지수함수 적분표

1. 개요

지수함수 적분표는 지수 함수를 포함하는 다양한 함수의 부정 적분과 정적분 공식을 정리한 문서이다. 부정 적분은 미분의 역연산으로, 지수 함수, 다항식, 삼각 함수, 오차 함수 등을 포함하는 함수의 적분 공식을 제공한다. 정적분은 주어진 구간에서 함수의 적분 값을 계산하며, 가우스 적분, 삼각 함수, 그리고 다양한 지수 함수와 관련된 적분 공식을 포함한다.

2. 부정적분

부정적분은 미분 연산의 역연산으로, 주어진 함수의 도함수를 가지는 모든 함수(역도함수)를 찾는 연산이다. 이 문서에서는 적분 상수 $C$는 생략한다.

2.1. 지수 함수만 포함하는 함수의 적분

다음은 지수 함수 적분표의 부정 적분 목록이다. 오른쪽에 오는 적분 상수는 생략했다.

:$\backslash int\; f\text{'}(x)e^\{f(x)\}\backslash ;\backslash mathrm\{d\}x\; =\; e^\{f(x)\}$

:$\backslash int\; e^\{cx\}\backslash ;\backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{c\}\; e^\{cx\}$

:$\backslash int\; a^\{cx\}\backslash ;\backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{c\backslash cdot\; \backslash ln\; a\}\; a^\{cx\}$ (a^영어 > 0, a^영어 ≠ 1)

:$\backslash int\; xe^\{cx\}\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{e^\{cx\}\}\{c^2\}(cx-1)$

:$\backslash int\; x^2\; e^\{cx\}\backslash ;\backslash mathrm\{d\}x\; =\; e^\{cx\}\backslash left(\backslash frac\{x^2\}\{c\}-\backslash frac\{2x\}\{c^2\}+\backslash frac\{2\}\{c^3\}\backslash right)$

:$\backslash int\; x^n\; e^\{cx\}\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{c\}\; x^n\; e^\{cx\}\; -\; \backslash frac\{n\}\{c\}\backslash int\; x^\{n-1\}\; e^\{cx\}\; \backslash mathrm\{d\}x\; =\; \backslash left(\; \backslash frac\{\backslash partial\}\{\backslash partial\; c\}\; \backslash right)^n\; \backslash frac\{e^\{cx\}\}\{c\}$

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

:$\backslash int\backslash frac\{e^\{cx\}\}\{x^n\}\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{n-1\}\backslash left(-\backslash frac\{e^\{cx\}\}\{x^\{n-1\}\}+c\backslash int\backslash frac\{e^\{cx\}\; \}\{x^\{n-1\}\}\backslash ,\backslash mathrm\{d\}x\backslash right)$ (n^영어≠ 1)

:$\backslash int\; e^\{cx\}\backslash ln\; x\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{c\}\backslash left(e^\{cx\}\backslash ln|x|-\backslash operatorname\{Ei\}\backslash ,(cx)\backslash right)$

:$\backslash int\; e^\{cx\}\backslash sin\; bx\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{e^\{cx\}\}\{c^2+b^2\}(c\backslash sin\; bx\; -\; b\backslash cos\; bx)$

:$\backslash int\; e^\{cx\}\backslash cos\; bx\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{e^\{cx\}\}\{c^2+b^2\}(c\backslash cos\; bx\; +\; b\backslash sin\; bx)$

:$\backslash int\; e^\{cx\}\backslash sin^n\; x\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{e^\{cx\}\backslash sin^\{n-1\}\; x\}\{c^2+n^2\}(c\backslash sin\; x-n\backslash cos\; x)+\backslash frac\{n(n-1)\}\{c^2+n^2\}\backslash int\; e^\{cx\}\backslash sin^\{n-2\}\; x\backslash ;\backslash mathrm\{d\}x$

:$\backslash int\; e^\{cx\}\backslash cos^n\; x\backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{e^\{cx\}\backslash cos^\{n-1\}\; x\}\{c^2+n^2\}(c\backslash cos\; x+n\backslash sin\; x)+\backslash frac\{n(n-1)\}\{c^2+n^2\}\backslash int\; e^\{cx\}\backslash cos^\{n-2\}\; x\backslash ;\backslash mathrm\{d\}x$

