지수함수의 통합 목록

다음은 지수함수의 통합 목록이다. 통합 기능의 전체 목록은 통합 목록을 참조하십시오.

무한정적분

무기한 통합은 반물질적 기능이다. 상수(통합의 상수)는 이러한 공식 중 어느 하나라도 오른손에 추가할 수 있지만 간결함을 위해 여기서는 억제되었다.

다항식 통합

${\displaystyle \int xe^{cx}\,dx=e^{cx}\lefted\frac {c-1}{c^{2}}}\오른쪽){\text{{}{}}}}{}c\neq 0;}$

${\displaystyle \int x^{2}e^{cx}\,dx=e^{cx}\left\frac {x^{2}}-{c}-{2x}{c^}}+{2}{2}}{c^{3}}\오른쪽)}$

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

${\displaystyle \int{\frac{e^{cx}}{x}\,dx=\ln x +\sum _{n=1}^{n1}{n}{n}{n\cdot n!}}}$

${\displaystyle \int {\frac {e^{cx}}{x^{n}}}\,dx={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,dx\right)\qquad {\text{(for }}n\neq 1{\text{)}}}$

지수 함수만 포함하는 통합

${\displaystyle \int f'(x)e^{f(x)}\,dx=e^{f(x)}}}$

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

${\displaystyle \int a^{cx}\,dx={\frac {1}{c\cdot \ln \l}a^{cx}\qquad {\text{}{{}a>0,\\neq 1}$

오류 함수와 관련된 통합

다음 공식에서 erf는 오류 함수, Ei는 지수 적분이다.

${\displaystyle \int e^{cx}\ln x\,dx={\frac {1}{1}{c}\왼쪽(e^{cx}\ln x -\operatorname {Ei}(cx)\right)}$

${\displaystyle \int xe^{cx^{2}}\,dx={\frac{1}{1}{2c}e^{cx^{2}}}$

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

${\displaystyle \int xe^{-cx^{2}}\,dx=-{\frac {1}{2c}e^{-cx^{2}}:}$

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

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

기타 통합

${\displaystyle \int e^{x^{2}}\,dx=e^{x^{2}}\left(\sum _{j=0}^{n-1}c_{2j}{\frac {1}{x^{2j+1}}}\right)+(2n-1)c_{2n-2}\int {\frac {e^{x^{2}}}{x^{2n}}}\,dx\quad {\text{valid for any }}n>0,}$

여기서 $c{\displaystyle {}_{}}$ ${\displaystyle {\mathrm{2j}}_{}}$= ${\displaystyle 1\frac{\mathrm{}}{}}$ ${\displaystyle \frac{\cdot }{}}$ ${\displaystyle \frac{3}{}}$ ${\displaystyle \frac{\cdot }{}}$ ${\displaystyle \frac{5}{}}$ ${\displaystyle \frac{\cdots }{}}$( ${\displaystyle \frac{2}{}}$ ${\displaystyle \frac{\mathrm{j}}{}}$- 1${\displaystyle \frac{}{}}$) 2 ${\displaystyle \frac{j^{}}{}\frac{{}^{}}{}}$+ ${\displaystyle \frac{1^{}}{}}$=${\displaystyle \frac{}{}}$( ${\displaystyle \frac{2}{}}$ ${\displaystyle \frac{j}{}}$)${\displaystyle \frac{}{}}$! ${\displaystyle \frac{}{}}$! ${\displaystyle \frac{{2\mathrm{}}^{}}{}}$ j${\displaystyle \frac{{}^{}}{}}$+ ${\displaystyle \frac{1^{}}{}\frac{{}^{}}{}}$. ${\displaystyle c_{2j}={\frac$ {$1\cdot 3\cdot$5$\cdot (2j-1){2^{j+1}={$2^{$frac$ {($2j!$$}}{j!2^{2j+1}}\ .}$

(식 값이 n의 값과 독립적이므로 적분에는 나타나지 않는다는 점에 유의하십시오.)

