다음은 지수 함수(exponential function)의 적분(integral)의 목록입니다. 적분 함수의 전체 목록에 대해, 적분의 목록(list of integrals)을 참조하십시오.
Indefinite integral
부정 적분은 역도함수(antiderivative)입니다. 상수 (적분의 상수(constant of integration))는 이들 공식의 임의의 것의 오른쪽 변에 더해질 수 있지만, 여기서는 간결의 목적으로 생략되었습니다.
Integrals of polynomials
\(\quad\displaystyle \int xe^{cx}\,dx = e^{cx}\left(\frac{cx-1}{c^{2}}\right)\)
\(\quad\displaystyle \int x^2 e^{cx}\,dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right)\)
\(\quad\displaystyle \begin{align}
\int x^n e^{cx}\,dx &= \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} \,dx \\
&= \left( \frac{\partial}{\partial c} \right)^n \frac{e^{cx}}{c} \\
&= e^{cx}\sum_{i=0}^n (-1)^i\frac{n!}{(n-i)!c^{i+1}}x^{n-i} \\
&= e^{cx}\sum_{i=0}^n (-1)^{n-i}\frac{n!}{i!c^{n-i+1}}x^i
\end{align}\)
\(\quad\displaystyle \int\frac{e^{cx}}{x}\,dx = \ln|x| +\sum_{n=1}^\infty\frac{(cx)^n}{n\cdot n!}\)
\(\quad\displaystyle \int\frac{e^{cx}}{x^n}\,dx = \frac{1}{n-1}\left(-\frac{e^{cx}}{x^{n-1}}+c\int\frac{e^{cx} }{x^{n-1}}\,dx\right) \qquad\text{(for }n\neq 1\text{)}\)
Integrals involving only exponential functions
\(\quad\displaystyle \int f'(x)e^{f(x)}\,dx = e^{f(x)}\)
\(\quad\displaystyle \int e^{cx}\,dx = \frac{1}{c} e^{cx}\)
\(\quad\displaystyle \int a^{cx}\,dx = \frac{1}{c\cdot \ln a} a^{cx}\qquad\text{ for }a > 0,\ a \ne 1\)
Integrals involving exponential and trigonometric functions
\(\quad\displaystyle \begin{align}
\int e^{cx}\sin bx\,dx &= \frac{e^{cx}}{c^2+b^2}(c\sin bx - b\cos bx) \\
&= \frac{e^{cx}}{\sqrt{c^2+b^2}}\sin(bx-\phi)\qquad \text{where }\cos(\phi) = \frac{c}{\sqrt{c^2+b^2}}
\end{align}\)
\(\quad\displaystyle \begin{align}
\int e^{cx}\cos bx\,dx &= \frac{e^{cx}}{c^2+b^2}(c\cos bx + b\sin bx) \\
&= \frac{e^{cx}}{\sqrt{c^2+b^2}}\cos(bx-\phi)\qquad \text{where }\cos(\phi) = \frac{c}{\sqrt{c^2+b^2}}
\end{align}\)
\(\quad\displaystyle \int e^{cx}\sin^n x\,dx = \frac{e^{cx}\sin^{n-1} x}{c^2+n^2}(c\sin x-n\cos x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}\sin^{n-2} x\,dx\)
\(\quad\displaystyle \int e^{cx}\cos^n x\,dx = \frac{e^{cx}\cos^{n-1} x}{c^2+n^2}(c\cos x+n\sin x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}\cos^{n-2} x\,dx\)
Integrals involving the error function
다음 공식에서, erf는 오차 함수(error function)이고 Ei는 지수 적분(exponential integral)입니다.
\(\quad\displaystyle \int e^{cx}\ln x\,dx = \frac{1}{c}\left(e^{cx}\ln|x|-\operatorname{Ei}(cx)\right)\)
\(\quad\displaystyle \int x e^{c x^2 }\,dx= \frac{1}{2c} e^{c x^2}\)
\(\quad\displaystyle \int e^{-c x^2 }\,dx= \sqrt{\frac{\pi}{4c}} \operatorname{erf}(\sqrt{c} x)\)
\(\quad\displaystyle \int xe^{-c x^2 }\,dx=-\frac{1}{2c}e^{-cx^2} \)
\(\quad\displaystyle \int\frac{e^{-x^2}}{x^2}\,dx = -\frac{e^{-x^2}}{x} - \sqrt{\pi} \operatorname{erf} (x) \)
\(\quad\displaystyle \int {\frac{1}{\sigma\sqrt{2\pi}} e^{ -\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2 }}\,dx= \frac{1}{2}\operatorname{erf}\left(\frac{x-\mu}{\sigma \sqrt{2}}\right)\)
Other integrals
\(\quad\displaystyle \int e^{x^2}\,dx = e^{x^2}\left( \sum_{j=0}^{n-1}c_{2j}\frac{1}{x^{2j+1}} \right )+(2n-1)c_{2n-2} \int \frac{e^{x^2}}{x^{2n}}\,dx \quad \text{valid for any } n > 0, \)
여기서 \(\quad\displaystyle c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1}}=\frac{(2j)!}{j!2^{2j+1}} \ . \)
(표현식의 값은 n의 값과 독립적이고, 그것이 적분에 나타나지 않는 이유임을 주목하십시오.)