:$\backslash int\; x\; e^\{c\; x^2\; \}\backslash ;\; \backslash mathrm\{d\}x=\; \backslash frac\{1\}\{2c\}\; \backslash ;\; e^\{c\; x^2\}$

:$\backslash int\; e^\{-c\; x^2\; \}\backslash ;\; \backslash mathrm\{d\}x=\; \backslash sqrt\{\backslash frac\{\backslash pi\}\{4c\}\}\; \backslash operatorname\{erf\}(\backslash sqrt\{c\}\; x)$ (erf^영어는 오차 함수)

:$\backslash int\; xe^\{-c\; x^2\; \}\backslash ;\; \backslash mathrm\{d\}x=-\backslash frac\{1\}\{2c\}e^\{-cx^2\}$

:$\backslash int\backslash frac\{e^\{-x^2\}\}\{x^2\}\backslash ;\; \backslash mathrm\{d\}x\; =\; -\backslash frac\{e^\{-x^2\}\}\{x\}\; -\; \backslash sqrt\{\backslash pi\}\; \backslash mathrm\{erf\}\; (x)$

:$\backslash int\; \{\backslash frac\{1\}\{\backslash sigma\backslash sqrt\{2\backslash pi\}\}\; e^\{\; -\backslash frac\{1\}\{2\}\backslash left(\backslash frac\{x-\backslash mu\}\{\backslash sigma\}\backslash right)^2\; \}\}\backslash ;\; \backslash mathrm\{d\}x=\; \backslash frac\{1\}\{2\}\; \backslash left(\backslash operatorname\{erf\}\backslash ,\backslash frac\{x-\backslash mu\}\{\backslash sigma\; \backslash sqrt\{2\}\}\backslash right)$

:$\backslash int\; e^\{x^2\}\backslash ,\backslash mathrm\{d\}x\; =\; e^\{x^2\}\backslash left(\; \backslash sum\_\{j=0\}^\{n-1\}c\_\{2j\}\backslash ,\backslash frac\{1\}\{x^\{2j+1\}\}\; \backslash right\; )+(2n-1)c\_\{2n-2\}\; \backslash int\; \backslash frac\{e^\{x^2\}\}\{x^\{2n\}\}\backslash ;\backslash mathrm\{d\}x$ (n^영어 > 0)
: 여기서 $c\_\{2j\}=\backslash frac\{\; 1\; \backslash cdot\; 3\; \backslash cdot\; 5\; \backslash cdots\; (2j-1)\}\{2^\{j+1\}\}=\backslash frac\{(2j)\backslash ,!\}\{j!\backslash ,\; 2^\{2j+1\}\}$ 로 한다.

:$\{\backslash int\; \backslash underbrace\{x^\{x^\{\backslash cdot^\{\backslash cdot^\{x\}\}\}\}\}\_m\; \backslash ,dx=\; \backslash int\; x\backslash uparrow\backslash uparrow\; m\; \backslash ,dx=\; \backslash int\; x\backslash to\; m\backslash to\; 2\; \backslash ,dx=\; \backslash sum\_\{n=0\}^m\backslash frac\{(-1)^n(n+1)^\{n-1\}\}\{n!\}\backslash Gamma(n+1,-\; \backslash ln\; x)\; +\; \backslash sum\_\{n=m+1\}^\backslash infty(-1)^na\_\{mn\}\backslash Gamma(n+1,-\backslash ln\; x)$ (for x^영어> 0)
:: 여기서
:: $\backslash Gamma(x,y)$는 감마 함수, $\backslash uparrow$는 크누스 화살표 표기, $\backslash to$는 콘웨이 사슬 표기

:$\backslash int\; \backslash frac\{1\}\{ae^\{\backslash lambda\; x\}\; +\; b\}\; \backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{x\}\{b\}\; -\; \backslash frac\{1\}\{b\; \backslash lambda\}\; \backslash ln\backslash left(a\; e^\{\backslash lambda\; x\}\; +\; b\; \backslash right)$ (b^영어 ≠ 0, λ^영어 ≠ 0, 그리고 $ae^\{\backslash lambda\; x\}\; +\; b\; >\; 0$)

