Formelsammlung Mathematik: Bestimmte Integrale: Form R(x,exp)

Zurück zu Bestimmte Integrale

0.1

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx=\sqrt{\pi }$

0.2

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^4}dx=2^{-1/4}{\pi }^{1/4}{\varpi }^{1/2}$

0.3

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{e^x-1}-\frac{1}{x{e}^x})\mathrm{d}x=\gamma$

0.4

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x}+x-1}{x(e^x+1)}\mathrm{d}x=\ln \left(\frac{4}{\pi }\right)$

0.5

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[\frac{1}{e^{2x}-1}-\frac{1}{x{e}^x(e^x+1)}]\mathrm{d}x=\frac{1}{2}\gamma -\frac{1}{2}\ln \left(\frac{4}{\pi }\right)$

0.6

$\underset{0}{\overset{\mathrm{\infty }}{\int }}[\frac{1}{e^{2x}-1}+\frac{x-1}{x(e^x+1)}]\mathrm{d}x=\frac{1}{2}\gamma +\frac{1}{2}\ln \left(\frac{4}{\pi }\right)$

1.1

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{z-1}{e}^{-x}dx=\mathrm{\Gamma }(z)\text{Re}(z)>0$

1.2

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\exp (-x^2-\frac{{\alpha }^2}{x^2})dx=\frac{\sqrt{\pi }}{2}{e}^{-2\alpha }\alpha >0$

1.3

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{x{e}^x}-\frac{e^{-x(z-1)}}{e^x-1})dx=\psi (z)\text{Re}(z)>0$

1.4

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{z-1}{x{e}^x}-\frac{1-e^{-x(z-1)}}{x(e^x-1)})dx=\log \mathrm{\Gamma }(z)\text{Re}(z)>0$

1.5

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{e^x}-\frac{1}{(1+x{)}^z})\frac{dx}{x}=\psi (z)\text{Re}(z)>0$

1.6

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{z-1}}{e^x-1}dx=\mathrm{\Gamma }(z)\zeta (z)\text{Re}(z)>1$

1.7

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{z-1}}{e^x+1}dx=\mathrm{\Gamma }(z)\eta (z)\text{Re}(z)>0$

1.8

$\frac{1}{\mathrm{\Gamma }(z)}=\frac{1}{2\pi i}\underset{C}{\int }{t}^{-z}{e}^{t}dtC$ ist eine Kurve in $ℂ\setminus {ℝ}^{\ge 0}$, die von $-\mathrm{\infty }-i\epsilon$ nach $-\mathrm{\infty }+i\epsilon$ läuft.

Für $\text{Re}(z)>0$ kann man als Integrationsweg $C$ auch die Gerade $a+iℝ$, mit $a>0$, hernehmen.

1.9

$J_{\nu }(x)=\frac{1}{2\pi i}\underset{C}{\int }{e}^{\frac{x}{2}(t-\frac{1}{t})}\frac{dt}{t^{\nu +1}}\text{Re}(x)>0C$ ist eine Kurve in $ℂ\setminus {ℝ}^{\ge 0}$, die von $-\mathrm{\infty }-i\epsilon$ nach $-\mathrm{\infty }+i\epsilon$ läuft.

1.10

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{x}-\frac{1}{e^x-1})e^{-zx}dx=\psi (z+1)-\log z\text{Re}(z)>0$

1.11

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{2}-\frac{1}{x}+\frac{1}{e^x-1})\frac{e^{-zx}}{x}dx=\log \left(\frac{z!e^z}{z^z\sqrt{2\pi z}}\right)\text{Re}(z)>0$

1.12

$J_{\nu }(z)=\frac{1}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }(\nu +\frac{1}{2})}{\left(\frac{z}{2}\right)}^{\nu }\underset{-1}{\overset{1}{\int }}{e}^{izx}(1-x^2{)}^{\nu -\frac{1}{2}}dx\text{Re}(\nu )>-\frac{1}{2}$

2.1

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-sx}\frac{a}{a^2+x^2}dx=\sin (as)\text{Ci}(as)-\cos (as)(\text{Si}(as)-\frac{\pi }{2})$

2.2

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-sx}\frac{x}{a^2+x^2}dx=\sin (as)(\frac{\pi }{2}-\text{Si}(as))-\cos (as)\text{Ci}(as)$

2.3

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{z-1}{e}^{-\mu x}dx=\frac{\mathrm{\Gamma }(z)}{{\mu }^z}\text{Re}(z),\text{Re}(\mu )>0$

2.4

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{z-1}{e}^{-i\mu x}dx=\frac{\mathrm{\Gamma }(z)}{(i\mu {)}^z}0<\mathrm{Re}(z)<1,\mu >0$

2.5

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{a^2}{(e^x-ax-b{)}^2+(a\pi {)}^2}dx=\frac{1}{1+W\left(\frac{1}{a}{e}^{-b/a}\right)}a>0,b\in ℝ$

4.1

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{kx}-e^{\lambda x}}{e^{\mu x}-e^{\nu x}}dx=\frac{1}{\mu -\nu }(\psi \left(\frac{\mu -\lambda }{\mu -\nu }\right)-\psi \left(\frac{\mu -k}{\mu -\nu }\right))$