
1
)
k
−
1
d
k
−
1
f
(
x
)
d
x
k
−
1
+
C
{\displaystyle \int {e^{x}\left(f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{x}\sum _{k=1}^{n}{\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C}
∫
e
−
x
(
f
(
x
)
−
d
n
f
(
x
)
d
x
n
)
d
x
=
−
e
−
x
∑
k
=
1
n
d
k
−
1
f
(
x
)
d
x
k
−
1
+
C
{\displaystyle \int {e^{-x}\left(f\left(x\right)-{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=-e^{-x}\sum _{k=1}^{n}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}+C}
∫
e
a
x
(
(
a
)
n
f
(
x
)
−
(
−
1
)
n
d
n
f
(
x
)
d
x
n
)
d
x
=
e
a
x
∑
k
=
1
n
(
a
)
n
−
k
(
−
1
)
k
−
1
d
k
−
1
f
(
x
)
d
x
k
−
1
+
C
{\displaystyle \int {e^{ax}\left(\left(a\right)^{n}f\left(x\right)-\left(-1\right)^{n}{\frac {d^{n}f\left(x\right)}{dx^{n}}}\right)\,dx}=e^{ax}\sum _{k=1}^{n}{\left(a\right)^{n-k}\left(-1\right)^{k-1}{\frac {d^{k-1}f\left(x\right)}{dx^{k-1}}}}+C}
∫
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
for
a
>
0
,
b
>
0
,
a
≠
b
{\displaystyle {\begin{aligned}\int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\,dx&=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\,dx\\&=\int _{0}^{1}a^{x}\cdot b^{1-x}\,dx\\&={\frac {a-b}{\ln a-\ln b}}\qquad {\text{for }}a>0,\ b>0,\ a\neq b\end{aligned}}}
The last expression is the logarithmic mean .
∫
0
∞
e
−
a
x
d
x
=
1
a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\,dx={\frac {1}{a}}\quad (\operatorname {Re} (a)>0)}
∫
0
∞
e
−
a
x
2
d
x
=
1
2
π ぱい a
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a}}\quad (a>0)}
Gaussian integral )
∫
−
∞
∞
e
−
a
x
2
d
x
=
π ぱい a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}\,dx={\sqrt {\pi \over a}}\quad (a>0)}
∫
−
∞
∞
e
−
a
x
2
e
−
b
x
2
d
x
=
π ぱい a
e
−
2
a
b
(
a
,
b
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{-{\frac {b}{x^{2}}}}\,dx={\sqrt {\frac {\pi }{a}}}e^{-2{\sqrt {ab}}}\quad (a,b>0)}
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
)
d
x
=
π ぱい a
e
b
2
4
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx)}\,dx={\sqrt {\pi \over a}}e^{\tfrac {b^{2}}{4a}}\quad (a>0)}
∫
−
∞
∞
e
−
(
a
x
2
+
b
x
+
c
)
d
x
=
π ぱい a
e
b
2
4
a
−
c
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-(ax^{2}+bx+c)}\,dx={\sqrt {\pi \over a}}e^{{\tfrac {b^{2}}{4a}}-c}\quad (a>0)}
∫
−
∞
∞
e
−
a
x
2
e
−
2
b
x
d
x
=
π ぱい a
e
b
2
a
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)}
Integral of a Gaussian function )
∫
−
∞
∞
x
e
−
a
(
x
−
b
)
2
d
x
=
b
π ぱい a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,dx=b{\sqrt {\frac {\pi }{a}}}\quad (\operatorname {Re} (a)>0)}
∫
−
∞
∞
x
e
−
a
x
2
+
b
x
d
x
=
π ぱい
b
2
a
3
/
2
e
b
2
4
a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{-\infty }^{\infty }xe^{-ax^{2}+bx}\,dx={\frac {{\sqrt {\pi }}b}{2a^{3/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
∫
−
∞
∞
x
2
e
−
a
x
2
d
x
=
1
2
π ぱい
a
3
(
a
>
0
)
{\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-ax^{2}}\,dx={\frac {1}{2}}{\sqrt {\pi \over a^{3}}}\quad (a>0)}
∫
−
∞
∞
x
2
e
−
(
a
x
2
+
b
x
)
d
x
=
π ぱい
(
2
a
+
b
2
)
4
a
5
/
2
e
b
2
4
a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(2a+b^{2})}{4a^{5/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
∫
−
∞
∞
x
3
e
−
(
a
x
2
+
b
x
)
d
x
=
π ぱい
(
6
a
+
b
2
)
b
8
a
7
/
2
e
b
2
4
a
(
Re
(
a
)
>
0
)
{\displaystyle \int _{-\infty }^{\infty }x^{3}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(6a+b^{2})b}{8a^{7/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}
∫
0
∞
x
n
e
−
a
x
2
d
x
=
{
Γ がんま
(
n
+
1
2
)
2
(
a
n
+
1
2
)
(
n
>
−
1
,
a
>
0
)
(
2
k
−
1
)
!
