Список на определени интеграли

Во математиката, определениот интеграл:

$\underset{a}{\overset{b}{\int }}f(x)dx$

е површината од областа на xy-рамнината ограничена од графиконот на f, x-оската и линиите x = a и x = b, така што површината над x-оската се додава на вкупниот збир, а под x-оската се одзема од вкупната сума.

Основната теорема на анализата воспоставува однос меѓу неопределените и определените интеграли и воведува техника за пресметување на определените интеграли.

Ако интервалот е бесконечен, определениот интеграл се нарекува несвојствен интеграл и се определува користејќи соодветни процедури на лимитирање. На пример:

$\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\underset{b\to \mathrm{\infty }}{\lim }[\underset{a}{\overset{b}{\int }}f(x)dx]$

Константа, како пи, која може да биде дефинирана со интеграла на алгебарска функција во алгебарски домен е позната како периода.

Следи списокот на најчестите определени интеграли. Списокот на неопределени интеграли е дадени во списокот на неопределени интеграли.

Определени интеграли од рационални или ирационални изрази

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2+a^2}=\frac{\pi }{2a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{x^n+a^n}=\frac{\pi a^{m-n+1}}{n\sin \left(\frac{m+1}{n}\pi \right)}\text{for }0<m+1<n$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{p-1}dx}{1+x}=\frac{\pi }{\sin (p\pi )}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{1+2x\cos \beta +x^2}=\frac{\pi }{\sin (m\pi )}\cdot \frac{\sin (m\beta )}{\sin (\beta )}$

$\underset{0}{\overset{a}{\int }}\frac{dx}{\sqrt{a^2-x^2}}=\frac{\pi }{2}$

$\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx=\frac{\pi a^2}{4}$

$\underset{0}{\overset{a}{\int }}{x}^m(a^n-x^n{)}^{p}dx=\frac{a^{m+1+np}\mathrm{\Gamma }\left(\frac{m+1}{n}\right)\mathrm{\Gamma }(p+1)}{n\mathrm{\Gamma }(\frac{m+1}{n}+p+1)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{m}dx}{({x^n+a^n)}^r}=\frac{(-1{)}^{r-1}\pi a^{m+1-nr}\mathrm{\Gamma }\left(\frac{m+1}{n}\right)}{n\sin \left(\frac{m+1}{n}\pi \right)(r-1)!\mathrm{\Gamma }(\frac{m+1}{n}-r+1)}\text{for }n(r-2)<m+1<nr$

Определени интеграли од тригонометриски функции

$\underset{0}{\overset{\pi }{\int }}\sin (mx)\sin (nx)dx=\{\begin{array}{cc}0& \text{ако }m\ne n\\ \\ \frac{\pi }{2}& \text{ако }m=n\end{array}\text{за }m,n\text{ позитивни цели броеви}$

$\underset{0}{\overset{\pi }{\int }}\cos (mx)\cos (nx)dx=\{\begin{array}{cc}0& \text{if }m\ne n\\ \\ \frac{\pi }{2}& \text{ако }m=n\end{array}\text{за }m,n\text{ позитивни цели броеви}$