${\displaystyle {\int \underbrace {x^{x^{\cdot^{\x}}} _{m}dx=\sum _{n=0}^{m}\frac {(-1)^{n}(n+1)^{n-1}{n!}}}\감마(n+1,-\ln x)+\sum _{n=m+1}^{\infit }^{n1}a_{mn}\감마(n+1,-\ln x)\qquad{{}(}x>0{\text{})}}}}$

여기서 ={ n = = ∑ j= - - ,j - 1 의 $displaystyle a_{mn}={\case}1&{\text}{{}n=0,\\\\\dfr${1$}{n!$$}}{{\text}{{}m=1,\\\{\dfrac {1}{n}\sum _{j=1}^{n1}ja_{m,n-j}a_{m-1,j-1}&{\text}}\case}}}}}}}}$

및 ((x,y)는 상부 불완전 감마함수다.

∫ 1x는 eλ+bdx=-1b− 1bλ ln ⁡(x+b는 eλ){\displaystyle \int{\frac{1}{ae^{\lambda)}+b}}\,dx={\frac{)}{b}}-{\frac{1}{b\lambda}}\ln \left(ae^{\lambda)}+b\right)} 때 b≠ 0{\displaystyle b\neq 0},λ ≠ 0{\displaystyle \lambda 0\neq}, x+b을은 eλ;0.{\displayst.$yle ae^{\bda x}+b>0.$$}$

${\displaystyle \int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\,dx={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]}$ when ${\displaystyle a\neq 0}$, ${\displa$$ystyle \lambda \neq 0$ 그리고 e + > $.$ {\$displaystyle ae^{\data$ x}+$b>0.$$}$

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

확정집적

${\displaystyle {\required}\int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\,dx&=\int _{0}^{0}{1}\frac {b}\right)^{x}\cdot b\,dx\\&=\int _{0}^{1}a^{x}\cdot b^{1-x}\,dx\\&={\frac {a-b}{\ln a-\ln b}}\qquad {\text{for }}a>0,\ b>0,\ a\neq b\end{aligned}}}$

마지막 표현식은 로그 평균이다.

${\displaystyle \int _{0}^{\\nf^{-ax}\,dx={\frac {1}{a}\quad (\operatorname {Re}(a)>0)}$

- d = 1 (> ) ${\displaystyle \int _{0}^{\\inflt \e^{-ax^{2}}\,dx={1}{1$}{1$}:{\sqrt {\pi$\}}}}}{{{1}{{{{$1}}}}}}}}}}}\sqrapi \quad$ (a>0$}}}}}}}}}})$

${\displaystyle \int_{-\infit _{-\infit }^{-ax^{2}}\,dx={\sqrt{\pi \over a}\}\cHB(a>0)}$

${\displaystyle \int_{-\infit _{-\inflit _{-ax^{2}}e^{-{\frac {b}{2}}\}\x={\sqrt {\pi{a}}e^{-2}{\sqrt{ab}}}}}\i(a,b)}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}$

${\displaystyle \int _{-\infit }^{-(ax^{2}+bx)}\,dx={\sqrt{\pi \over a}e^{\b^{2}}:{4a}\i(a>0)}$

${\displaystyle \int_{-\infit _{-}^{-(ax^{2}+bx+c)}\,dx={\sqrt{\pi \over a}e^{\b^{b^{2}}:{4a}-c}\i(a>0)}$

${\displaystyle \int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,dx={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)}$ (see Integral of a Gaussian function)

${\displaystyle \int_{-\infit }^{-a(x-b)^{2}}\,dx=b{\sqrt{\frac {}{a}}}\quad(\operatorname {Re}(a)>0)}$

