지수함수 적분표

틀:위키데이터 속성 추적 아래 목록은 지수함수의 적분이다.

부정적분

각 적분식에서 적분상수 $C$는 생략하였다.

지수함수만 포함하는 함수의 적분

$\int f^{\prime }(x)e^{f(x)}dx=e^{f(x)}$

$\int e^{cx}dx=\frac{1}{c}{e}^{cx}$

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

다항식을 포함하는 함수의 적분

$\int x{e}^{cx}dx=e^{cx}\left(\frac{cx-1}{c^2}\right)$

$\int x^2{e}^{cx}dx=e^{cx}(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3})$

$\begin{array}{cc}\int x^n{e}^{cx}dx& =\frac{1}{c}{x}^n{e}^{cx}-\frac{n}{c}\int x^{n-1}{e}^{cx}dx\\ & ={\left(\frac{\partial }{\partial c}\right)}^n\frac{e^{cx}}{c}\\ & =e^{cx}\sum _{i=0}^n(-1{)}^i\frac{n!}{(n-i)!c^{i+1}}{x}^{n-i}\\ & =e^{cx}\sum _{i=0}^n(-1{)}^{n-i}\frac{n!}{i!c^{n-i+1}}{x}^i\end{array}$

$\int \frac{e^{cx}}{x}dx=\ln |x|+\sum _{n=1}^{\mathrm{\infty }}\frac{(cx{)}^n}{n\cdot n!}$

$\int \frac{e^{cx}}{x^n}dx=\frac{1}{n-1}(-\frac{e^{cx}}{x^{n-1}}+c\int \frac{e^{cx}}{x^{n-1}}dx)$ $(n\ne 1)$

삼각함수를 포함하는 함수의 적분

$\begin{array}{cc}\int e^{cx}\sin bxdx& =\frac{e^{cx}}{c^2+b^2}(c\sin bx-b\cos bx)\\ & =\frac{e^{cx}}{\sqrt{c^2+b^2}}\sin (bx-\varphi )\end{array}$ (이때 $\cos (\varphi )=\frac{c}{\sqrt{c^2+b^2}}$)

$\begin{array}{cc}\int e^{cx}\cos bxdx& =\frac{e^{cx}}{c^2+b^2}(c\cos bx+b\sin bx)\\ & =\frac{e^{cx}}{\sqrt{c^2+b^2}}\cos (bx-\varphi )\end{array}$ (이때 $\cos (\varphi )=\frac{c}{\sqrt{c^2+b^2}}$)

$\int e^{cx}{\sin }^{n}xdx=\frac{e^{cx}{\sin }^{n-1}x}{c^2+n^2}(c\sin x-n\cos x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}{\sin }^{n-2}xdx$

$\int e^{cx}{\cos }^{n}xdx=\frac{e^{cx}{\cos }^{n-1}x}{c^2+n^2}(c\cos x+n\sin x)+\frac{n(n-1)}{c^2+n^2}\int e^{cx}{\cos }^{n-2}xdx$

오차함수와 관련된 함수의 적분

다음 식들에서 틀:Math는 오차 함수이고, 틀:Math는 지수 적분 함수이다.

$\int e^{cx}\ln xdx=\frac{1}{c}(e^{cx}\ln |x|-\mathrm{Ei}(cx))$

$\int x{e}^{c{x}^2}dx=\frac{1}{2c}{e}^{c{x}^2}$

$\int e^{-c{x}^2}dx=\sqrt{\frac{\pi }{4c}}\mathrm{erf}(\sqrt{c}x)$

$\int x{e}^{-c{x}^2}dx=-\frac{1}{2c}{e}^{-c{x}^2}$

$\int \frac{e^{-x^2}}{x^2}dx=-\frac{e^{-x^2}}{x}-\sqrt{\pi }\mathrm{erf}(x)$

$\int \frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}dx=\frac{1}{2}\mathrm{erf}\left(\frac{x-\mu }{\sigma \sqrt{2}}\right)$

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 $n>0$에 대해 성립한다.)

$\int \underset{m}{\underbrace{x^{x^{{\cdot }^{{\cdot }^x}}}}}dx=\sum _{n=0}^m\frac{(-1{)}^n(n+1{)}^{n-1}}{n!}\mathrm{\Gamma }(n+1,-\ln x)+\sum _{n=m+1}^{\mathrm{\infty }}(-1{)}^n{a}_{mn}\mathrm{\Gamma }(n+1,-\ln x)\text{(for }x>0\text{)}$

(이때 $a_{mn}=\{\begin{array}{cc}1& \text{if }n=0,\\ \\ \frac{1}{n!}& \text{if }m=1,\\ \\ \frac{1}{n}\sum _{j=1}^{n}j{a}_{m,n-j}{a}_{m-1,j-1}& \text{otherwise}\end{array}$이고, 틀:Math는 불완전 감마 함수이다.)

$\int \frac{1}{a{e}^{\lambda x}+b}dx=\frac{x}{b}-\frac{1}{b\lambda }\ln (a{e}^{\lambda x}+b)$ (이때 $b\ne 0$, $\lambda \ne 0$이고 $a{e}^{\lambda x}+b>0$이다.)