\(\quad\displaystyle {\int \underbrace{x^{x^{\cdot^{\cdot^{x}}}}}_mdx= \sum_{n=0}^m\frac{(-1)^n(n+1)^{n-1}}{n!}\Gamma(n+1,- \ln x) + \sum_{n=m+1}^\infty(-1)^na_{mn}\Gamma(n+1,-\ln x) \qquad\text{(for }x> 0\text{)}}\)
여기서 \(\quad\displaystyle a_{mn}=\begin{cases}1 &\text{if } n = 0, \\ \\ \dfrac{1}{n!} &\text{if } m=1, \\ \\ \dfrac{1}{n}\sum_{j=1}^{n}ja_{m,n-j}a_{m-1,j-1} &\text{otherwise} \end{cases}\)
그리고 Γ(x,y)는 위쪽 불완전 감마 함수(upper incomplete gamma function)입니다.
\(\quad\displaystyle \int \frac{1}{ae^{\lambda x} + b} \,dx = \frac{x}{b} - \frac{1}{b \lambda} \ln\left(a e^{\lambda x} + b \right) \) when \(\quad\displaystyle b \neq 0\), \(\quad\displaystyle \lambda \neq 0\), and \(\quad\displaystyle ae^{\lambda x} + b > 0.\)
\(\quad\displaystyle \int \frac{e^{2\lambda x}}{ae^{\lambda x} + b} \,dx = \frac{1}{a^2 \lambda} \left[a e^{\lambda x} + b - b \ln\left(a e^{\lambda x} + b \right) \right] \) when \(\quad\displaystyle a \neq 0\), \(\quad\displaystyle \lambda \neq 0\), and \(\quad\displaystyle ae^{\lambda x} + b > 0.\)
\(\quad\displaystyle \int \frac{ae^{cx}-1}{be^{cx}-1}\,dx=\frac{(a-b)\log(1-be^{cx})}{bc}+x.\)
Definite integrals
\(\quad\displaystyle \begin{align}
\int_0^1 e^{x\cdot \ln a + (1-x)\cdot \ln b}\,dx
&= \int_0^1 \left(\frac{a}{b}\right)^{x}\cdot b\,dx \\
&= \int_0^1 a^{x}\cdot b^{1-x}\,dx \\
&= \frac{a-b}{\ln a - \ln b} \qquad\text{for } a > 0,\ b > 0,\ a \neq b
\end{align}\)
마지막 표현은 로그 평균(logarithmic mean)입니다.
\(\quad\displaystyle \int_0^{\infty} e^{-ax}\,dx=\frac{1}{a} \quad (\operatorname{Re}(a)>0)\)
\(\quad\displaystyle \int_0^{\infty} e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a} \quad (a>0)\) (the Gaussian integral)
\(\quad\displaystyle \int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} e^{-ax^2} e^{-\frac{b}{x^2}}\,dx=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}} \quad (a,b>0) \)
\(\quad\displaystyle \int_{-\infty}^{\infty} e^{-ax^2 + bx}\,dx= \sqrt{\pi \over a}e^{\tfrac{b^2}{4a}} \quad(a > 0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} e^{-ax^2} e^{-2bx}\,dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}} \quad (a>0)\) (see Integral of a Gaussian function)
\(\quad\displaystyle \int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,dx= b \sqrt{\frac{\pi}{a}} \quad (\operatorname{Re}(a)>0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} x e^{-ax^2+bx}\,dx= \frac{ \sqrt{\pi} b }{2a^{3/2}} e^{\frac{b^2}{4a}} \quad (\operatorname{Re}(a)>0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3} \quad (a>0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} x^2 e^{-ax^2+bx}\,dx=\frac{\sqrt{\pi}(2a+b^2)}{4a^{5/2}} e^{\frac{b^2}{4a}} \quad (\operatorname{Re}(a)>0)\)
\(\quad\displaystyle \int_{-\infty}^{\infty} x^3 e^{-ax^2+bx}\,dx=\frac{\sqrt{\pi}(6a+b^2)b}{8a^{7/2}} e^{\frac{b^2}{4a}} \quad (\operatorname{Re}(a)>0)\)
\(\quad\displaystyle \int_0^{\infty} x^{n} e^{-ax^2}\,dx =
\begin{cases}
\dfrac{\Gamma \left(\frac{n+1}{2}\right)}{2\left(a^\frac{n+1}{2}\right) } & (n>-1,\ a>0) \\ \\
\dfrac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\dfrac{\pi}{a}} & (n=2k,\ k \text{ integer},\ a>0) \ \text{(!! is the double factorial)} \\ \\
