Definite Integrals

Definition of a Definite Integral

Let $f(x)$ be a defined integral in an interval $a\leq x\leq b$. Divide the interval into $n$ equal parts of length $\Delta x = \frac{b-a}{n}$. Then the definite integral of $F(x)$ between $x = a$ and $x = b$ is defined as
$\int^b_a f(x)\ dx=$ $\lim_{n\to\infty}f(a)\Delta x+f(a+\Delta x)\Delta x+f(a+2\Delta x)\Delta x+\cdots$ $+f(a+(n-1)\Delta x)\Delta x$

The limit will certainly exist if $f(x)$ is piecewise continuous.

If $f(x)=\frac{d}{dx}g(x)$, then by the fundamental theorem of the integral calculus the above definite integral can be evaluated by using the result
$\int^b_a f(x)\ dx=\int^b_a \frac{d}{dx}g(x)\ dx= g(x)|^b_a=g(b)-g(a)$
If the interval is infinite or if $f(x)$ has a singularity at some point in the interval, the definite integral is called an improper integral and can be defined by using appropriate limiting procedures. For example,

$\int_a^\infty f(x)\ dx=\lim_{b\to\infty}\int_a^b f(x)\ dx$

$\int_{-\infty}^\infty f(x)\ dx=\lim_{\substack{a\to-\infty \\ b\to\infty}}\int_a^b f(x)\ dx$

$\int_a^b f(x)\ dx=\lim_{\epsilon\to 0}\int_a^{b-\epsilon} f(x)\ dx$   if $b$ is a singular point

$\int_a^b f(x)\ dx=\lim_{\epsilon\to 0}\int_{a+\epsilon}^b f(x)\ dx$   if $a$ is a singular point

General Formulas Involving Definite Integrals

$\int_a^b\{f(x)\pm g(x)\pm h(x)\pm \cdots\}\ dx=$ $\int_a^b f(x)\ dx\pm\int_a^b g(x)\ dx\pm\int_a^b h(x)\ dx\pm\cdots$

$\int_a^b cf(x)\ dx=c\int_a^b f(x)\ dx$   where $c$ is any constant

$\int_a^a f(x)\ dx=0$

$\int_a^b f(x)\ dx=-\int_b^a f(x)\ dx$

$\int_a^b f(x)\ dx=\int_a^c f(x)\ dx+\int_c^b f(x)\ dx$

$\int_a^b f(x)\ dx=(b-c)f(c)$   where $c$ is between $a$ and $b$

This is called the mean value theorem for definite integrals and is valid if $f(x)$ is continuous in $a \leq x \leq b$.

$\int_a^b f(x)g(x)\ dx=f(c)\int_a^b g(x)\ dx$   where $c$ is between $a$ and $b$

This is a generalization of the previous one and is valid if $f(x)$ and $g(x)$ are continuous in $a\leq x\leq b$ and $g(x)\geq 0$.

Leibnitz's Rule for Differentiation of Integrals

$\frac{d}{d\alpha}\int_{\phi_1(\alpha)}^{\phi_2(\alpha)}F(x,\alpha)\ dx=$ $\int_{\phi_1(\alpha)}^{\phi_2(\alpha)}\frac{\partial F}{\partial\alpha}dx+F(\phi_2,\alpha)\frac{d\phi_1}{d\alpha}-F(\phi_1,\alpha)\frac{d\phi_2}{d\alpha}$

Approximate Formulas for Definite Integrals

In the following the interval from $x = a$ to $x = b$ is subdivided into $n$ equal parts by the points $a=x_0, x_2, . . ., x_{n - 1}, x_n=b$ and we let $y_0=f(x_0), y_1=f(x_1), y_2=f(x_2),...,$ $y_n=f(x_n), h=\frac{b-a}{n}$.
Rectangular formula
$\int_a^b f(x)\ dx\approx h(y_0+y_1+y_2+\cdots+y_{n-1})$
Trapezoidal formula
$\int_a^b f(x)\ dx\approx \frac{h}{2}(y_0+2y_1+2y_2+\cdots+2y_{n-1}+y_n)$
Simpson’s formula (or parabolic formula) for $n$ even
$\int_a^b f(x)\ dx\approx \frac{h}{3}(y_0+4y_1+2y_2+4y_3+\cdots+2y_{n-2}+4y_{n-1}+y_n)$