:$\backslash int\; \backslash frac\{e^\{2\backslash lambda\; x\}\}\{ae^\{\backslash lambda\; x\}\; +\; b\}\; \backslash ;\; \backslash mathrm\{d\}x\; =\; \backslash frac\{1\}\{a^2\; \backslash lambda\}\; \backslash left[a\; e^\{\backslash lambda\; x\}\; +\; b\; -\; b\; \backslash ln\backslash left(a\; e^\{\backslash lambda\; x\}\; +\; b\; \backslash right)\; \backslash right]$ (a^영어 ≠ 0, λ^영어 ≠ 0, 그리고 $ae^\{\backslash lambda\; x\}\; +\; b\; >\; 0$)

2.2. 다항식을 포함하는 함수의 적분

:$\backslash int\; xe^\{cx\}\backslash ,dx\; =\; e^\{cx\}\backslash left(\backslash frac\{cx-1\}\{c^\{2\}\}\backslash right)$

:$\backslash int\; x^2\; e^\{cx\}\backslash ,dx\; =\; e^\{cx\}\backslash left(\backslash frac\{x^2\}\{c\}-\backslash frac\{2x\}\{c^2\}+\backslash frac\{2\}\{c^3\}\backslash right)$

:$\backslash begin\{align\}$
\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}

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

:$\backslash int\backslash frac\{e^\{cx\}\}\{x^n\}\backslash ,dx\; =\; \backslash frac\{1\}\{n-1\}\backslash left(-\backslash frac\{e^\{cx\}\}\{x^\{n-1\}\}+c\backslash int\backslash frac\{e^\{cx\}\; \}\{x^\{n-1\}\}\backslash ,dx\backslash right)$ (단, $n\backslash neq\; 1$)

2.3. 삼각함수를 포함하는 함수의 적분

$\backslash int\; e^\{cx\}\backslash sin\; bx\backslash ,dx\; =\; \backslash frac\{e^\{cx\}\}\{c^2+b^2\}(c\backslash sin\; bx\; -\; b\backslash cos\; bx)\; =\; \backslash frac\{e^\{cx\}\}\{\backslash sqrt\{c^2+b^2\}\}\backslash sin(bx-\backslash phi)$ (이때 $\backslash cos(\backslash phi)\; =\; \backslash frac\{c\}\{\backslash sqrt\{c^2+b^2\}\}$)

$\backslash int\; e^\{cx\}\backslash cos\; bx\backslash ,dx\; =\; \backslash frac\{e^\{cx\}\}\{c^2+b^2\}(c\backslash cos\; bx\; +\; b\backslash sin\; bx)\; =\; \backslash frac\{e^\{cx\}\}\{\backslash sqrt\{c^2+b^2\}\}\backslash cos(bx-\backslash phi)$ (이때 $\backslash cos(\backslash phi)\; =\; \backslash frac\{c\}\{\backslash sqrt\{c^2+b^2\}\}$)

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

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

2.4. 오차 함수와 관련된 함수의 적분

다음은 오차 함수(erf) 및 지수 적분 함수(Ei)와 관련된 함수의 적분 공식이다. (적분 상수는 생략)

:$\backslash int\; e^\{cx\}\backslash ln\; x\backslash ,dx\; =\; \backslash frac\{1\}\{c\}\backslash left(e^\{cx\}\backslash ln|x|-\backslash operatorname\{Ei\}(cx)\backslash right)$
:$\backslash int\; x\; e^\{c\; x^2\; \}\backslash ,dx=\; \backslash frac\{1\}\{2c\}\; e^\{c\; x^2\}$
:$\backslash int\; e^\{-c\; x^2\; \}\backslash ,dx=\; \backslash sqrt\{\backslash frac\{\backslash pi\}\{4c\}\}\; \backslash operatorname\{erf\}(\backslash sqrt\{c\}\; x)$
:$\backslash int\; xe^\{-c\; x^2\; \}\backslash ,dx=-\backslash frac\{1\}\{2c\}e^\{-cx^2\}$
:$\backslash int\backslash frac\{e^\{-x^2\}\}\{x^2\}\backslash ,dx\; =\; -\backslash frac\{e^\{-x^2\}\}\{x\}\; -\; \backslash sqrt\{\backslash pi\}\; \backslash operatorname\{erf\}\; (x)$
:$\backslash int\; \{\backslash frac\{1\}\{\backslash sigma\backslash sqrt\{2\backslash pi\}\}\; e^\{\; -\backslash frac\{1\}\{2\}\backslash left(\backslash frac\{x-\backslash mu\}\{\backslash sigma\}\backslash right)^2\; \}\}\backslash ,dx=\; \backslash frac\{1\}\{2\}\backslash operatorname\{erf\}\backslash left(\backslash frac\{x-\backslash mu\}\{\backslash sigma\; \backslash sqrt\{2\}\}\backslash right)$

