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Improper Integral

The definite integral we have studied so far can only apply to the continuous functions. Now we extend this definition to the function with finite number of discontinuity.

Improper Integral of the 1st kind
[1] If $f(x)$ is continuous on $(a,b]$ and discontinuous at $x = a$. Then $f(x)$ is continuous on the interval $[a + \varepsilon, b]$. Thus we can think of the following definite integral

$\displaystyle \int_{a+\varepsilon}^{b}f(x)dx$

Then

$\displaystyle \int_a^b f(x)dx = \lim_{\varepsilon \to 0+} \int_{a + \varepsilon}^{b} f(x)dx. $

[2] If $f(x)$ is continuous on $[a,b)$ and discontinuous at $x = b$. Then $f(x)$ is continuous on the interval $[a, b - \varepsilon]$. Thus we can think of the following definite integral

$\displaystyle \int_{a}^{b-\varepsilon}f(x)dx$

Then

$\displaystyle \int_a^b f(x)dx = \lim_{\varepsilon \to 0+} \int_{a + \varepsilon}^{b} f(x)dx. $

Example 3..18   Evaluate the following improper integrals.

1. $\displaystyle{\int_{0}^{1}\frac{dx}{\sqrt{1-x}}}$2. $\displaystyle{\int_{0}^{1}\frac{1}{x}dx}$

SOLUTION 1. Let $u = \sqrt{1-x}$. Then $u^2 = 1-x$ and $2udu = -dx$. \begin{displaymath}\begin{array}{c\vert ccc} x & 0 & \to & 1\\ \hline u & 1 & \to & 0 \end{array}\end{displaymath},
$\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x}}$ $\displaystyle =$ $\displaystyle \int_1^0\frac{-2udu}{u} = \int_0^1 \frac{2}{u}du = \lim_{\varepsi... ...\varepsilon \to 0+}\left[2t\right]_\varepsilon^1 = 2\ensuremath{\ \blacksquare}$  

2. $\displaystyle{f(x) = \frac{1}{x}}$ is continuous on $(0,1]$ and antiderivative of $f(x)$ is $\log{\vert x\vert}$.
$\displaystyle \int_0^1 \frac{1}{x}dx$ $\displaystyle =$ $\displaystyle \lim_{\varepsilon \to 0+}\int_{\varepsilon}^{1}\frac{1}{x}dx = \l... ...arepsilon}^{1} = \log{1} - \lim_{\varepsilon \to 0+}\log{\vert\varepsilon\vert}$  
  $\displaystyle =$ $\displaystyle 0 - (-\infty) = \infty \ensuremath{\ \blacksquare}$  

Exercise 3..18   Evaluate the following improper integrals.
1. $\displaystyle{\int_{1}^e \frac{1}{x\log{x}}dx}$2. $\displaystyle{\int_{-1}^{1} \frac{1}{\sqrt{1 - x^2}}dx}$

SOLUTION 1. $\frac{1}{x\log{x}}$ is continuous on $(0,e]$, but discontinuous at $x = 0$. Let $t = \log{x}$. Then $dt = \frac{1}{x}dx$, \begin{displaymath}\begin{array}{c\vert ccc} x & 1 & \to & e\\ \hline t & 0 & \to & 1 \end{array}\end{displaymath},

$\displaystyle \int_{1}^e \frac{1}{x\log{x}}dx = \int_0^1 \frac{1}{t}dt.$

Note that $f(t) = \frac{1}{t}$ is not continuous at $t = 0$.
$\displaystyle \int_0^1 \frac{1}{t}dt$ $\displaystyle =$ $\displaystyle \lim_{\varepsilon \to 0+}\int_\varepsilon^1 \frac{1}{t}dt = \lim_... ...}\right]_{\varepsilon}^1 = \log{1} - \lim_{\varepsilon \to 0+}\log{\varepsilon}$  
  $\displaystyle =$ $\displaystyle 1 - (-\infty) = \infty\ensuremath{\ \blacksquare}$  

