فهرست انتگرال تابع‌های نمایی

در ادامه فهرستی از انتگرال تابع‌های نمایی آمده‌است؛ برای آگاهی از فهرستی کامل تر، صفحهٔ فهرست انتگرال‌ها را نگاه کنید.

انتگرال نامعین

در این روابط باید در سمت راست یعنی پاد مشتق مقدار ثابت انتگرال لحاظ شود، ولی در برخی از روابط زیر برای ساده‌نویسی نوشته نشده‌است.

$\int e^{x}\;\mathrm {d} x=e^{x}$$\int e^{x}\;\mathrm {d} x=e^{x}$

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

$\int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}$$\int a^{cx}\;\mathrm {d} x={\frac {1}{c\cdot \ln a}}a^{cx}$ for $a>0,\ a\neq 1$$a>0,\ a\neq 1$

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

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

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

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

$\int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}$$\int {\frac {e^{cx}}{x^{n}}}\;\mathrm {d} x={\frac {1}{n-1}}\left(-{\frac {e^{cx}}{x^{n-1}}}+c\int {\frac {e^{cx}}{x^{n-1}}}\,\mathrm {d} x\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}$

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

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

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

$\int e^{cx}\sin ^{n}x\;\mathrm {d} x={\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\;\mathrm {d} x$$\int e^{cx}\sin ^{n}x\;\mathrm {d} x={\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\;\mathrm {d} x$

$\int e^{cx}\cos ^{n}x\;\mathrm {d} x={\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\;\mathrm {d} x$$\int e^{cx}\cos ^{n}x\;\mathrm {d} x={\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\;\mathrm {d} x$

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

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

که ${\mbox{erf}}$${\mbox{erf}}$ همان تابع خطا یا Error function است.

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

$\int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;\mathrm {d} x=-{\frac {1}{2}}\left({\mbox{erf}}\,{\frac {-x+\mu }{\sigma {\sqrt {2}}}}\right)$$\int {1 \over \sigma {\sqrt {2\pi }}}\,e^{-{(x-\mu )^{2}/2\sigma ^{2}}}\;\mathrm {d} x=-{\frac {1}{2}}\left({\mbox{erf}}\,{\frac {-x+\mu }{\sigma {\sqrt {2}}}}\right)$

$\int e^{x^{2}}\,\mathrm {d} x=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}}}\;\mathrm {d} x\quad {\mbox{valid for }}n>0,$$\int e^{x^{2}}\,\mathrm {d} x=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}}}\;\mathrm {d} x\quad {\mbox{valid for }}n>0,$

که $c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)\,!}{j!\,2^{2j+1}}}\ .$$c_{2j}={\frac {1\cdot 3\cdot 5\cdots (2j-1)}{2^{j+1}}}={\frac {(2j)\,!}{j!\,2^{2j+1}}}\ .$

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

where $a_{mn}={\begin{cases}1&{\text{if }}n=0,\\{\frac {1}{n!}}&{\text{if }}m=1,\\{\frac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}$$a_{mn}={\begin{cases}1&{\text{if }}n=0,\\{\frac {1}{n!}}&{\text{if }}m=1,\\{\frac {1}{n}}\sum _{j=1}^{n}ja_{m,n-j}a_{m-1,j-1}&{\text{otherwise}}\end{cases}}$

$\int {\frac {1}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)\,$$\int {\frac {1}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {x}{b}}-{\frac {1}{b\lambda }}\ln \left(ae^{\lambda x}+b\right)\,$ هرگاه $b\neq 0$$b\neq 0$, $\lambda \neq 0$$\lambda \neq 0$, و $ae^{\lambda x}+b>0\,.$$ae^{\lambda x}+b>0\,.$

$\int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]\,$$\int {\frac {e^{2\lambda x}}{ae^{\lambda x}+b}}\;\mathrm {d} x={\frac {1}{a^{2}\lambda }}\left[ae^{\lambda x}+b-b\ln \left(ae^{\lambda x}+b\right)\right]\,$ هرگاه $a\neq 0$$a\neq 0$, $\lambda \neq 0$$\lambda \neq 0$, and $ae^{\lambda x}+b>0\,.$$ae^{\lambda x}+b>0\,.$

انتگرال معین

$\int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\;\mathrm {d} x=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\;\mathrm {d} x=\int _{0}^{1}a^{x}\cdot b^{1-x}\;\mathrm {d} x={\frac {a-b}{\ln a-\ln b}}$$\int _{0}^{1}e^{x\cdot \ln a+(1-x)\cdot \ln b}\;\mathrm {d} x=\int _{0}^{1}\left({\frac {a}{b}}\right)^{x}\cdot b\;\mathrm {d} x=\int _{0}^{1}a^{x}\cdot b^{1-x}\;\mathrm {d} x={\frac {a-b}{\ln a-\ln b}}$ برای $a>0,\ b>0,\ a\neq b$$a>0,\ b>0,\ a\neq b$, که همان میانگین لگاریتمی است.

$\int _{0}^{\infty }e^{-ax}\,\mathrm {d} x={\frac {1}{a}}(a>0)$$\int _{0}^{\infty }e^{-ax}\,\mathrm {d} x={\frac {1}{a}}(a>0)$

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

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

$\int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)$$\int _{-\infty }^{\infty }e^{-ax^{2}}e^{-2bx}\,\mathrm {d} x={\sqrt {\frac {\pi }{a}}}e^{\frac {b^{2}}{a}}\quad (a>0)$ (نگاه کنید به انتگرال تابع‌های گوسی)

$\int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}$$\int _{-\infty }^{\infty }xe^{-a(x-b)^{2}}\,\mathrm {d} x=b{\sqrt {\frac {\pi }{a}}}$

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

$\int _{0}^{\infty }x^{n}e^{-ax^{2}}\,\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}}$$\int _{0}^{\infty }x^{n}e^{-ax^{2}}\,\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}}$ (!! is the double factorial)

$\int _{0}^{\infty }x^{n}e^{-ax}\,\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}}$$\int _{0}^{\infty }x^{n}e^{-ax}\,\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}}$

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

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

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

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

$\int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)$$\int _{0}^{2\pi }e^{x\cos \theta }d\theta =2\pi I_{0}(x)$ ($I_{0}$$I_{0}$ تابع بسل اصلاح شده از نوع نخست است.)

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

منابع

• Wolfram Mathematica Online Integrator
• V. H. Moll, The Integrals in Gradshteyn and Ryzhik