$\int \frac{e^{2\lambda x}}{a{e}^{\lambda x}+b}dx=\frac{1}{a^2\lambda }[a{e}^{\lambda x}+b-b\ln (a{e}^{\lambda x}+b)]$ (이때 $a\ne 0$, $\lambda \ne 0$이고 $a{e}^{\lambda x}+b>0$이다.)

$\int \frac{a{e}^{cx}-1}{b{e}^{cx}-1}dx=\frac{(a-b)\log (1-b{e}^{cx})}{bc}+x.$

정적분

$\begin{array}{cc}\underset{0}{\overset{1}{\int }}{e}^{x\cdot \ln a+(1-x)\cdot \ln b}dx& =\underset{0}{\overset{1}{\int }}{\left(\frac{a}{b}\right)}^x\cdot bdx\\ & =\underset{0}{\overset{1}{\int }}{a}^x\cdot b^{1-x}dx\\ & =\frac{a-b}{\ln a-\ln b}(a>0,b>0,a\ne b)\end{array}$

위 적분식의 마지막 값은 로그 평균을 뜻한다.

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}dx=\frac{1}{a}(\mathrm{Re}(a)>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}(a>0)$ (가우스 적분)

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\sqrt{\frac{\pi }{a}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}{e}^{-\frac{b}{x^2}}dx=\sqrt{\frac{\pi }{a}}{e}^{-2\sqrt{ab}}(a,b>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx)}dx=\sqrt{\frac{\pi }{a}}{e}^{\frac{b^2}{4a}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}{e}^{-2bx}dx=\sqrt{\frac{\pi }{a}}{e}^{\frac{b^2}{a}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-a(x-b{)}^2}dx=b\sqrt{\frac{\pi }{a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}x{e}^{-a{x}^2+bx}dx=\frac{\sqrt{\pi }b}{2a^{3/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a^3}}(a>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-(a{x}^2+bx)}dx=\frac{\sqrt{\pi }(2a+b^2)}{4a^{5/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^3{e}^{-(a{x}^2+bx)}dx=\frac{\sqrt{\pi }(6a+b^2)b}{8a^{7/2}}{e}^{\frac{b^2}{4a}}(\mathrm{Re}(a)>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-a{x}^2}dx=\{\begin{array}{cc}\frac{\mathrm{\Gamma }\left(\frac{n+1}{2}\right)}{2\left(a^{\frac{n+1}{2}}\right)}& (n>-1,a>0)\\ \\ \frac{(2k-1)!!}{2^{k+1}{a}^k}\sqrt{\frac{\pi }{a}}& (n=2k,a>0)\\ \\ \frac{k!}{2(a^{k+1})}& (n=2k+1,a>0)\end{array}$

(이때 $k$는 정수, $!!$는 이중계승이다.)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-ax}dx=\{\begin{array}{cc}\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}& (n>-1,\mathrm{Re}(a)>0)\\ \\ \frac{n!}{a^{n+1}}& (n=0,1,2,\dots ,\mathrm{Re}(a)>0)\end{array}$

$\underset{0}{\overset{1}{\int }}{x}^n{e}^{-ax}dx=\frac{n!}{a^{n+1}}[1-e^{-a}\sum _{i=0}^n\frac{a^i}{i!}]$

$\underset{0}{\overset{b}{\int }}{x}^n{e}^{-ax}dx=\frac{n!}{a^{n+1}}[1-e^{-ab}\sum _{i=0}^n\frac{(ab{)}^i}{i!}]$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^b}dx=\frac{1}{b}{a}^{-\frac{1}{b}}\mathrm{\Gamma }\left(\frac{1}{b}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-a{x}^b}dx=\frac{1}{b}{a}^{-\frac{n+1}{b}}\mathrm{\Gamma }\left(\frac{n+1}{b}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\sin bxdx=\frac{b}{a^2+b^2}(a>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\cos bxdx=\frac{a}{a^2+b^2}(a>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}x{e}^{-ax}\sin bxdx=\frac{2ab}{(a^2+b^2{)}^2}(a>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}x{e}^{-ax}\cos bxdx=\frac{a^2-b^2}{(a^2+b^2{)}^2}(a>0)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}\sin bx}{x}dx=\arctan \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}dx=\ln \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}\sin pxdx=\arctan \frac{b}{p}-\arctan \frac{a}{p}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}\cos pxdx=\frac{1}{2}\ln \frac{b^2+p^2}{a^2+p^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}(1-\cos x)}{x^2}dx=\mathrm{arccot}a-\frac{a}{2}\ln (a^2+1)$

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta }d\theta =2\pi I_0(x)$ (틀:Math는 제1종 변형 베셀 함수이다.)

$\underset{0}{\overset{2\pi }{\int }}{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\left(\sqrt{x^2+y^2}\right)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{s-1}}{e^x/z-1}dx={\mathrm{Li}}_s(z)\mathrm{\Gamma }(s),$ (${\mathrm{Li}}_s(z)$는 다중로그이다.)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth \frac{m}{2}-\frac{1}{2m}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}\ln xdx=-\gamma ,$ ($\gamma$는 오일러-마스케로니 상수)

같이 보기

• 지수함수
• 적분표