2.5. 기타 적분

:$\backslash int\; e^\{x^2\}\backslash ,dx\; =\; e^\{x^2\}\backslash left(\; \backslash sum\_\{j=0\}^\{n-1\}c\_\{2j\}\backslash frac\{1\}\{x^\{2j+1\}\}\; \backslash right\; )+(2n-1)c\_\{2n-2\}\; \backslash int\; \backslash frac\{e^\{x^2\}\}\{x^\{2n\}\}\backslash ,dx$ (이때 $c\_\{2j\}=\backslash frac\{\; 1\; \backslash cdot\; 3\; \backslash cdot\; 5\; \backslash cdots\; (2j-1)\}\{2^\{j+1\}\}=\backslash frac\{(2j)!\}\{j!2^\{2j+1\}\}$이고, 모든 $n>0$에 대해 성립한다.)

:$\{\backslash int\; \backslash underbrace\{x^\{x^\{\backslash cdot^\{\backslash cdot^\{x\}\}\}\}\}\_mdx=\; \backslash sum\_\{n=0\}^m\backslash frac\{(-1)^n(n+1)^\{n-1\}\}\{n!\}\backslash Gamma(n+1,-\; \backslash ln\; x)\; +\; \backslash sum\_\{n=m+1\}^\backslash infty(-1)^na\_\{mn\}\backslash Gamma(n+1,-\backslash ln\; x)\; \backslash qquad\backslash text\{(for\; \}x>\; 0\backslash text\{)\}\}$
: (이때 이고, $\backslash Gamma(x,y)$는 불완전 감마 함수이다.)

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

:$\backslash int\; \backslash frac\{e^\{2\backslash lambda\; x\}\}\{ae^\{\backslash lambda\; x\}\; +\; b\}\; \backslash ,dx\; =\; \backslash frac\{1\}\{a^2\; \backslash lambda\}\; \backslash left[a\; e^\{\backslash lambda\; x\}\; +\; b\; -\; b\; \backslash ln\backslash left(a\; e^\{\backslash lambda\; x\}\; +\; b\; \backslash right)\; \backslash right]$ (이때 $a\; \backslash neq\; 0$, $\backslash lambda\; \backslash neq\; 0$이고 $ae^\{\backslash lambda\; x\}\; +\; b\; >\; 0$이다.)

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

3. 정적분

다음은 지수함수의 정적분 공식들이다.

:: $\backslash int\_0^1\; e^\{x\backslash cdot\; \backslash ln\; a\; +\; (1-x)\backslash cdot\; \backslash ln\; b\}\backslash ,dx$
= \frac{a-b}{\ln a - \ln b} (a > 0, b > 0, a ≠ b)

위 적분식의 마지막 값은 로그 평균을 뜻한다.

:: $\backslash int\_0^\{\backslash infty\}\; e^\{-ax\}\backslash ,dx=\backslash frac\{1\}\{a\}\; \backslash quad\; (\backslash operatorname\{Re\}(a)>0)$

:: $\backslash int\_0^\{\backslash infty\}\; e^\{-ax^2\}\backslash ,dx=\backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash pi\; \backslash over\; a\}\; \backslash quad\; (a>0)$ (가우스 적분)

:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-ax^2\}\backslash ,dx=\backslash sqrt\{\backslash pi\; \backslash over\; a\}\; \backslash quad\; (a>0)$
:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-ax^2\}\; e^\{-\backslash frac\{b\}\{x^2\}\}\backslash ,dx=\backslash sqrt\{\backslash frac\{\backslash pi\}\{a\}\}e^\{-2\backslash sqrt\{ab\}\}\; \backslash quad\; (a,b>0)$
:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-(ax^2\; +\; bx)\}\backslash ,dx=\; \backslash sqrt\{\backslash pi\; \backslash over\; a\}e^\{\backslash tfrac\{b^2\}\{4a\}\}\; \backslash quad(a\; >\; 0)$

:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-ax^2\}\; e^\{-2bx\}\backslash ,dx=\backslash sqrt\{\backslash frac\{\backslash pi\}\{a\}\}e^\{\backslash frac\{b^2\}\{a\}\}\; \backslash quad\; (a>0)$

:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; x\; e^\{-a(x-b)^2\}\backslash ,dx=\; b\; \backslash sqrt\{\backslash frac\{\backslash pi\}\{a\}\}\; \backslash quad\; (\backslash operatorname\{Re\}(a)>0)$

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

:: $\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; x^2\; e^\{-ax^2\}\backslash ,dx=\backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash pi\; \backslash over\; a^3\}\; \backslash quad\; (a>0)$

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

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

:: $\backslash int\_0^\{\backslash infty\}\; x^\{n\}\; e^\{-ax^2\}\backslash ,dx\; =$
\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.)

:: $\backslash int\_0^\{\backslash infty\}\; x^n\; e^\{-ax\}\backslash ,dx\; =$
\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}

:: $\backslash int\_0^\{1\}\; x^n\; e^\{-ax\}\backslash ,dx\; =$
\frac{n!}{a^{n+1}}\left[
1-e^{-a}\sum_{i=0}^{n} \frac{a^i}{i!}
\right]

:: $\backslash int\_0^\{b\}\; x^n\; e^\{-ax\}\backslash ,dx\; =$
\frac{n!}{a^{n+1}}\left[
1-e^{-ab}\sum_{i=0}^{n} \frac{(ab)^i}{i!}
\right]

:: $\backslash int\_0^\backslash infty\; e^\{-ax^b\}\; dx\; =\; \backslash frac\{1\}\{b\}\backslash \; a^\{-\backslash frac\{1\}\{b\}\}\backslash Gamma\backslash left(\backslash frac\{1\}\{b\}\backslash right)$

:: $\backslash int\_0^\backslash infty\; x^n\; e^\{-ax^b\}\; dx\; =\; \backslash frac\{1\}\{b\}\backslash \; a^\{-\backslash frac\{n+1\}\{b\}\}\backslash Gamma\backslash left(\backslash frac\{n+1\}\{b\}\backslash right)$

:: $\backslash int\_0^\{\backslash infty\}\; e^\{-ax\}\backslash sin\; bx\backslash ,dx\; =\; \backslash frac\{b\}\{a^2+b^2\}\; \backslash quad\; (a>0)$

:: $\backslash int\_0^\{\backslash infty\}\; e^\{-ax\}\backslash cos\; bx\backslash ,dx\; =\; \backslash frac\{a\}\{a^2+b^2\}\; \backslash quad\; (a>0)$

:: $\backslash int\_0^\{\backslash infty\}\; xe^\{-ax\}\backslash sin\; bx\backslash ,dx\; =\; \backslash frac\{2ab\}\{(a^2+b^2)^2\}\; \backslash quad\; (a>0)$

:: $\backslash int\_0^\{\backslash infty\}\; xe^\{-ax\}\backslash cos\; bx\backslash ,dx\; =\; \backslash frac\{a^2-b^2\}\{(a^2+b^2)^2\}\; \backslash quad\; (a>0)$
:: $\backslash int\_0^\{\backslash infty\}\; \backslash frac\{e^\{-ax\}\backslash sin\; bx\}\{x\}\backslash ,dx=\backslash arctan\; \backslash frac\{b\}\{a\}$
:: $\backslash int\_0^\{\backslash infty\}\; \backslash frac\{e^\{-ax\}-e^\{-bx\}\}\{x\}\backslash ,dx=\backslash ln\; \backslash frac\{b\}\{a\}$
:: $\backslash int\_0^\{\backslash infty\}\; \backslash frac\{e^\{-ax\}-e^\{-bx\}\}\{x\}\; \backslash sin\; px\; \backslash ,\; dx=\backslash arctan\; \backslash frac\{b\}\{p\}\; -\; \backslash arctan\; \backslash frac\{a\}\{p\}$
:: $\backslash int\_0^\{\backslash infty\}\; \backslash frac\{e^\{-ax\}-e^\{-bx\}\}\{x\}\; \backslash cos\; px\; \backslash ,\; dx=\backslash frac\{1\}\{2\}\; \backslash ln\; \backslash frac\{b^2+p^2\}\{a^2+p^2\}$
:: $\backslash int\_0^\{\backslash infty\}\; \backslash frac\{e^\{-ax\}\; (1-\backslash cos\; x)\}\{x^2\}\backslash ,dx=\backslash arccot\; a\; -\; \backslash frac\{a\}\{2\}\backslash ln\; (a^2+1)$