${\displaystyle \int _{-\infty }^{\infty }xe^{-ax^{2}+bx}\,dx={\frac {{\sqrt {\pi }}b}{2a^{3/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{-\infit _{-\}^{2}e^{-ax^{2}}\,dx={\frac{1}{1}2}}:{\sqrt{{{3}}}\pi \cover (a>0)}$

${\displaystyle \int _{-\infty }^{\infty }x^{2}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(2a+b^{2})}{4a^{5/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{-\infty }^{\infty }x^{3}e^{-(ax^{2}+bx)}\,dx={\frac {{\sqrt {\pi }}(6a+b^{2})b}{8a^{7/2}}}e^{\frac {b^{2}}{4a}}\quad (\operatorname {Re} (a)>0)}$

${\displaystyle \int _{0}^{\infty }x^{n}e^{-ax^{2}}\,dx={\begin{cases}{\dfrac {\Gamma \left({\frac {n+1}{2}}\right)}{2\left(a^{\frac {n+1}{2}}\right)}}&(n>-1,\ a>0)\\\\{\dfrac {(2k-1)!!}}{2^{k+1}a^{k}}}{\sqrt {\dfrac {\pi }}}&(n=2k,\k={\text{정수,\>0)\ {\text{(!!는 이중 요인)}}}\\\{\dfrac {k!}}{2(a^{k+1}})&(n=2k+1,\k{\text 정수},\a>0)\end{case}}}}$

(조작자${\displaystyle !)}$ ${\displaystyle!!$${}$은(는) 이중 요인)

${\displaystyle \int _{0}^{n}^{n}e^{-ax}\,dx={\begin{base}{\dfrac {\dfrac}{\dfrac {\\\\\dfrac}{\dfrac}{\\\\\\dfrac}}\\\\\\\\dfrac {\\\\dfrac}\\\\\\\\\\\\\\\\\\dfrac {\\dfrac}\\}}{a^{n+1}}&(n=0,1,2,\ldots,\\\operatorname {Re}(a)>0)\case}}$

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

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

${\displaystyle \int_{0}^{\inflit }e^{-ax^{b}dx={\frac{1}{b}\a^{-{\frac{1}{b}}\Mama \left({\frac {1}{b}\오른쪽)}$

${\displaystyle \int_{0}^{n}^{n}e^{-ax^{b}}dx={\frac {1}{b}\^{\frac {1}{b}}}\ma \left({\frac{n+1}{b\rig)}}$

${\displaystyle \int_{0}^{0}^{\in^{-ax}\sin bx\,dx={\frac {b}{a^{a^{2}+b^{2}}}}\message (a>0)}$

${\displaystyle \int_{0}^{0}^{\nf^{-ax}\cos bx\,dx={\frac {a}{a^{2}+b^{2}}}}\message (a>0)}$

${\displaystyle \int_{0}^{0}^{\infl^{-ax}\sin bx\,dx={\frac {2ab}{(a^{2}+b^{2})^{2}}}}{2}}\cHB(a>0)}$

${\displaystyle \int_{0}^{0}^{\xe^{-ax}\cos bx\,dx={\frac {a^{2}-b^{2}}:{a^{2}+b^{2}}^{2}}:}\cape (a>0)}$

${\displaystyle \int_{0}^{0}^{\inflac {e^{-ax}\sin bx}{x}}\,dx=\arctan {\frac{b}{a}}}}}}}}$

${\displaystyle \int_{0}^{0}^{\inflt}{e^{-ax}-e^{-x}}\,dx=\ln {\frac}{a}}}}}}$

${\displaystyle \int_{0}^{0}^{\infrac {e^{-ax}-e^{-bx}}}}{x}}\sin px\,dx=\arctan {b}-\fractan {\p}}}}}$

${\displaystyle \int _{0}^{0}^{\frac{-ax}-e^{-bx}}}{x}}}\cospx\,dx={1}{1}{1}{1}}}}\frac {b^{2}+p^{2}{2}}}}{a^{2}+p^{2}}}}}$

