Thảo luận:Nhập môn Giải tích/Tích phân/Tích phân bất định/Tích phân hàm số mũ

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

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

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

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

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

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

$\int \frac{e^{cx}}{x^n}\mathrm{d}x=\frac{1}{n-1}(-\frac{e^{cx}}{x^{n-1}}+c\int \frac{e^{cx}}{x^{n-1}}\mathrm{d}x)\text{(for }n\ne 1\text{)}$

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

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

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

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

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

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

$\int e^{-c{x}^2}\mathrm{d}x=\sqrt{\frac{\pi }{4c}}\text{erf}(\sqrt{c}x)$ ($\text{erf}$ is the Error function)

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

$\int \frac{1}{\sigma \sqrt{2\pi }}{e}^{-(x-\mu {)}^2/2{\sigma }^2}\mathrm{d}x=\frac{1}{2}(1+\text{erf}\frac{x-\mu }{\sigma \sqrt{2}})$

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

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

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

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

Definite integrals

${\int }_0^1{e}^{x\cdot \ln a+(1-x)\cdot \ln b}\mathrm{d}x={\int }_0^1{\left(\frac{a}{b}\right)}^x\cdot b\mathrm{d}x={\int }_0^1{a}^x\cdot b^{1-x}\mathrm{d}x=\frac{a-b}{\ln a-\ln b}$ for $a>0,b>0,a\ne b$, which is the logarithmic mean

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\mathrm{d}x=\frac{1}{a}$

${\int }_0^{\mathrm{\infty }}{e}^{-a{x}^2}\mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{a}}(a>0)$ (the Gaussian integral)

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

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

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

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

${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-a{x}^2}\mathrm{d}x=\{\begin{array}{cc}\frac{1}{2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}}& (n>-1,a>0)\\ \frac{(2k-1)!!}{2^{k+1}{a}^k}\sqrt{\frac{\pi }{a}}& (n=2k,k\text{integer},a>0)\\ \frac{k!}{2a^{k+1}}& (n=2k+1,k\text{integer},a>0)\end{array}$ (!! is the double factorial)

${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-ax}\mathrm{d}x=\{\begin{array}{cc}\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}& (n>-1,a>0)\\ \frac{n!}{a^{n+1}}& (n=0,1,2,\dots ,a>0)\end{array}$

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin bx\mathrm{d}x=\frac{b}{a^2+b^2}(a>0)$

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\cos bx\mathrm{d}x=\frac{a}{a^2+b^2}(a>0)$

${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\sin bx\mathrm{d}x=\frac{2ab}{(a^2+b^2{)}^2}(a>0)$

${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\cos bx\mathrm{d}x=\frac{a^2-b^2}{(a^2+b^2{)}^2}(a>0)$

${\int }_0^{2\pi }{e}^{x\cos \theta }d\theta =2\pi I_0(x)$ ($I_0$ is the modified Bessel function of the first kind)

${\int }_0^{2\pi }{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\left(\sqrt{x^2+y^2}\right)$

The following is a list of integrals (antiderivative functions) of exponential functions. For a complete list of Integral functions, please see the list of integrals.

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

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

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

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

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

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

$\int \frac{e^{cx}}{x^n}\mathrm{d}x=\frac{1}{n-1}(-\frac{e^{cx}}{x^{n-1}}+c\int \frac{e^{cx}}{x^{n-1}}\mathrm{d}x)\text{(for }n\ne 1\text{)}$

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

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

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

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

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

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

$\int e^{-c{x}^2}\mathrm{d}x=\sqrt{\frac{\pi }{4c}}\text{erf}(\sqrt{c}x)$ ($\text{erf}$ is the Error function)

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

$\int \frac{1}{\sigma \sqrt{2\pi }}{e}^{-(x-\mu {)}^2/2{\sigma }^2}\mathrm{d}x=\frac{1}{2}(1+\text{erf}\frac{x-\mu }{\sigma \sqrt{2}})$

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

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

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

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

Definite integrals

${\int }_0^1{e}^{x\cdot \ln a+(1-x)\cdot \ln b}\mathrm{d}x={\int }_0^1{\left(\frac{a}{b}\right)}^x\cdot b\mathrm{d}x={\int }_0^1{a}^x\cdot b^{1-x}\mathrm{d}x=\frac{a-b}{\ln a-\ln b}$ for $a>0,b>0,a\ne b$, which is the logarithmic mean

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\mathrm{d}x=\frac{1}{a}$

${\int }_0^{\mathrm{\infty }}{e}^{-a{x}^2}\mathrm{d}x=\frac{1}{2}\sqrt{\frac{\pi }{a}}(a>0)$ (the Gaussian integral)

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

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

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

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

${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-a{x}^2}\mathrm{d}x=\{\begin{array}{cc}\frac{1}{2}\mathrm{\Gamma }\left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}}& (n>-1,a>0)\\ \frac{(2k-1)!!}{2^{k+1}{a}^k}\sqrt{\frac{\pi }{a}}& (n=2k,k\text{integer},a>0)\\ \frac{k!}{2a^{k+1}}& (n=2k+1,k\text{integer},a>0)\end{array}$ (!! is the double factorial)

${\int }_0^{\mathrm{\infty }}{x}^n{e}^{-ax}\mathrm{d}x=\{\begin{array}{cc}\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}& (n>-1,a>0)\\ \frac{n!}{a^{n+1}}& (n=0,1,2,\dots ,a>0)\end{array}$

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\sin bx\mathrm{d}x=\frac{b}{a^2+b^2}(a>0)$

${\int }_0^{\mathrm{\infty }}{e}^{-ax}\cos bx\mathrm{d}x=\frac{a}{a^2+b^2}(a>0)$

${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\sin bx\mathrm{d}x=\frac{2ab}{(a^2+b^2{)}^2}(a>0)$

${\int }_0^{\mathrm{\infty }}x{e}^{-ax}\cos bx\mathrm{d}x=\frac{a^2-b^2}{(a^2+b^2{)}^2}(a>0)$

${\int }_0^{2\pi }{e}^{x\cos \theta }d\theta =2\pi I_0(x)$ ($I_0$ is the modified Bessel function of the first kind)

${\int }_0^{2\pi }{e}^{x\cos \theta +y\sin \theta }d\theta =2\pi I_0\left(\sqrt{x^2+y^2}\right)$