$\underset{0}{\overset{\pi }{\int }}\sin (mx)\cos (nx)dx=\{\begin{array}{cc}0& \text{if }m+n\text{ even}\\ \\ \frac{2m}{m^2-n^2}& \text{ако }m+n\text{ непарен}\end{array}\text{за }m,n\text{ цели броеви}.$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^2(x)dx=\frac{\pi }{4}$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2m}(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2m}(x)dx=\frac{1\times 3\times 5\times \cdots \times (2m-1)}{2\times 4\times 6\times \cdots \times 2m}\cdot \frac{\pi }{2}\text{for }m=1,2,3\dots$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2m+1}(x)dx=\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\cos }^{2m+1}(x)dx=\frac{2\times 4\times 6\times \cdots \times 2m}{1\times 3\times 5\times \cdots \times (2m+1)}\text{for }m=1,2,3\dots$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^{2p-1}(x){\cos }^{2q-1}(x)dx=\frac{\mathrm{\Gamma }(p)\mathrm{\Gamma }(q)}{2\mathrm{\Gamma }(p+q)}=\frac{1}{2}\text{B}(p,q)$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (px)}{x}dx=\{\begin{array}{cc}\frac{\pi }{2}& \text{ако }p>0\\ \\ 0& \text{ако }p=0\\ \\ -\frac{\pi }{2}& \text{ако }p<0\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\cos qx}{x}dx=\{\begin{array}{cc}0& \text{ ако }q>p>0\\ \\ \frac{\pi }{2}& \text{ ако }0<q<p\\ \\ \frac{\pi }{4}& \text{ ако }p=q>0\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin px\sin qx}{x^2}dx=\{\begin{array}{cc}\frac{\pi p}{2}& \text{ ако }0<p\le q\\ \\ \frac{\pi q}{2}& \text{ ако }0<q\le p\end{array}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{{\sin }^{2}px}{x^2}dx=\frac{\pi p}{2}$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos px-\cos qx}{x}dx=\ln \frac{q}{p}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos px-\cos qx}{x^2}dx=\frac{\pi (q-p)}{2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos mx}{x^2+a^2}dx=\frac{\pi }{2a}{e}^{-ma}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x\sin mx}{x^2+a^2}dx=\frac{\pi }{2}{e}^{-ma}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{x(x^2+a^2)}dx=\frac{\pi }{2a^2}(1-e^{-ma})$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{a+b\sin x}=\frac{2\pi }{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{a+b\cos x}=\frac{2\pi }{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{dx}{a+b\cos x}=\frac{{\cos }^{-1}\left(\frac{b}{a}\right)}{\sqrt{a^2-b^2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{(a+b\sin x{)}^2}=\underset{0}{\overset{2\pi }{\int }}\frac{dx}{(a+b\cos x{)}^2}=\frac{2\pi a}{(a^2-b^2{)}^{3/2}}$

$\underset{0}{\overset{2\pi }{\int }}\frac{dx}{1-2a\cos x+a^2}=\frac{2\pi }{1-a^2}\text{for }0<a<1$

$\underset{0}{\overset{\pi }{\int }}\frac{x\sin xdx}{1-2a\cos x+a^2}=\{\begin{array}{cc}\frac{\pi }{a}\ln |1+a|& \text{ако }|a|<1\\ \\ \frac{\pi }{a}\ln |1+\frac{1}{a}|& \text{ако }|a|>1\end{array}$

$\underset{0}{\overset{\pi }{\int }}\frac{\cos mxdx}{1-2a\cos x+a^2}=\frac{\pi a^m}{1-a^2}\text{for }{a}^2<1,m=0,1,2,\dots$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^n=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }\left(\frac{1}{n}\right)\sin \frac{\pi }{2n}\text{for }n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^n=\frac{1}{n{a}^{1/n}}\mathrm{\Gamma }\left(\frac{1}{n}\right)\cos \frac{\pi }{2n}\text{for }n>1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{\sqrt{x}}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{\sqrt{x}}dx=\sqrt{\frac{\pi }{2}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\sin \left(\frac{p\pi }{2}\right)}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos x}{x^p}dx=\frac{\pi }{2\mathrm{\Gamma }(p)\cos \left(\frac{p\pi }{2}\right)}\text{for }0<p<1$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sin a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}-\sin \frac{b^2}{a})$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\cos a{x}^2\cos 2bxdx=\frac{1}{2}\sqrt{\frac{\pi }{2a}}(\cos \frac{b^2}{a}+\sin \frac{b^2}{a})$