${\displaystyle \int _{0}^{\inflit }{e^{-ax}(1-\cos x)}{x^{2}}:\,dx=\name {arccot} a-{\frac {a}{a}{2}}\ln(a^{2}+1)}}}$

${\displaystyle {\int }_{}^{}}$ 0 ${\displaystyle {\mathrm{}}_{}^{}}$ ${\displaystyle {\pi }^{}}$ ${\displaystyle {}_{}^{}{}^{}}$ ${\displaystyle {\cos }^{}}$ ${\displaystyle {}^{}}$ ${\displaystyle {}^{}}$ ${\displaystyle \theta }$= ${\displaystyle 2}$ ${\displaystyle \pi }$ I ${\displaystyle {0\mathrm{}}_{}}$( ${\displaystyle x}$) ${\displaystyle \int _{0}^{2\pi }e^{x\cos \theta }d\$pi =2$\pi I_{0}(x)}($나는 $$첫 번째 종류의₀ 수정된 베셀 함수)

${\displaystyle \int_{0}^{0}^{2\pi }e^{x\cos \theta +y\sin \d}d\theta =2\pi I_{0}\left({\sqrt {x^{2}+y^{2}{2}){2}}}\오른쪽)}$

${\displaystyle \int_{0}^{0}^{\inflit }{x^{s-1}{e^{x}/z-1}\,dx=\operatorname {Li} _{s}(z)\Gamma(s),}$

${\displaystyle \mathrm{여기}}$서 ${\displaystyle {\mathrm{Li}}_{}}$ ${\displaystyle s_{}}$ ${\displaystyle {}_{}}$( z${\displaystyle }$) ${\displaystyle \operatorname {Li} _{s}(z)}$은 Polylogarithm이다.

${\displaystyle \int_{0}^{0}^{\nfrac {\sin mx}{e^{2\pi x}-1}\,dx={\frac {1}{4}}}}}\frac {1}{1}{2m}}}}}}}}}$

${\displaystyle \int _{0}^{\\infully }e^{-x}\ln x\,dx=-\property ,}$

여기서 $\gamma {\displaystyle }$ ${\displaystyle \gamma}$은(는) 오일러-마스케로니 상수로, 다수의 확정 통합 값과 동일하다.

마지막으로, 잘 알려진 결과는,

${\displaystyle {\int }_{}^{}}$ ${\displaystyle {0\mathrm{}}_{}^{}}$ ${\displaystyle {\mathrm{}}_{}^{}}$ ${\displaystyle {\pi }^{}}$ e ${\displaystyle {}_{}^{}{}^{}}$ ${\displaystyle {(m}^{}}$- ${\displaystyle n^{}}$) ${\displaystyle {\varphi }^{}}$ ${\displaystyle {}^{}}$ ${\displaystyle \varphi }$= ${\displaystyle 2_{}}$ ${\displaystyle {}_{}}$ ${\displaystyle {m,}_{}}$ n ${\$정수 m,n)$$

여기서 ${\displaystyle {\Delta m}_{}}$, ${\displaystyle n_{}}$ ${\$은 크론커 델타다.

참고 항목

• 그라스틴과 리지크

추가 읽기

• Moll, Victor Hugo (2014-11-12). Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume I. Series: Monographs and Research Notes in Mathematics. I (1 ed.). Chapman and Hall/CRC Press. ISBN 978-1-48225-651-2. Retrieved 2016-02-12.
• Moll, Victor Hugo (2015-10-27). Special Integrals of Gradshteyn and Ryzhik: the Proofs – Volume II. Series: Monographs and Research Notes in Mathematics. II (1 ed.). Chapman and Hall/CRC Press. ISBN 978-1-48225-653-6. Retrieved 2016-02-12.

외부 링크

• 울프람 매스매티카 온라인 통합업체
• Moll, Victor Hugo. "List with the formulas and proofs in GR". Retrieved 2016-02-12.