2. $\displaystyle{f(x) = \frac{dx}{\sqrt{1-x^2}}}$ is continuous on $-1 < x < 1$, but not continuous at $x = \pm 1$. Then we write

$\displaystyle \lim_{\varepsilon, \varepsilon' \to 0+}\int_{-1+\varepsilon}^{1-\varepsilon'}\frac{dx}{\sqrt{1 -x^2}}$

Then

$\displaystyle \lim_{\varepsilon, \varepsilon' \to 0+}\int_{-1+\varepsilon}^{1-\... ...arepsilon'} = \frac{\pi}{2} - (\frac{-\pi}{2}) = \pi\ensuremath{\ \blacksquare}$

Improper Integral of the 1st kind
[3] $f(x)$ is discontinuous at $c_{1},c_{2},\ldots,c_{n} \in [a,b ]$. Then divide the interval $[a,b]$ into subintervals $[a,c_{1}],[c_{1},c_{2}],\ldots,[c_{n},b]$. Now consider the improper imtegral on each subintervals. If all improper integrals exist, then we define the sum of improper integrals as improper integral of $f(x)$ on $[a,b]$.

$\displaystyle \int_{a}^{b}f(x)dx = \int_{a}^{c_{1}}f(x)dx + \int_{c_{1}}^{c_{2}}f(x)dx + \cdots + \int_{c_{n}}^{b}f(x)dx $

Example 3..19   Evaluate the following improper integral.

$\displaystyle \int_{-1}^{1}\frac{1}{x^2}\:dx$

SOLUTION $f(x) = \frac{1}{x^2}$ is not continuous at $x = 0$. Then we write

$\displaystyle \int_{-1}^{0}\frac{1}{x^2}\:dx + \int_{0}^{1}\frac{1}{x^2}\:dx.$

Now

$\displaystyle \int_{0}^{1}\frac{1}{x^2}\:dx = \lim_{\varepsilon \to 0+}\int_{\v... ...on \to 0+}\left[-\frac{1}{x}\right]_{\varepsilon}^{1} = -(1 - \infty) = \infty.$

Thus no improper integral exists. $\ \blacksquare$

Exercise 3..19   Evaluate the following improper integral

$\displaystyle \int_{-1}^{1}f(x)\:dx, $

where $\displaystyle{f(x) = \left\{\begin{array}{ll} \frac{1}{x^{2/3}} & (-1 < x < 0)\\ \frac{1}{x^2} & (0 < x < 1) \end{array}\right. }$

SOLUTION $f(x)$ is not continuous at $x = 0$. Then we write the integral as follows:

$\displaystyle \int_{-1}^{0}\frac{1}{x^{2/3}}\:dx + \int_{0}^{1}\frac{1}{x^2}\:dx.$

Now

$\displaystyle \int_{-1}^{0}\frac{1}{x^{2/3}}\:dx = \lim_{\varepsilon \to 0-}\in... ...im_{\varepsilon \to 0-}\left[3x^{1/3}\right]_{-1}^{\varepsilon} = 0 - (-3) = 3.$

But $\displaystyle{\int_{0}^{1}\frac{1}{x^2}\:dx = \infty}$. Thus we can conclude that no improper integral exists $\ \blacksquare$
Improper Integral of the 2nd kind
$f(x)$ is continuous on $[a,\infty)$. Then $f(x)$ is continuous on $[a,b]$, where $a < b < \infty$. Then we define the infinite integral using the limit of $\displaystyle{\int_a^b f(x)dx}$.

$\displaystyle \int_{a}^{\infty}f(x)dx = \lim_{b \rightarrow \infty}\int_{a}^{b}f(x)dx .$

Similarly for the case where $f(x)$ is continuous on $(-\infty,b]$.

$\displaystyle \int_{a}^{\infty}f(x)dx = \lim_{a \rightarrow -\infty}\int_{a}^{b}f(x)dx .$

Example 3..20   Evaluate the following improper integral.