Definite Integrals Involving Rational or Irrational Expressions

$\int_0^\infty\frac{dx}{x^2+a^2}=\frac{\pi}{2a}$

$\int_0^\infty\frac{x^{p-1}dx}{1+x}=\frac{\pi}{\sin p\pi}$,   $0< p<1$

$\int_0^\infty\frac{x^mdx}{x^n+a^n}=\frac{\pi a^{m+1-n}}{n\sin[(m+1)\frac{\pi}{n}]}$,   $0< m+1< n$

$\int_0^\infty\frac{x^mdx}{1+2x\cos\beta+x^2}=\frac{\pi}{\sin m\pi}\frac{\sin m\beta}{\sin\beta}$

$\int_0^a\frac{dx}{\sqrt{a^2-x^2}}=\frac{\pi}{2}$

$\int_0^a\sqrt{a^2-x^2}\ dx=\frac{\pi a^2}{4}$

$\int_0^a x^m(a^n-x^n)^p\ dx=\frac{a^{m+1+np}\Gamma\left[\frac{m+1}{n}\right]\Gamma(p+1)}{n\Gamma\left[\frac{m+1}{n}+p+1\right]}$

$\int_0^a \frac{x^m dx}{(x^n+a^n)^r}=\frac{(-1)^{r-1}\pi a^{m+1-nr}\Gamma\left[\frac{m+1}{n}\right]}{n\sin\left[\frac{(m+1)\pi}{n}\right](r-1)!\Gamma\left[\frac{m+1}{n}-r+1\right]}$   $0< m+1< nr$

Definite Integrals Involving Trigonometric Functions

All letters are considered positive unless otherwise indicated.

$\int_0^\pi\sin mx\sin nx\ dx=\left\{\begin{array}{lr}0\quad m,n\ \text{integers and}\ m\neq n\\ \frac{\pi}{2}\quad m,n\ \text{integers and}\ m=n\end{array}\right.$

$\int_0^\pi\cos mx\cos nx\ dx=\left\{\begin{array}{lr}0\quad m,n\ \text{integers and}\ m\neq n\\ \frac{\pi}{2}\quad m,n\ \text{integers and}\ m=n\end{array}\right.$

$\int_0^\pi\sin mx\cos nx\ dx=\left\{\begin{array}{lr}0\quad m,n\ \text{integers and}\ m+n\ \text{odd}\\ \frac{2m}{m^2-n^2}\quad m,n\ \text{integers and}\ m+n\ \text{even}\end{array}\right.$

$\int_0^\frac{\pi}{2}\sin^2x\ dx=\int_0^\frac{\pi}{2}\cos^2x\ dx=\frac{\pi}{4}$

$\int_0^\frac{\pi}{2}\sin^{2m}x\ dx=\int_0^\frac{\pi}{2}\cos^{2m}x\ dx=\frac{1\cdot3\cdot5\cdots2m-1}{2\cdot4\cdot6\cdots2m}\frac{\pi}{2}$,   $m=1,2,\cdots$

$\int_0^\frac{\pi}{2}\sin^{2m+1}x\ dx=\int_0^\frac{\pi}{2}\cos^{2m+1}x\ dx=\frac{2\cdot4\cdot6\cdots2m}{1\cdot3\cdot5\cdots2m+1}$,   $m=1,2,\cdots$

$\int_0^\frac{\pi}{2}\sin^{2p-1}x\cos^{2q-1}x\ dx=\frac{\Gamma(p)\Gamma(q)}{2\Gamma(p+q)}$

$\int_0^\infty\frac{\sin px}{x}dx=\left\{\begin{array}{lr}\frac{\pi}{2}\quad p>0\\ 0\quad p=0\\ -\frac{\pi}{2}\quad p<0\end{array}\right.$

$\int_0^\infty\frac{\sin px\cos qx}{x}dx=\left\{\begin{array}{lr} 0 \qquad p>q>0\\ \frac{\pi}{2}\quad 0< p< q\\ \frac{\pi}{4}\quad p=q>0\end{array}\right.$

$\int_0^\infty\frac{\sin px\sin qx}{x^2}dx=\left\{\begin{array}{lr}\frac{\pi p}{2}\quad0< p\leq q\\ \frac{\pi q}{2}\quad p\geq q>0 \end{array}\right.$

