Popis integrala eksponencijalnih funkcija

Slijedi popis integrala (antiderivacija funkcija) eksponencijalnih funkcija. Za potpun popis integrala funkcija, pogledati tablica integrala i popis integrala.

[math]\displaystyle{ \int e^{cx}\;dx = \frac{1}{c} e^{cx} + C }[/math], ali [math]\displaystyle{ \int e^{2x}\;dx = 2e^{2x} + C }[/math]

[math]\displaystyle{ \int a^{cx}\;dx = \frac{1}{c \ln a} a^{cx} + C \qquad\mbox{(za } a \gt 0,\mbox{ }a \ne 1\mbox{)} }[/math]

[math]\displaystyle{ \int xe^{cx}\; dx = \frac{e^{cx}}{c^2}(cx-1) + C }[/math]

[math]\displaystyle{ \int x^2 e^{cx}\;dx = e^{cx}\left(\frac{x^2}{c}-\frac{2x}{c^2}+\frac{2}{c^3}\right) + C }[/math]

[math]\displaystyle{ \int x^n e^{cx}\; dx = \frac{1}{c} x^n e^{cx} - \frac{n}{c}\int x^{n-1} e^{cx} dx + C }[/math]

[math]\displaystyle{ \int\frac{e^{cx}}{x}\; dx = \ln|x| +\sum_{i=1}^\infty\frac{(cx)^i}{i\cdot i!} + C }[/math]

[math]\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) + C \qquad\mbox{(za }n\neq 1\mbox{)} }[/math]

[math]\displaystyle{ \int e^{cx}\ln x\; dx = \frac{1}{c}e^{cx}\ln|x|-\operatorname{Ei}\,(cx) + C }[/math]

[math]\displaystyle{ \int e^{cx}\sin bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\sin bx - b\cos bx) + C }[/math]

[math]\displaystyle{ \int e^{cx}\cos bx\; dx = \frac{e^{cx}}{c^2+b^2}(c\cos bx + b\sin bx) + C }[/math]

[math]\displaystyle{ \int e^{cx}\sin^n x\; dx = \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\;dx + C }[/math]

[math]\displaystyle{ \int e^{cx}\cos^n x\; dx = \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\;dx + C }[/math]

[math]\displaystyle{ \int x e^{c x^2 }\; dx= \frac{1}{2c} \; e^{c x^2} + C }[/math]

[math]\displaystyle{ \int e^{-c x^2 }\; dx= \sqrt{\frac{\pi}{4c}} \operatorname{erf}(\sqrt{c} x) + C }[/math]([math]\displaystyle{ \mbox{erf} }[/math] je funkcija grješke (error function))

[math]\displaystyle{ \int xe^{-c x^2 }\; dx=-\frac{1}{2c}e^{-cx^2} + C }[/math]

[math]\displaystyle{ \int {1 \over \sigma\sqrt{2\pi} }\,e^{-{(x-\mu )^2 / 2\sigma^2}}\; dx= \frac{1}{2} (1 + \mbox{erf}\,\frac{x-\mu}{\sigma \sqrt{2}}) + C }[/math]

[math]\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 + C \quad \mbox{valjano za } n \gt 0, }[/math]

pri čemu je [math]\displaystyle{ c_{2j}=\frac{ 1 \cdot 3 \cdot 5 \cdots (2j-1)}{2^{j+1}}=\frac{(2j)\,!}{j!\, 2^{2j+1}} \ . }[/math]

Određeni integrali

[math]\displaystyle{ \int_{0}^{\infty} e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a} \quad (a\gt 0) }[/math] (Gaussov integral)

[math]\displaystyle{ \int_{-\infty}^{\infty} e^{-ax^2}\,dx=\sqrt{\pi \over a} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{-\infty}^{\infty} e^{-ax^2} e^{2bx}\,dx=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{-\infty}^{\infty} x e^{-a(x-b)^2}\,dx=b \sqrt{\pi \over a} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{-\infty}^{\infty} x^2 e^{-ax^2}\,dx=\frac{1}{2} \sqrt{\pi \over a^3} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{0}^{\infty} x^{n} e^{-ax^2}\,dx = \begin{cases} \frac{1}{2}\Gamma \left(\frac{n+1}{2}\right)/a^{\frac{n+1}{2}} & (n\gt -1,a\gt 0) \\ \frac{(2k-1)!!}{2^{k+1}a^k}\sqrt{\frac{\pi}{a}} & (n=2k, k \;\text{cijeli broj}, a\gt 0) \\ \frac{k!}{2a^{k+1}} & (n=2k+1,k \;\text{cijeli broj}, a\gt 0) \end{cases} }[/math] (!! je dvostruka faktorijela)

[math]\displaystyle{ \int_{0}^{\infty} x^n e^{-ax}\,dx = \begin{cases} \frac{\Gamma(n+1)}{a^{n+1}} & (n\gt -1,a\gt 0) \\ \frac{n!}{a^{n+1}} & (n=0,1,2,\ldots,a\gt 0) \\ \end{cases} }[/math]

[math]\displaystyle{ \int_{0}^{\infty} e^{-ax}\sin bx \, dx = \frac{b}{a^2+b^2} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{0}^{\infty} e^{-ax}\cos bx \, dx = \frac{a}{a^2+b^2} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{0}^{\infty} xe^{-ax}\sin bx \, dx = \frac{2ab}{(a^2+b^2)^2} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{0}^{\infty} xe^{-ax}\cos bx \, dx = \frac{a^2-b^2}{(a^2+b^2)^2} \quad (a\gt 0) }[/math]

[math]\displaystyle{ \int_{0}^{2 \pi} e^{x \cos \theta} d \theta = 2 \pi I_{0}(x) }[/math] ([math]\displaystyle{ I_{0} }[/math] je modificirana Besselova funkcija prve vrste)

[math]\displaystyle{ \int_{0}^{2 \pi} e^{x \cos \theta + y \sin \theta} d \theta = 2 \pi I_{0} \left( \sqrt{x^2 + y^2} \right) }[/math]