$\displaystyle \int_{1}^{\infty}\frac{dx}{x^{p}},\ p > 0$

SOLUTION For $p \neq 1$,
$\displaystyle \int_{1}^{\infty}\frac{dx}{x^{p}}$ $\displaystyle =$ $\displaystyle \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x^{p}}\:dx = \lim_{b \to ... ...}\int_1^b x^{-p}\:dx = \lim_{b \to \infty}\frac{1}{1-p}\left[x^{1-p}\right]_1^b$  
  $\displaystyle =$ $\displaystyle \lim_{b \to \infty} \frac{1}{1-p}(b^{1-p} - 1) = \left\{\begin{array}{cl} \frac{1}{p-1}, & p > 1\\ \infty, & p < 1 \end{array}\right].$  

For $p = 1$,

$\displaystyle \int_{1}^{\infty}\frac{dx}{x^{p}} = \lim_{b \to \infty}\int_{1}^{... ...im_{b \to \infty} \log{\vert x\vert} - 1 = \infty \ \ensuremath{\ \blacksquare}$

If $f(x)$ is continuous on $(-\infty,\infty)$ and $\lim_{a \rightarrow -\infty,b \rightarrow \infty}\int_{a}^{b}f(x)dx$ exists, then we express this limit $\int_{-\infty}^{\infty}f(x)dx$.

Exercise 3..20   Evaluate the following improper integral.

$\displaystyle I = \int_{-\infty}^{\infty}\frac{dx}{1+x^{2}} $

SOLUTION
$\displaystyle I$ $\displaystyle =$ $\displaystyle \lim_{{a \rightarrow -\infty}, {b \rightarrow \infty}}\int_{a}^{b... ...{1+x^{2}} = \lim_{a \to -\infty, b \to \infty}\left[\tan^{-1}{x}\right]_{a}^{b}$  
  $\displaystyle =$ $\displaystyle \lim_{b \rightarrow \infty} \tan^{-1}{b} - \lim_{a \rightarrow - ... ...an^{-1}{a} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi \ensuremath{\ \blacksquare}$  

Gamma Function
Let $\displaystyle{\Gamma(\lambda) = \int_{0}^{\infty}x^{\lambda-1}e^{-x}dx}$. Then

$\displaystyle{1. \ \Gamma(n + 1) = n\Gamma(n) \ (n > 0)}$

$2. \ n$ is natural number  $\displaystyle{\Gamma(n) = (n-1)!}$

$\displaystyle{3. \ \Gamma(\frac{1}{2}) = \sqrt{\pi}}$

Proof 1.
$\displaystyle \Gamma(n+1)$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}x^{n}e^{-x} dx$  

$\displaystyle \int_{0}^{\infty}x^{n}e^{-x} dx = \lim_{b \to \infty}\int_{0}^{b}x^{n}e^{-x} dx.$

Using integration by parts, $\left\{\begin{array}{ll} f = x^{n} & g' = e^{-x}\\ f' = nx^{n -1} & g = - e^{-x} \end{array}\right.$

$\displaystyle \lim_{b \to \infty}\int_{0}^{b}x^{n}e^{-x} dx$ $\displaystyle =$ $\displaystyle \lim_{b \to \infty}\left[x^{n}(-e^{-x}) \right ]_{0}^{b} - \lim_{b \to \infty}\int_{0}^{b}nx^{n-1}(-e^{-x}) dx$  
  $\displaystyle =$ $\displaystyle \lim_{b \to \infty}\left[-x^{n}(e^{-x}) \right ]_{0}^{b} + n\lim_{b \to \infty}\int_{0}^{b}x^{n-1}(e^{-x}) dx$  
  $\displaystyle =$ $\displaystyle \lim_{b \rightarrow \infty}\frac{-b^{n}}{e^{b}} + n\Gamma(n)$  

To find $\lim_{b \to \infty}\frac{-b^n}{e^b}$, we use the Maclaurin series expansion of $e^x$.