!
2
k
+
1
a
k
π ぱい a
(
n
=
2
k
,
k
integer
,
a
>
0
)
k
!
2
(
a
k
+
1
)
(
n
=
2
k
+
1
,
k
integer
,
a
>
0
)
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,dx={\begin{cases}{\dfrac {\Gamma \left({\frac {n+1}{2}}\right)}{2\left(a^{\frac {n+1}{2}}\right)}}&(n>-1,\ a>0)\\{\dfrac {(2k-1)!!}{2^{k+1}a^{k}}}{\sqrt {\dfrac {\pi }{a}}}&(n=2k,\ k{\text{ integer}},\ a>0)\\{\dfrac {k!}{2(a^{k+1})}}&(n=2k+1,\ k{\text{ integer}},\ a>0)\end{cases}}}
(the operator
!
!
{\displaystyle !!}
Double factorial )
∫
0
∞
x
n
e
−
a
x
d
x
=
{
Γ がんま (
n
+
1
)
a
n
+
1
(
n
>
−
1
,
Re
(
a
)
>
0
)
n
!
a
n
+
1
(
n
=
0
,
1
,
2
,
…
,
Re
(
a
)
>
0
)
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax}\,dx={\begin{cases}{\dfrac {\Gamma (n+1)}{a^{n+1}}}&(n>-1,\ \operatorname {Re} (a)>0)\\\\{\dfrac {n!}{a^{n+1}}}&(n=0,1,2,\ldots ,\ \operatorname {Re} (a)>0)\end{cases}}}
∫
0
1
x
n
e
−
a
x
d
x
=
n
!
a
n
+
1
[
1
−
e
−
a
∑
i
=
0
n
a
i
i
!
]
{\displaystyle \int _{0}^{1}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-a}\sum _{i=0}^{n}{\frac {a^{i}}{i!}}\right]}
∫
0
b
x
n
e
−
a
x
d
x
=
n
!
a
n
+
1
[
1
−
e
−
a
b
∑
i
=
0
n
(
a
b
)
i
i
!
]
{\displaystyle \int _{0}^{b}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-ab}\sum _{i=0}^{n}{\frac {(ab)^{i}}{i!}}\right]}
∫
0
∞
e
−
a
x
b
d
x
=
1
b
a
−
1
b
Γ がんま
(
1
b
)
{\displaystyle \int _{0}^{\infty }e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {1}{b}}}\Gamma \left({\frac {1}{b}}\right)}
∫
0
∞
x
n
e
−
a
x
b
d
x
=
1
b
a
−
n
+
1
b
Γ がんま
(
n
+
1
b
)
{\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{b}}dx={\frac {1}{b}}\ a^{-{\frac {n+1}{b}}}\Gamma \left({\frac {n+1}{b}}\right)}
∫
0
∞
e
−
a
x
sin
b
x
d
x
=
b
a
2
+
b
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\sin bx\,dx={\frac {b}{a^{2}+b^{2}}}\quad (a>0)}
∫
0
∞
e
−
a
x
cos
b
x
d
x
=
a
a
2
+
b
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }e^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}\quad (a>0)}
∫
0
∞
x
e
−
a
x
sin
b
x
d
x
=
2
a
b
(
a
2
+
b
2
)
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }xe^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
∫
0
∞
x
e
−
a
x
cos
b
x
d
x
=
a
2
−
b
2
(
a
2
+
b
2
)
2
(
a
>
0
)
{\displaystyle \int _{0}^{\infty }xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2}}{(a^{2}+b^{2})^{2}}}\quad (a>0)}
∫
0
∞
e
−
a
x
sin
b
x
x
d
x
=
arctan
b
a
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}\sin bx}{x}}\,dx=\arctan {\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
−
e
−
b
x
x
sin
p
x
d
x
=
arctan
b
p
−
arctan
a
p
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}-e^{-bx}}{x}}\sin px\,dx=\arctan {\frac {b}{p}}-\arctan {\frac {a}{p}}}
∫
0
∞
e
−
a
x
−
e
−
b
x
x
cos
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}}\cos px\,dx={\frac {1}{2}}\ln {\frac {b^{2}+p^{2}}{a^{2}+p^{2}}}}
∫
0
∞
e
−
a
x
(
1
−
cos
x
)
x
2
d
x
=
arccot
a
−
a
2
ln
(
1
a
2
+
1
)
{\displaystyle \int _{0}^{\infty }{\frac {e^{-ax}(1-\cos x)}{x^{2}}}\,dx=\operatorname {arccot} a-{\frac {a}{2}}\ln {\Big (}{\frac {1}{a^{2}}}+1{\Big )}}
∫
−
∞
∞
e
a
x
4
+
b
x
3
+
c
x
2
+
d
x
+
f
d
x
=
e
f
∑
n
,
m
,
p
=
0
∞
b
4
n
(
4
n
)
!
