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 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 ∫ 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 ∫ 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 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 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 erf ( x ) {\displaystyle \int {\frac {e^{-x^{2}{x^{2}\,dx=-{\frac {e^{-x^{2}{x}-{\sqrt {\pi }\operatorname {erf} (x)} ∫ 1 σ 2 π e − 1 2 ( x − μ σ ) 2 d x = 1 2 erf ( x − μ σ 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)}
Other integrals (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 ( n + 1 , − ln x ) + ∑ n = m + 1 ∞ ( − 1 ) n a m n Γ ( 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(x ,y ) is the upper incomplete gamma function .
∫ 1 a e λ x + b d x = x b − 1 b λ ln ( a e λ 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)} when b ≠ 0 {\displaystyle b\neq 0} , λ ≠ 0 {\displaystyle \lambda \neq 0} , and a e λ x + b > 0. {\displaystyle ae^{\lambda x}+b>0.} ∫ e 2 λ x a e λ x + b d x = 1 a 2 λ [ a e λ x + b − b ln ( a e λ 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]} when a ≠ 0 {\displaystyle a\neq 0} , λ ≠ 0 {\displaystyle \lambda \neq 0} , and a e λ 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}
Definite integrals ∫ 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 a ( a > 0 ) {\displaystyle \int _{0}^{\infty }e^{-ax^{2}\,dx={\frac {1}{2}{\sqrt {\pi \over a}\quad (a>0)} (the Gaussian integral ) ∫ − ∞ ∞ e − a x 2 d x = π 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 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 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 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 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)} (see Integral of a Gaussian function ) ∫ − ∞ ∞ x e − a ( x − b ) 2 d x = b π 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 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 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 ( 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 ( 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 ( n + 1 2 ) 2 ( a n + 1 2 ) ( n > − 1 , a > 0 ) ( 2 k − 1 ) ! ! 2 k + 1 a k π 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 !!} is the Double factorial )
∫ 0 ∞ x n e − a x d x = { Γ ( 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 ( 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 ( 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 ( 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} (appears in several models of extended superstring theory in higher dimensions) ∫ 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 is the modified Bessel function of the first kind) ∫ 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)} ∫ 0 ∞ x s − 1 e x / z − 1 d x = Li s ( z ) Γ ( 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)} is the Polylogarithm .
∫ 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 ∞ e − x ln x d x = − γ , {\displaystyle \int _{0}^{\infty }e^{-x}\ln x\,dx=-\gamma ,} where γ {\displaystyle \gamma } is the Euler–Mascheroni constant which equals the value of a number of definite integrals.
Finally, a well known result,
∫ 0 2 π e i ( m − n ) ϕ d ϕ = 2 π δ 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} }
where δ m , n {\displaystyle \delta _{m,n} is the Kronecker delta .
See also
References
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]
Further reading
External links