$\int_0^\infty\frac{\sin^2 px}{x^2}\ dx=\frac{\pi p}{2}$

$\int_0^\infty\frac{1-\cos px}{x^2}\ dx=\frac{\pi p}{2}$

$\int_0^\infty\frac{\cos px-\cos qx}{x}\ dx=\ln\frac{q}{p}$

$\int_0^\infty\frac{\cos px-\cos qx}{x^2}\ dx=\frac{\pi(q-p)}{2}$

$\int_0^\infty\frac{\cos mx}{x^2+a^2}\ dx=\frac{\pi}{2a}e^{-ma}$

$\int_0^\infty\frac{x\sin mx}{x^2+a^2}\ dx=\frac{\pi}{2}e^{-ma}$

$\int_0^\infty\frac{\sin mx}{x(x^2+a^2)}\ dx=\frac{\pi}{2a^2}(1-e^{-ma})$

$\int_0^{2\pi}\frac{dx}{a+b\sin x}=\frac{2\pi}{\sqrt{a^2-b^2}}$

$\int_0^{2\pi}\frac{dx}{a+b\cos x}=\frac{2\pi}{\sqrt{a^2-b^2}}$

$\int_0^\frac{\pi}{2}\frac{dx}{a+b\cos x}=\frac{\cos^{-1}\left(\frac{b}{a}\right)}{\sqrt{a^2-b^2}}$

$\int_0^{2\pi}\frac{dx}{(a+b\sin x)^2}=\int_0^{2\pi}\frac{dx}{(a+b\cos x)^2}=\frac{2\pi a}{(a^2-b^2)^\frac{3}{2}}$

$\int_0^{2\pi}\frac{dx}{1-2a\cos x+a^2}=\frac{2\pi}{1-a^2},\qquad 0< a<1$

$\int_0^{\pi}\frac{x\sin x\ dx}{1-2a\cos x+a^2}=\left\{\begin{array}{lr}\left(\frac{\pi}{a}\right)\ln(1+a)\quad |a|<1\\ \pi\ln\left(1+\frac{1}{a}\right)\quad |a|>1\end{array}\right.$

$\int_0^{\pi}\frac{\cos mx\ dx}{1-2a\cos x+a^2}=\frac{\pi a^m}{1-a^2},\quad a^2<1,\quad m=0,1,2,\cdots$

$\int_0^\infty\sin ax^2\ dx=\int_0^\infty\cos ax^2\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}$

$\int_0^\infty\sin ax^n\ dx=\frac{1}{na^{\frac{1}{n}}}\Gamma\left(\frac{1}{n}\right)\sin\frac{\pi}{2n}$,   $n>1$

$\int_0^\infty\cos ax^n\ dx=\frac{1}{na^{\frac{1}{n}}}\Gamma\left(\frac{1}{n}\right)\cos\frac{\pi}{2n}$,   $n>1$

$\int_0^\infty\frac{\sin x}{\sqrt{x}}dx=\int_0^\infty\frac{\cos x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$

$\int_0^\infty\frac{\sin x}{x^p}dx=\frac{\pi}{2\Gamma(p)\sin\left(\frac{p \pi}{2}\right)}$,   $0< p<1$

$\int_0^\infty\frac{\cos x}{x^p}dx=\frac{\pi}{2\Gamma(p)\cos\left(\frac{p \pi}{2}\right)}$,   $0< p<1$

$\int_0^\infty\sin ax^2\cos2bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos\frac{b^2}{a}-\sin\frac{b^2}{a}\right)$

$\int_0^\infty\cos ax^2\cos2bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{2a}}\left(\cos\frac{b^2}{a}+\sin\frac{b^2}{a}\right)$

$\int_0^\infty\frac{\sin^3 x}{x^3}dx=\frac{3\pi}{8}$

$\int_0^\infty\frac{\sin^4 x}{x^4}dx=\frac{\pi}{3}$

$\int_0^\infty\frac{\tan x}{x}dx=\frac{\pi}{2}$

$\int_0^\frac{\pi}{2}\frac{dx}{1+\tan^mx}=\frac{\pi}{4}$

$\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=2\left\{\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots\right\}$

$\int_0^1\frac{\tan^{-1}x}{x}dx=\frac{1}{1^2}-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\cdots$

