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

Zurück zu Bestimmte Integrale

0.1

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{x}{\sin x}dx=2G$

0.2

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{x^2}{\sin x}dx=2\pi G-\frac{7}{2}\zeta (3)$

0.3

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{x^2}{{\sin }^{2}x}dx=G+\frac{\pi }{4}\log 2-\frac{{\pi }^2}{16}$

0.4

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{x^3}{{\sin }^{2}x}dx=\frac{3}{4}\pi G-\frac{{\pi }^3}{64}+\frac{3}{32}{\pi }^2\log 2-\frac{105}{64}\zeta (3)$

0.5

$\underset{0}{\overset{1}{\int }}\sin (\pi x)x^x(1-x{)}^{1-x}dx=\frac{\pi e}{24}$

0.6

$\underset{0}{\overset{1}{\int }}\frac{\sin (\pi x)}{x^x(1-x{)}^{1-x}}dx=\frac{\pi }{e}$

1.1

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n+1}xdx=\frac{n}{n+1}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{n-1}xdx$

1.2

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2n}xdx=\frac{1}{2^{2n}}(\genfrac{}{}{0ex}{}{2n}{n})\frac{\pi }{2}$

1.3

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2n+1}xdx=\frac{1}{2n+1}{\left[\frac{1}{2^{2n}}(\genfrac{}{}{0ex}{}{2n}{n})\right]}^{-1}$

1.4

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin \alpha x}{x}dx=\pi \alpha >0$

1.5

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\alpha \sin x}{{\alpha }^2+x^2}dx=\text{Shi}(\alpha )\cosh (\alpha )-\text{Chi}(\alpha )\sinh (\alpha )\text{Re}(\alpha )>0$

1.6

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

1.7

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\sin (x^2+\frac{{\alpha }^2}{x^2})dx=\sqrt{\frac{\pi }{2}}(\cos 2\alpha +\sin 2\alpha )$

1.8

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{x\sin \alpha x}{1+x^2}dx=\pi e^{-\alpha }\alpha >0$

1.9

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{x\sin \alpha x}{1-x^2}dx=-\pi \cos \alpha \alpha >0$

1.10

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{|\sin \alpha x|}{1+x^2}dx=4\sinh \alpha \text{artanh}{e}^{-\alpha }\alpha >0$

1.11

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{k\sin x}{\sqrt{1-k^2{\sin }^{2}x}}dx=\text{artanh}k$

1.12

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}|\sin x{|}^{\alpha -1}\frac{\sin x}{x}dx=2^{\alpha -1}\frac{{\mathrm{\Gamma }}^2\left(\frac{\alpha }{2}\right)}{\mathrm{\Gamma }(\alpha )}\text{Re}(\alpha )>0$

1.13

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\text{sinc}(x)\cdot \text{sinc}(u-x)dx=\pi \cdot \text{sinc}(u)$    oder für $f(x)=\text{sinc}(\pi x)$ gilt $f*f=f$.

2.1

$\underset{0}{\overset{\pi }{\int }}\sin nx\sin mxdx={\delta }_{mn}\frac{\pi }{2}n,m\in {\mathbb{Z}}^{\ge 1}$

2.2

$\frac{1}{\pi }\underset{0}{\overset{\pi }{\int }}{x}^2\sin nx\sin mxdx=\{\begin{array}{ccc}\frac{{\pi }^2}{6}\pm \frac{1}{4nm}& ,& n=m\\ \\ \frac{(-1{)}^{n-m}}{(n-m{)}^2}\pm \frac{(-1{)}^{n+m}}{(n+m{)}^2}& ,& n\ne m\end{array}n,m\in {\mathbb{Z}}^{>0}$

2.3

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2n}x}{x^{2m}}dx=\frac{\pi \cdot (-1{)}^{n-m}}{2^{2n}(2m-1)!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{2n}{k})(2n-2k{)}^{2m-1}n,m\in {\mathbb{Z}}^{>0}n\ge m$

2.4

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2n+1}x}{x^{2m+1}}dx=\frac{\pi \cdot (-1{)}^{n-m}}{2^{2n+1}(2m)!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{2n+1}{k})(2n+1-2k{)}^{2m}n,m\in {\mathbb{Z}}^{>0}n\ge m$

2.5

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2n+1}x}{x^{2m}}dx=\frac{(-1{)}^{n-m}}{2^{2n}(2m-1)!}\sum _{k=0}^{n-1}(-1{)}^k(\genfrac{}{}{0ex}{}{2n+1}{k})(2n+1-2k{)}^{2m-1}\log (2n+1-2k)n,m\in {\mathbb{Z}}^{>0}n\ge m$

2.6

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2n}x}{x^{2m+1}}dx=\frac{(-1{)}^{n-m-1}}{2^{2n-1}(2m)!}\sum _{k=0}^{n-1}(-1{)}^k(\genfrac{}{}{0ex}{}{2n}{k})(2n-2k{)}^{2m}\log (2n-2k)n,m\in {\mathbb{Z}}^{\ge 0}n>m$

2.7

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{|\sin \alpha x|-|\sin \beta x|}{x}dx=\frac{2}{\pi }\log \frac{\alpha }{\beta }\alpha ,\beta >0$

2.8

$J_{\nu }(z)=\frac{1}{2\pi }\underset{C}{\int }{e}^{-iz\sin t+i\nu t}dt\text{Re}(z)>0,\nu \in ℂ$

$C$ ist hierbei die Kurve, die gradlinig von $-\pi +i\mathrm{\infty }$ über $-\pi ,\pi$ nach $\pi +i\mathrm{\infty }$ läuft.

3.1

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin (\alpha t^{\frac{1}{z}}+\beta )dt=\frac{\mathrm{\Gamma }(z+1)}{{\alpha }^z}\sin (\frac{\pi z}{2}+\beta )0<z<1,\alpha >0,\beta \in ℂ$

n.1

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\prod _{k=0}^n\frac{\sin (a_{k}x)}{x}dx=\frac{\pi }{2}\cdot \prod _{k=1}^n{a}_k{a}_k>0$     und     $a_0>\sum _{k=1}^n{a}_k$