c
2
m
(
2
m
)
!
d
4
p
(
4
p
)
!
Γ がんま (
3
n
+
m
+
p
+
1
4
)
a
3
n
+
m
+
p
+
1
4
{\displaystyle \int _{-\infty }^{\infty }e^{ax^{4}+bx^{3}+cx^{2}+dx+f}\,dx=e^{f}\sum _{n,m,p=0}^{\infty }{\frac {b^{4n}}{(4n)!}}{\frac {c^{2m}}{(2m)!}}{\frac {d^{4p}}{(4p)!}}{\frac {\Gamma (3n+m+p+{\frac {1}{4}})}{a^{3n+m+p+{\frac {1}{4}}}}}}
superstring theory in higher dimensions)
∫
0
2
π ぱい
e
x
cos
θ しーた
d
θ しーた =
2
π ぱい
I
0
(
x
)
{\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)}
I 0 modified Bessel function of the first kind)
∫
0
2
π ぱい
e
x
cos
θ しーた +
y
sin
θ しーた
d
θ しーた =
2
π ぱい
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)}
∫
0
∞
x
s
−
1
e
x
/
z
−
1
d
x
=
Li
s
(
z
)
Γ がんま (
s
)
,
{\displaystyle \int _{0}^{\infty }{\frac {x^{s-1}}{e^{x}/z-1}}\,dx=\operatorname {Li} _{s}(z)\Gamma (s),}
where
Li
s
(
z
)
{\displaystyle \operatorname {Li} _{s}(z)}
Polylogarithm .
∫
0
∞
sin
m
x
e
2
π ぱい 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
∞
e
−
x
ln
x
d
x
=
−
γ がんま ,
{\displaystyle \int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,}
where
γ がんま
{\displaystyle \gamma }
Euler–Mascheroni constant which equals the value of a number of definite integrals.
Finally, a well known result,
∫
0
2
π ぱい
e
i
(
m
−
n
)
ϕ
d
ϕ
=
2
π ぱい
δ でるた
m
,
n
for
m
,
n
∈
Z
{\displaystyle \int _{0}^{2\pi }e^{i(m-n)\phi }d\phi =2\pi \delta _{m,n}\qquad {\text{for }}m,n\in \mathbb {Z} }
δ でるた
m
,
n
{\displaystyle \delta _{m,n}}
Kronecker delta .
Toyesh Prakash Sharma , Etisha Sharma , "Putting Forward Another Generalization Of The Class Of Exponential Integrals And Their Applications.," International Journal of Scientific Research in Mathematical and Statistical Sciences, Vol.10, Issue.2, pp.1-8, 2023.[1]
![{\displaystyle \int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\,dx={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1fb1fd68afed8ab416094845d04ac2ed1c3bc788)
![{\displaystyle \int _{0}^{1}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-a}\sum _{i=0}^{n}{\frac {a^{i}}{i!}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92482c2e5d7502755c6da9b6f088ff00721580e1)
![{\displaystyle \int _{0}^{b}x^{n}e^{-ax}\,dx={\frac {n!}{a^{n+1}}}\left[1-e^{-ab}\sum _{i=0}^{n}{\frac {(ab)^{i}}{i!}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f17026bf5a6616142c6b1a8f0392f3ceb373cbbd)
