지수함수 적분표

아래 목록은 지수함수의 적분이다.
# 부정적분
각 적분식에서 적분상수 C는 생략하였다.
## 지수함수만 포함하는 함수의 적분
: \int f'(x)e^{f(x)}\,dx = e^{f(x)} : \int e^{cx}\,dx = \frac{1}{c} e^{cx} : \int a^{cx}\,dx = \frac{1}{c\cdot \ln a} a^{cx} (a > 0,\ a \ne 1)
## 다항식을 포함하는 함수의 적분
: \int xe^{cx}\,dx = e^{cx}\left(\frac{cx-1}{c^{2\right) : \int x^2 e^{cx}\,dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right) : \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+1x^{n-i} \\\ &= e^{cx}\sum_{i=0}^n (-1)^{n-i}\frac{n!}{i!c^{n-i+1x^i \end{align} : \int\frac{e^{cx{x}\,dx = \ln|x| +\sum_{n=1}^\infty\frac{(cx)^n}{n\cdot n!} : \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) (n\neq 1)
## 삼각함수를 포함하는 함수의 적분
: \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)\end{align} (이때 \cos(\phi) = \frac{c}{\sqrt{c^2+b^2) : \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)\end{align} (이때 \cos(\phi) = \frac{c}{\sqrt{c^2+b^2) : \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 : \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
## 오차함수와 관련된 함수의 적분
다음 식들에서 erf는 오차 함수이고, Ei는 지수 적분 함수이다. : \int e^{cx}\ln x\,dx = \frac{1}{c}\left(e^{cx}\ln|x|-\operatorname{Ei}(cx)\right) : \int x e^{c x^2 }\,dx= \frac{1}{2c} e^{c x^2} : \int e^{-c x^2 }\,dx= \sqrt{\frac{\pi}{4c \operatorname{erf}(\sqrt{c} x) : \int xe^{-c x^2 }\,dx=-\frac{1}{2c}e^{-cx^2} : \int\frac{e^{-x^2{x^2}\,dx = -\frac{e^{-x^2{x} - \sqrt{\pi} \operatorname{erf} (x) : \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)
## 기타 적분
: \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 :: (이때 c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1=\frac{(2j)!}{j!2^{2j+1 이고, 모든 n>0에 대해 성립한다.) : {\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{) :: (이때 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}이고, Γ(*x*,*y*)는 불완전 감마 함수이다.) : \int \frac{1}{ae^{\lambda x} + b} \,dx = \frac{x}{b} - \frac{1}{b \lambda} \ln\left(a e^{\lambda x} + b \right) (이때 b \neq 0, \lambda \neq 0이고 ae^{\lambda x} + b > 0이다.) : \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] (이때 a \neq 0, \lambda \neq 0이고 ae^{\lambda x} + b > 0이다.) : \int \frac{ae^{cx}-1}{be^{cx}-1}\,dx=\frac{(a-b)\log(1-be^{cx})}{bc}+x.
# 정적분
: \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}(a > 0,\ b > 0,\ a \neq b)\end{align} 위 적분식의 마지막 값은 로그 평균을 뜻한다. : \int_0^{\infty} e^{-ax}\,dx=\frac{1}{a} \quad (\operatorname{Re}(a)>0) : \int_0^{\infty} e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a} \quad (a>0) (가우스 적분) : \int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0) : \int_{-\infty}^{\infty} e^{-ax^2} e^{-\frac{b}{x^2\,dx=\sqrt{\frac{\pi}{ae^{-2\sqrt{ab \quad (a,b>0) : : \int_{-\infty}^{\infty} e^{-(ax^2 + bx)}\,dx= \sqrt{\pi \over a}e^{\tfrac{b^2}{4a \quad(a > 0) : \int_{-\infty}^{\infty} e^{-ax^2} e^{-2bx}\,dx=\sqrt{\frac{\pi}{ae^{\frac{b^2}{a \quad (a>0) : \int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,dx= b \sqrt{\frac{\pi}{a \quad (\operatorname{Re}(a)>0) : \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) : \int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3} \quad (a>0) : \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) : \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) : \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,\ a>0) \\\ \\\ \dfrac{k!}{2(a^{k+1})} & (n=2k+1,\ a>0) \end{cases} : (이때 k 는 정수, !!는 이중계승이다.) : \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} : \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] : \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] : \int_0^\infty e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{1}{b\Gamma\left(\frac{1}{b}\right) : \int_0^\infty x^n e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{n+1}{b\Gamma\left(\frac{n+1}{b}\right) : \int_0^{\infty} e^{-ax}\sin bx\,dx = \frac{b}{a^2+b^2} \quad (a>0) : \int_0^{\infty} e^{-ax}\cos bx\,dx = \frac{a}{a^2+b^2} \quad (a>0) : \int_0^{\infty} xe^{-ax}\sin bx\,dx = \frac{2ab}{(a^2+b^2)^2} \quad (a>0) : \int_0^{\infty} xe^{-ax}\cos bx\,dx = \frac{a^2-b^2}{(a^2+b^2)^2} \quad (a>0) : \int_0^{\infty} \frac{e^{-ax}\sin bx}{x}\,dx=\arctan \frac{b}{a} : \int_0^{\infty} \frac{e^{-ax}-e^{-bx{x}\,dx=\ln \frac{b}{a} : \int_0^{\infty} \frac{e^{-ax}-e^{-bx{x} \sin px \, dx=\arctan \frac{b}{p} - \arctan \frac{a}{p} : \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} : \int_0^{\infty} \frac{e^{-ax} (1-\cos x)}{x^2}\,dx=\arccot a - \frac{a}{2}\ln (a^2+1) : \int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_0(x) (*I*0는 제1종 변형 베셀 함수이다.) : \int_0^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_0 \left( \sqrt{x^2 + y^2} \right) : \int_0^\infty\frac{x^{s-1{e^x/z-1} \,dx = \operatorname{Li}_{s}(z)\Gamma(s), (\operatorname{Li}_{s}(z)는 다중로그이다.) : \int_0^\infty\frac{\sin mx}{e^{2 \pi x}-1} \,dx = \frac{1}{4} \coth \frac{m}{2} - \frac{1}{2m} : \int_0^\infty e^{-x} \ln x\, dx = - \gamma, (\gamma는 오일러-마스케로니 상수)