\dfrac{k!}{2(a^{k+1})} & (n=2k+1,\ k \text{ integer},\ a>0)
\end{cases} \)
(연산자 \(!!\)는 두-배 팩토리얼(double factorial)입니다)
\(\quad\displaystyle \int_0^{\infty} x^n e^{-ax}\,dx =
\begin{cases}
\dfrac{\Gamma(n+1)}{a^{n+1}} & (n>-1,\ \operatorname{Re}(a)>0) \\ \\
\dfrac{n!}{a^{n+1}} & (n=0,1,2,\ldots,\ \operatorname{Re}(a)>0)
\end{cases}\)
\(\quad\displaystyle \int_0^{1} x^n e^{-ax}\,dx =
\frac{n!}{a^{n+1}}\left[
1-e^{-a}\sum_{i=0}^{n} \frac{a^i}{i!}
\right]\)
\(\quad\displaystyle \int_0^\infty e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{1}{b}}\Gamma\left(\frac{1}{b}\right)\)
\(\quad\displaystyle \int_0^\infty x^n e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{n+1}{b}}\Gamma\left(\frac{n+1}{b}\right)\)
\(\quad\displaystyle \int_0^{\infty} e^{-ax}\sin bx\,dx = \frac{b}{a^2+b^2} \quad (a>0)\)
\(\quad\displaystyle \int_0^{\infty} e^{-ax}\cos bx\,dx = \frac{a}{a^2+b^2} \quad (a>0)\)
\(\quad\displaystyle \int_0^{\infty} xe^{-ax}\sin bx\,dx = \frac{2ab}{(a^2+b^2)^2} \quad (a>0)\)
\(\quad\displaystyle \int_0^{\infty} xe^{-ax}\cos bx\,dx = \frac{a^2-b^2}{(a^2+b^2)^2} \quad (a>0)\)
\(\quad\displaystyle \int_0^{\infty} \frac{e^{-ax}\sin bx}{x}\,dx=\arctan \frac{b}{a}\)
\(\quad\displaystyle \int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\,dx=\ln \frac{b}{a}\)
\(\quad\displaystyle \int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x} \sin px \, dx=\arctan \frac{b}{p} - \arctan \frac{a}{p}\)
\(\quad\displaystyle \int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x} \cos px \, dx=\frac{1}{2} \ln \frac{b^2+p^2}{a^2+p^2}\)
\(\quad\displaystyle \int_0^{\infty} \frac{e^{-ax} (1-\cos x)}{x^2}\,dx=\text{arccot} a - \frac{a}{2}\ln (a^2+1)\)
\(\quad\displaystyle \int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_0(x)\) (\(I_0\) is the modified Bessel function of the first kind)
\(\quad\displaystyle \int_0^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_0 \left( \sqrt{x^2 + y^2} \right)\)
\(\quad\displaystyle \int_0^\infty\frac{x^{s-1}}{e^x/z-1} \,dx = \operatorname{Li}_{s}(z)\Gamma(s), \)
여기서 \(\quad\displaystyle \operatorname{Li}_{s}(z)\)는 다중로그(Polylogarithm)입니다.
\(\quad\displaystyle \int_0^\infty\frac{\sin mx}{e^{2 \pi x}-1} \,dx = \frac{1}{4} \coth \frac{m}{2} - \frac{1}{2m} \)
\(\quad\displaystyle \int_0^\infty e^{-x} \ln x\, dx = - \gamma, \)
여기서 \(\gamma\)는 오일러–마스케로니 상수(Euler–Mascheroni constant)이며 다수의 부정 적분의 값과 같습니다.
마지막으로, 하나의 잘 알려진 결과,
\(\quad\displaystyle \int_0^{2 \pi} e^{i(m-n)\phi} d\phi = 2 \pi \delta_{m,n}\) (정수 m, n에 대해)
여기서 \(\delta_{m,n}\)는 크로네크 델타(Kronecker delta)입니다.
urther reading
- Moll, Victor Hugo (2014-11-12). Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume I. Vol. I (1 ed.). Chapman and Hall/CRC Press. ISBN 978-1-48225-651-2. Retrieved 2016-02-12. {{cite book}}: |work= ignored (help)
- Moll, Victor Hugo (2015-10-27). Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume II. Vol. II (1 ed.). Chapman and Hall/CRC Press. ISBN 978-1-48225-653-6. Retrieved 2016-02-12. {{cite book}}: |work= ignored (help)
External links
- Wolfram Mathematica Online Integrator
- Moll, Victor Hugo. "List with the formulas and proofs in GR". Retrieved 2016-02-12.