Определени интеграли од експоненцијални функции

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\sqrt{x}{e}^{-x}dx=\frac{1}{2}\sqrt{\pi }$ (види и Гама функција)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\cos bxdx=\frac{a}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-ax}\sin bxdx=\frac{b}{a^2+b^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}\sin bx}{x}dx={\tan }^{-1}\frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x}dx=\ln \frac{b}{a}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}\text{for }a>0$ (Гаусов интеграл)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-a{x}^2}\cos bxdx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{-b^2}{4a}\right)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{b^2-4ac}{4a}\right)}\cdot \mathrm{erfc}\frac{b}{2\sqrt{a}},\text{ каде }\mathrm{erfc}(p)=\frac{2}{\sqrt{\pi }}\underset{p}{\overset{\mathrm{\infty }}{\int }}{e}^{-x^2}dx$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{e}^{-(a{x}^2+bx+c)}dx=\sqrt{\frac{\pi }{a}}{e}^{\left(\frac{b^2-4ac}{4a}\right)}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^n{e}^{-ax}dx=\frac{\mathrm{\Gamma }(n+1)}{a^{n+1}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-a{x}^2}dx=\frac{1}{4}\sqrt{\frac{\pi }{a^3}}\text{for }a>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n}{e}^{-a{x}^2}dx=\frac{2n-1}{2a}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2(n-1)}{e}^{-a{x}^2}dx=\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}=\frac{(2n)!}{n!2^{2n+1}}\sqrt{\frac{\pi }{a^{2n+1}}}\text{for }a>0,n=1,2,3\dots$ (каде!! е двоен факториел)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^3{e}^{-a{x}^2}dx=\frac{1}{2a^2}\text{for }a>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n+1}{e}^{-a{x}^2}dx=\frac{n}{a}\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^{2n-1}{e}^{-a{x}^2}dx=\frac{n!}{2a^{n+1}}\text{for }a>0,n=0,1,2\dots$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^m{e}^{-a{x}^2}dx=\frac{\mathrm{\Gamma }\left(\frac{m+1}{2}\right)}{2a^{\left(\frac{m+1}{2}\right)}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{(-a{x}^2-\frac{b}{x^2})}dx=\frac{1}{2}\sqrt{\frac{\pi }{a}}{e}^{-2\sqrt{ab}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x-1}dx=\zeta (2)=\frac{{\pi }^2}{6}$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{e^x+1}dx=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\dots =\frac{{\pi }^2}{12}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth \frac{m}{2}-\frac{1}{2m}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}(\frac{1}{1+x}-e^{-x})\frac{dx}{x}=\gamma$ (каде $\gamma$ е Ојлерова-Маскерониева константа)

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-x^2}-e^{-x}}{x}dx=\frac{\gamma }{2}$

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

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x\sec px}dx=\frac{1}{2}\ln \frac{b^2+p^2}{a^2+p^2}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}-e^{-bx}}{x\csc px}dx={\tan }^{-1}\frac{b}{p}-{\tan }^{-1}\frac{a}{p}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{e^{-ax}(1-\cos x)}{x^2}dx={\cot }^{-1}a-\frac{a}{2}\ln \left|\frac{a^2+1}{a^2}\right|$

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

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{x}^{2(n+1)}{e}^{-\frac{1}{2}{x}^2}dx=\frac{(2n+1)!}{2^{n}n!}\sqrt{2\pi }\text{for }n=0,1,2,\dots$

Определени интеграли од логаритамски функции

$\underset{0}{\overset{1}{\int }}{x}^m(\ln x{)}^{n}dx=\frac{(-1{)}^{n}n!}{(m+1{)}^{n+1}}\text{for }m>-1,n=0,1,2,\dots$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1+x}dx=-\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln x}{1-x}dx=-\frac{{\pi }^2}{6}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1+x)}{x}dx=\frac{{\pi }^2}{12}$

$\underset{0}{\overset{1}{\int }}\frac{\ln (1-x)}{x}dx=-\frac{{\pi }^2}{6}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\ln (a^2+x^2)}{b^2+x^2}dx=\frac{\pi }{b}\ln (a+b)\text{for }a,b>0$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\ln x}{x^2+a^2}dx=\frac{\pi \ln a}{2a}\text{for }a>0$

Определени интеграли од хиперболични функции

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin ax}{\sinh bx}dx=\frac{\pi }{2b}\tanh \frac{a\pi }{2b}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\cos ax}{\cosh bx}dx=\frac{\pi }{2b}\cdot \frac{1}{\cosh \frac{a\pi }{2b}}$

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x}{\sinh ax}dx=\frac{{\pi }^2}{4a^2}$

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\cosh x}dx=\pi$

Фруланиеви интеграли

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{f(ax)-f(bx)}{x}dx=(\underset{x\to 0}{\lim }f(x)-\underset{x\to \mathrm{\infty }}{\lim }f(x))\ln \left(\frac{b}{a}\right)$ важи доколку интегралот постои и $f^{\prime }(x)$ е континуална.

Поврзано

• Список на интеграли
• Бесконечна сума
• Гама функција
• Список на лимеси

Наводи

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