:: $\backslash int\_0^\{2\; \backslash pi\}\; e^\{x\; \backslash cos\; \backslash theta\}\; d\; \backslash theta\; =\; 2\; \backslash pi\; I\_0(x)$ (I₀^영어는 제1종 변형 베셀 함수이다.)

:: $\backslash int\_0^\{2\; \backslash pi\}\; e^\{x\; \backslash cos\; \backslash theta\; +\; y\; \backslash sin\; \backslash theta\}\; d\; \backslash theta\; =\; 2\; \backslash pi\; I\_0\; \backslash left(\; \backslash sqrt\{x^2\; +\; y^2\}\; \backslash right)$

:: $\backslash int\_0^\backslash infty\backslash frac\{x^\{s-1\}\}\{e^x/z-1\}\; \backslash ,dx\; =\; \backslash operatorname\{Li\}\_\{s\}(z)\backslash Gamma(s),$ ($\backslash operatorname\{Li\}\_\{s\}(z)$는 다중로그이다.)

:: $\backslash int\_0^\backslash infty\backslash frac\{\backslash sin\; mx\}\{e^\{2\; \backslash pi\; x\}-1\}\; \backslash ,dx\; =\; \backslash frac\{1\}\{4\}\; \backslash coth\; \backslash frac\{m\}\{2\}\; -\; \backslash frac\{1\}\{2m\}$

:: $\backslash int\_0^\backslash infty\; e^\{-x\}\; \backslash ln\; x\backslash ,\; dx\; =\; -\; \backslash gamma,$ ($\backslash gamma$는 오일러-마스케로니 상수이다.)

:: $$
\int_0^1 e^{x\cdot \ln a + (1-x)\cdot \ln b}\;\mathrm{d}x =
\frac{a-b}{\ln a - \ln b} ( $a\; >\; 0,\backslash \; b\; >\; 0,\backslash \; a\; \backslash ne\; b$ )

::

::$\backslash int\_\{0\}^\{\backslash infty\}\; e^\{-ax^2\}\backslash ,\backslash mathrm\{d\}x=\backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash pi\; \backslash over\; a\}\; \backslash quad\; (a>0)$ ( 가우스 적분 )

::$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-ax^2\}\backslash ,\backslash mathrm\{d\}x=\backslash sqrt\{\backslash pi\; \backslash over\; a\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; e^\{-ax^2\}\; e^\{-2bx\}\backslash ,\backslash mathrm\{d\}x=\backslash sqrt\{\backslash frac\{\backslash pi\}\{a\}\}e^\{\backslash frac\{b^2\}\{a\}\}\; \backslash quad\; (a>0)$ ( 가우스 함수의 적분 )

::$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; x\; e^\{-a(x-b)^2\}\backslash ,\backslash mathrm\{d\}x=\; b\; \backslash sqrt\{\backslash frac\{\backslash pi\}\{a\}\}\; \backslash quad\; (\backslash operatorname\{Re\}(a)>0)$

::$\backslash int\_\{-\backslash infty\}^\{\backslash infty\}\; x^2\; e^\{-ax^2\}\backslash ,\backslash mathrm\{d\}x=\backslash frac\{1\}\{2\}\; \backslash sqrt\{\backslash pi\; \backslash over\; a^3\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{0\}^\{\backslash infty\}\; x^\{n\}\; e^\{-ax^2\}\backslash ,\backslash mathrm\{d\}x\; =$
\begin{cases}
\frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n>-1,a>0) \\
\frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{integer}, a>0) \\
\frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{integer}, a>0)
\end{cases} ( !! 는 이중 계승 )