$\int_0^1\frac{\sin^{-1}x}{x}dx=\frac{\pi}{2}\ln2$

$\int_0^1\frac{1-\cos x}{x}dx-\int_1^\infty\frac{\cos x}{x}dx=\gamma$

$\int_0^\infty\left(\frac{1}{1+x^2}-\cos x\right)\frac{dx}{x}=\gamma$

$\int_0^\infty\frac{\tan^{-1}px-\tan^{-1}qx}{x}dx=\frac{\pi}{2}\ln\frac{p}{q}$

Definite Integrals Involving Exponential Functions

$\int_0^\infty e^{-ax}\cos bx\ dx=\frac{a}{a^2+b^2}$

$\int_0^\infty e^{-ax}\sin bx\ dx=\frac{b}{a^2+b^2}$

$\int_0^\infty \frac{e^{-ax}\sin bx}{x}\ dx=\tan^{-1}\frac{b}{a}$

$\int_0^\infty \frac{e^{-ax}-e^{-bx}}{x}\ dx=\ln\frac{b}{a}$

$\int_0^\infty e^{-ax^2}\ dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}$

$\int_0^\infty e^{-ax^2}\cos bx\ dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-\frac{b^2}{4a}}$

$\int_0^\infty e^{-(ax^2+bx+c)} dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{\frac{(b^2-4ac)}{4a}}\ \text{erfc}\frac{b}{2\sqrt{a}}$,   $\text{erfc}(p)=\frac{2}{\pi}\int_p^\infty e^{-x^2}dx$

$\int_{-\infty}^\infty e^{-(ax^2+bx+c)} dx=\sqrt{\frac{\pi}{a}}e^{\frac{(b^2-4ac)}{4a}}$

$\int_0^\infty x^n e^{-ax}\ dx=\frac{\Gamma(n+1)}{a^n+1}$

$\int_0^\infty x^m e^{-ax^2}\ dx=\frac{\Gamma\left[\frac{m+1}{2}\right]}{2a^\frac{m+1}{2}}$

$\int_0^\infty e^{-(ax^2+\frac{b}{x^2})} dx=\frac{1}{2}\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$

$\int_0^\infty\frac{x\ dx}{e^x-1}=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\cdots=\frac{\pi^2}{6}$

$\int_0^\infty\frac{x^{n-1}\ dx}{e^x-1}=\Gamma(n)\left(\frac{1}{1^n}+\frac{1}{2^n}+\frac{1}{3^n}+\cdots\right)$

For even $n$ this can be summed in terms of Bernoulli numbers.
$\int_0^\infty\frac{x\ dx}{e^x+1}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+\cdots=\frac{\pi^2}{12}$
$\int_0^\infty\frac{x^{n-1}\ dx}{e^x+1}=\Gamma(n)\left(\frac{1}{1^n}-\frac{1}{2^n}+\frac{1}{3^n}-\cdots\right)$

For some positive integer values of $n$ the series can be summed.
$\int_0^\infty\frac{\sin mx}{e^{2\pi x}-1}dx=\frac{1}{4}\coth\frac{m}{2}-\frac{1}{2m}$
$\int_0^\infty\left(\frac{1}{1+x}-e^{-x}\right)\frac{dx}{x}=\gamma$

$\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}dx=\frac{1}{2}\gamma$

$\int_0^\infty\left(\frac{1}{e^x-1}-\frac{e^{-x}}{x}\right)dx=\gamma$

$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x\sec px}dx=\frac{1}{2}\ln\left(\frac{b^2+p^2}{a^2+p^2}\right)$

$\int_0^\infty\frac{e^{-ax}-e^{-bx}}{x\csc px}dx=\tan^{-1}\frac{b}{p}-\tan^{-1}\frac{a}{p}$

$\int_0^\infty\frac{e^{-ax}(1-\cos x)}{x^2}dx=\cot^{-1}a-\frac{a}{2}\ln(a^2+1)$

Definite Integrals Involving Logarithmic Functions

$\int_0^1x^m(\ln x)^n\ dx=\frac{(-1)^n n!}{(m+1)^{n+1}}$   $m>-1, n=0,1,2,\cdots$

If $n\neq 0,1,2,\cdots$ replace $n!$ by $\Gamma(n + 1)$.