$\displaystyle e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!} + \cdots$

Then

$\displaystyle \lim_{b \to \infty}\frac{-b^n}{e^b} = \lim_{b \to \infty}\frac{-b^n}{1 + b+ \cdots + \frac{b^n}{n!} + \frac{b^{n+1}}{(n+1)!} + \cdots} = 0.$

2. By 1. for $n > 0$, we have the recurrence relation $\Gamma(n+1) = \Gamma(n)$.

$\displaystyle \Gamma(n) = (n-1)\Gamma(n-1) = (n-1)(n-2)\Gamma(n-2) = \cdots = (n-1)(n-2)\cdots 2\Gamma(1). $

Now
$\displaystyle \Gamma(1)$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}e^{-x} dx = \lim_{b \to \infty}\int_0^b e^{-x}dx... ...lim_{b \to \infty}\left[-e^{-x}\right]_0^b = -\lim_{b \to \infty}e^{-b} + 1 = 1$  

Thus

$\displaystyle \Gamma(n) = (n-1)!. $

3. Note that $\displaystyle{\Gamma(\frac{1}{2}) = \int_{0}^{\infty}x^{-\frac{1}{2}}e^{-x}\:dx}$. Then let $t = \sqrt{x}$. Then $t^2 = x$, $dx = 2t \:dt$. \begin{displaymath}\begin{array}{c\vert ccc} x & 0 & \to & \infty\\ \hline t & 0 & \to & \infty \end{array}\end{displaymath}. Thus,
$\displaystyle \Gamma(\frac{1}{2})$ $\displaystyle =$ $\displaystyle \int_{0}^{\infty}x^{-\frac{1}{2}}e^{-x}\:dx = \int_{0}^{\infty}e^{-t^2}\frac{1}{t}(2tdt) = 2\int_0^\infty e^{-t^2}dt$  

In Example5.5, we have shown $\displaystyle{\int_0^\infty e^{-t^2}dt} = \frac{\sqrt{\pi}}{2}$. Thus

$\displaystyle \Gamma(\frac{1}{2}) = \sqrt{\pi}\ensuremath{\ \blacksquare}$

Exercise A


1.
Evaluate the improper integrals that converge?D

(a) $\displaystyle{\int_{0}^{\infty}e^{-x}\ dx}$ (b) $\displaystyle{\int_{1}^{\infty}\frac{dx}{x}}$ (c) $\displaystyle{\int_{1}^{\infty}\frac{dx}{x^{2}}}$ (d) $\displaystyle{\int_{1}^{\infty}\cos{\pi x}\ dx}$ (e) $\displaystyle{\int_{0}^{1}\frac{dx}{\sqrt{x}}}$ (f) $\displaystyle{\int_{0}^{1}\frac{dx}{x^{2}}}$

Exercise B


1.
Evaluate the improper integrals that converge?D

(a) $\displaystyle{\int_{0}^{1}\frac{dx}{\sqrt{1-x}}}$ (b) $\displaystyle{\int_{0}^{1}\frac{dx}{x}}$ (c) $\displaystyle{\int_{0}^{1}\log{x}dx}$

2.
Evaluate the improper integrals that converge.

(a) $\displaystyle{\int_{0}^{\infty}xe^{-x}dx}$ (b) $\displaystyle{\int_{2}^{\infty}\frac{1}{x(\log{x})^{\alpha}}dx}$ (c) $\displaystyle{\int_{2}^{\infty}\frac{1}{x^{\alpha} \log{x}} dx}$

3.
Determine which of the following integrals converge?D

(a) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sqrt{\cos{x}}}}$ (b) $\displaystyle{\int_{0}^{1}\frac{\log{x}}{\sqrt{x}}}$

4.
enshu:3-9-4 $K(k) = \int_{0}^{1}\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}dx$ is called the complete elliptic integral of the 1st kind?DLet $agm(a_{0},b_{0}), a_{0} \geq b_{0} > 0$ be the arithmetic mean. Then the following is known?D

$\displaystyle \frac{\pi}{2} = agm(1, \sqrt{1 - k^2})\int_{0}^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$

Make sure that the above equation is true for $k^2 = 0, 0.75$?D



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