::$\backslash int\_\{0\}^\{\backslash infty\}\; x^n\; e^\{-ax\}\backslash ,\backslash mathrm\{d\}x\; =$
\begin{cases}
\frac{\Gamma(n+1)}{a^{n+1}} & (n>-1,a>0) \\
\frac{n!}{a^{n+1}} & (n=0,1,2,\ldots,a>0) \\
\end{cases}

::$\backslash int\_0^\backslash infty\; e^\{-ax^b\}\; dx\; =\; \backslash frac\{1\}\{b\}\backslash \; a^\{-\backslash frac\{1\}\{b\}\}\; \backslash ,\; \backslash Gamma\backslash left(\backslash frac\{1\}\{b\}\backslash right)$

::$\backslash int\_0^\backslash infty\; x^n\; e^\{-ax^b\}\; dx\; =\; \backslash frac\{1\}\{b\}\backslash \; a^\{-\backslash frac\{n+1\}\{b\}\}\; \backslash ,\; \backslash Gamma\backslash left(\backslash frac\{n+1\}\{b\}\backslash right)$

::$\backslash int\_\{0\}^\{\backslash infty\}\; e^\{-ax\}\backslash sin\; bx\; \backslash ,\; \backslash mathrm\{d\}x\; =\; \backslash frac\{b\}\{a^2+b^2\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{0\}^\{\backslash infty\}\; e^\{-ax\}\backslash cos\; bx\; \backslash ,\; \backslash mathrm\{d\}x\; =\; \backslash frac\{a\}\{a^2+b^2\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{0\}^\{\backslash infty\}\; xe^\{-ax\}\backslash sin\; bx\; \backslash ,\; \backslash mathrm\{d\}x\; =\; \backslash frac\{2ab\}\{(a^2+b^2)^2\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{0\}^\{\backslash infty\}\; xe^\{-ax\}\backslash cos\; bx\; \backslash ,\; \backslash mathrm\{d\}x\; =\; \backslash frac\{a^2-b^2\}\{(a^2+b^2)^2\}\; \backslash quad\; (a>0)$

::$\backslash int\_\{0\}^\{2\; \backslash pi\}\; e^\{x\; \backslash cos\; \backslash theta\}\; d\; \backslash theta\; =\; 2\; \backslash pi\; I\_\{0\}(x)$ ( $I\_\{0\}$ 는 변형 베셀 함수 )

::$\backslash int\_\{0\}^\{2\; \backslash pi\}\; e^\{x\; \backslash cos\; \backslash theta\; +\; y\; \backslash sin\; \backslash theta\}\; d\; \backslash theta\; =\; 2\; \backslash pi\; I\_\{0\}\; \backslash left(\; \backslash sqrt\{x^2\; +\; y^2\}\; \backslash right)$

:: $\backslash int\_0^\backslash infty\backslash frac\{x^\{s-1\}\}\{e^x/z-1\}\; \backslash ,dx\; =\; \backslash operatorname\{Li\}\_\{s\}(z)\backslash Gamma(s),$
::여기서 $\backslash operatorname\{Li\}\_\{s\}(z)$는 폴리로그 함수이다.

:: $\backslash int\_0^\backslash infty\backslash frac\{\backslash sin\; mx\}\{e^\{2\; \backslash pi\; x\}-1\}\; \backslash ,dx\; =\; \backslash frac\{1\}\{4\}\; \backslash coth\; \backslash frac\{m\}\{2\}\; -\; \backslash frac\{1\}\{2m\}$

:: $\backslash int\_0^\backslash infty\; e^\{-x\}\; \backslash ln\; x\backslash ,\; dx\; =\; -\; \backslash gamma,$
::여기서 $\backslash gamma$는 오일러-마스케로니 상수이다.

마지막으로, 잘 알려진 결과는 다음과 같다.
:$\backslash int\_0^\{2\; \backslash pi\}\; e^\{i(m-n)\backslash phi\}\; d\backslash phi\; =\; 2\; \backslash pi\; \backslash delta\_\{m,n\}\; \backslash qquad\backslash text\{for\; \}m,n\backslash in\backslash mathbb\{Z\}$
::여기서 $\backslash delta\_\{m,n\}$는 크로네커 델타이다.