$\int_0^1\frac{\ln x}{1+x}dx=-\frac{\pi^2}{12}$

$\int_0^1\frac{\ln x}{1-x}dx=-\frac{\pi^2}{6}$

$\int_0^1\frac{\ln(1+x)}{x}dx=\frac{\pi^2}{12}$

$\int_0^1\frac{\ln(1-x)}{x}dx=-\frac{\pi^2}{6}$

$\int_0^1\ln x\ln(1+x)\ dx=2-2\ln2-\frac{\pi^2}{12}$

$\int_0^1\ln x\ln(1-x)\ dx=2-\frac{\pi^2}{6}$

$\int_0^\infty\frac{x^{p-1}\ln x}{1+x}dx=-\pi^2\csc p\pi\cot p\pi$   $0< p<1$

$\int_0^1\frac{x^m-x^n}{\ln x}dx=\ln\frac{m+1}{n+1}$

$\int_0^\infty e^{-x}\ln x\ dx=-\gamma$

$\int_0^\infty e^{-x^2}\ln x\ dx=-\frac{\sqrt{\pi}}{4}(\gamma+2\ln2)$

$\int_0^\infty\ln\left(\frac{e^x+1}{e^x-1}\right)dx=\frac{\pi^2}{4}$

$\int_0^\frac{\pi}{2}\ln\sin x\ dx=\int_0^\frac{\pi}{2}\ln\cos x\ dx=-\frac{\pi}{2}\ln2$

$\int_0^\frac{\pi}{2}(\ln\sin x)^2\ dx=\int_0^\frac{\pi}{2}(\ln\cos x)^2\ dx=\frac{\pi}{2}(\ln2)^2+\frac{\pi^3}{24}$

$\int_0^\pi x\ln\sin x\ dx=-\frac{\pi^2}{2}\ln2$

$\int_0^\frac{\pi}{2} \sin x\ln\sin x\ dx=\ln2-1$

$\int_0^{2\pi} \ln(a+b\sin x)\ dx=\int_0^{2\pi} \ln(a+b\cos x)\ dx=2\pi\ln(a+\sqrt{a^2-b^2})$

$\int_0^\pi \ln(a+b\cos x)\ dx=x\ln\left(\frac{a+\sqrt{a^2-b^2}}{2}\right)$

$\int_0^\pi \ln(a^2-2ab\cos x+b^2)\ dx=\left\{\begin{array}{lr}2\pi\ln a,\quad a\geq b>0\\ 2\pi\ln b,\quad b\geq a>0\end{array}\right.$

$\int_0^\frac{\pi}{4} \ln(1+\tan x)\ dx=\frac{\pi}{8}\ln2$

$\int_0^\frac{\pi}{2} \sec x\ln\left(\frac{1+b\cos x}{1+a\cos x}\right)\ dx=\frac{1}{2}\left\{(\cos^{-1}a)^2-(\cos^{-1}b)^2\right\}$

$\int_0^a\ln\left(2\sin\frac{x}{2}\right)\ dx=-\left(\frac{\sin a}{1^2}+\frac{\sin2a}{2^2}+\frac{\sin3a}{3^2}+\cdots\right)$

Definite Integrals Involving Hyperbolic Functions

$\int_0^\infty\frac{\sin ax}{\sinh bx}dx=\frac{\pi}{2b}\tanh\frac{a\pi}{2b}$

$\int_0^\infty\frac{\cos ax}{\cosh bx}dx=\frac{\pi}{2b}sech\frac{a\pi}{2b}$

$\int_0^\infty\frac{x\ dx}{\sinh ax}=\frac{\pi^2}{4a^2}$

$\int_0^\infty\frac{x^n\ dx}{\sinh ax}=\frac{2^{n+1}-1}{2^na^{n+1}}\Gamma(n+1)\left\{\frac{1}{1^{n+1}}+\frac{1}{2^{n+1}}+\frac{1}{3^{n+1}}+\cdots\right\}$

If $n$ is an odd positive integer, the series can be summed.

$\int_0^\infty\frac{\sinh ax}{e^{bx}+1}dx=\frac{\pi}{2b}\csc\frac{a\pi}{b}-\frac{1}{2a}$

$\int_0^\infty\frac{\sinh ax}{e^{bx}-1}dx=\frac{1}{2a}-\frac{\pi}{2b}\cot\frac{a\pi}{b}$

Miscellaneous Definite Integrals

$\int_0^\infty\frac{f(ax)-f(bx)}{x}dx=\{f(0)-f(\infty)\}\ln\frac{b}{a}$

This is called Frullani’s integral. It holds if $f'(x)$ is continuous and $\int_1^\infty\frac{f(x)- f(\infty)}{x}\ dx$ converges.

$\int_0^1\frac{dx}{x^x}=\frac{1}{1^1}+\frac{1}{2^2}+\frac{1}{3^3}+\cdots$

$\int_{-a}^a (a+x)^{m-1}(a-x)^{n-1}dx=(2a)^{m+n-1}\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$


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