지수함수 적분표 문서 원본 보기

{{위키데이터 속성 추적}}
아래 목록은 [[지수함수]]의 적분이다.

== 부정적분 ==
각 적분식에서 [[적분상수]] <math>C</math>는 생략하였다.

=== 지수함수만 포함하는 함수의 적분 ===

: <math>\int  f'(x)e^{f(x)}\,dx = e^{f(x)}</math>

: <math>\int e^{cx}\,dx = \frac{1}{c} e^{cx}</math>

: <math>\int a^{cx}\,dx = \frac{1}{c\cdot \ln a} a^{cx}</math> <math>(a > 0,\ a \ne 1)</math>

=== 다항식을 포함하는 함수의 적분 ===

: <math>\int xe^{cx}\,dx = e^{cx}\left(\frac{cx-1}{c^{2}}\right)</math>

: <math>\int x^2 e^{cx}\,dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right)</math>

: <math>\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}</math>

: <math>\int\frac{e^{cx}}{x}\,dx = \ln|x| +\sum_{n=1}^\infty\frac{(cx)^n}{n\cdot n!}</math>

: <math>\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)</math> <math>(n\neq 1)</math>

=== 삼각함수를 포함하는 함수의 적분 ===

: <math>\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}</math> (이때 <math>\cos(\phi) = \frac{c}{\sqrt{c^2+b^2}}</math>)

: <math>\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}</math> (이때 <math>\cos(\phi) = \frac{c}{\sqrt{c^2+b^2}}</math>)

: <math>\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</math>

: <math>\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</math>

=== 오차함수와 관련된 함수의 적분 ===
다음 식들에서 {{math|erf}}는 [[오차 함수]]이고, {{math|Ei}}는 [[지수 적분 함수]]이다.

: <math>\int e^{cx}\ln x\,dx = \frac{1}{c}\left(e^{cx}\ln|x|-\operatorname{Ei}(cx)\right)</math>

: <math>\int x e^{c x^2 }\,dx= \frac{1}{2c} e^{c x^2}</math>

: <math>\int e^{-c x^2 }\,dx= \sqrt{\frac{\pi}{4c}} \operatorname{erf}(\sqrt{c} x)</math>

: <math>\int xe^{-c x^2 }\,dx=-\frac{1}{2c}e^{-cx^2} </math>

: <math>\int\frac{e^{-x^2}}{x^2}\,dx = -\frac{e^{-x^2}}{x} - \sqrt{\pi} \operatorname{erf} (x) </math>

: <math>\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)</math>

=== 기타 적분 ===

: <math>\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 </math>
:: (이때 <math> c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1}}=\frac{(2j)!}{j!2^{2j+1}} </math>이고, 모든 <math>n>0</math>에 대해 성립한다.)

: <math> {\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{)}}</math>
:: (이때 <math>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}</math>이고, {{math|Γ(''x'',''y'')}}는 [[불완전 감마 함수]]이다.)

: <math>\int \frac{1}{ae^{\lambda x} + b} \,dx = \frac{x}{b} - \frac{1}{b \lambda} \ln\left(a e^{\lambda x} + b \right) </math> (이때 <math>b \neq 0</math>, <math>\lambda \neq 0</math>이고 <math>ae^{\lambda x} + b > 0</math>이다.)

: <math>\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] </math> (이때 <math>a \neq 0</math>, <math>\lambda \neq 0</math>이고 <math>ae^{\lambda x} + b > 0</math>이다.)

: <math>\int \frac{ae^{cx}-1}{be^{cx}-1}\,dx=\frac{(a-b)\log(1-be^{cx})}{bc}+x.</math>

== 정적분 ==

: <math>\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}</math>

위 적분식의 마지막 값은 [[로그 평균]]을 뜻한다.

: <math>\int_0^{\infty} e^{-ax}\,dx=\frac{1}{a} \quad (\operatorname{Re}(a)>0)</math>

: <math>\int_0^{\infty} e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a} \quad  (a>0)</math> ([[가우스 적분]])

: <math>\int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a>0)</math>
: <math>\int_{-\infty}^{\infty} e^{-ax^2} e^{-\frac{b}{x^2}}\,dx=\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}} \quad (a,b>0) </math>
:
: <math>\int_{-\infty}^{\infty} e^{-(ax^2 + bx)}\,dx= \sqrt{\pi \over a}e^{\tfrac{b^2}{4a}} \quad(a > 0)</math>

: <math>\int_{-\infty}^{\infty} e^{-ax^2} e^{-2bx}\,dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}} \quad (a>0)</math>

: <math>\int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,dx= b \sqrt{\frac{\pi}{a}} \quad (\operatorname{Re}(a)>0)</math>

: <math>\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)</math>

: <math>\int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3} \quad (a>0)</math>

: <math>\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)</math>

: <math>\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)</math>

: <math>\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} </math>
: (이때 <math>k </math>는 정수, <math>!!</math>는 [[계승#이중계승|이중계승]]이다.)

: <math>\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}</math>

: <math>\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]</math>

: <math>\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]</math>

: <math>\int_0^\infty e^{-ax^b} dx =  \frac{1}{b}\ a^{-\frac{1}{b}}\Gamma\left(\frac{1}{b}\right)</math>

: <math>\int_0^\infty x^n e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{n+1}{b}}\Gamma\left(\frac{n+1}{b}\right)</math>

: <math>\int_0^{\infty} e^{-ax}\sin bx\,dx = \frac{b}{a^2+b^2} \quad (a>0)</math>

: <math>\int_0^{\infty} e^{-ax}\cos bx\,dx = \frac{a}{a^2+b^2} \quad (a>0)</math>

: <math>\int_0^{\infty} xe^{-ax}\sin bx\,dx = \frac{2ab}{(a^2+b^2)^2} \quad (a>0)</math>

: <math>\int_0^{\infty} xe^{-ax}\cos bx\,dx = \frac{a^2-b^2}{(a^2+b^2)^2} \quad (a>0)</math>
: <math>\int_0^{\infty} \frac{e^{-ax}\sin bx}{x}\,dx=\arctan \frac{b}{a}</math>
: <math>\int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\,dx=\ln \frac{b}{a}</math>
: <math>\int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x} \sin px \, dx=\arctan \frac{b}{p} - \arctan \frac{a}{p}</math>
: <math>\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}</math>
: <math>\int_0^{\infty} \frac{e^{-ax} (1-\cos x)}{x^2}\,dx=\arccot a - \frac{a}{2}\ln (a^2+1)</math>

: <math>\int_0^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_0(x)</math> ({{math|''I''<sub>0</sub>}}는 [[베셀 함수#제1종 변형 베셀 함수|제1종 변형 베셀 함수]]이다.)

: <math>\int_0^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_0 \left( \sqrt{x^2 + y^2} \right)</math>

: <math>\int_0^\infty\frac{x^{s-1}}{e^x/z-1} \,dx = \operatorname{Li}_{s}(z)\Gamma(s), </math> (<math>\operatorname{Li}_{s}(z)</math>는 [[다중로그]]이다.)

: <math>\int_0^\infty\frac{\sin mx}{e^{2 \pi x}-1} \,dx = \frac{1}{4} \coth \frac{m}{2} - \frac{1}{2m} </math>

: <math>\int_0^\infty  e^{-x} \ln x\, dx = - \gamma, </math> (<math>\gamma</math>는 [[오일러-마스케로니 상수]])

== 같이 보기 ==
* [[지수 함수|지수함수]]
* [[적분표]]

[[분류:적분]]
[[분류:거듭제곱]]