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.

1st kind Improper Integral of the 1st kind applies to the case where the function has discontinuous points in the limit of integral.

If the integrand is continuous on $(a,b]$ and discontinuous at $x=a$ , then integrate from $a + \varepsilon$ to $b$ . Then find the limit as $\varepsilon$ approaches 0 from the right.

**Figure 3.11:** Discontinuous at
$\includegraphics[width=3.5cm]{SOFTFIG-3/ImproperInt1.eps}$

Note that the integral from $a + \varepsilon$ to $b$ is the definite integral.

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

is continuous on $[a,b)$

and discontinuous at $x = b$

. Then

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

First use u-substitution to make the integrand simple function. Then solve improper integral.

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 $\log{\vert x\vert}$ .

$\displaystyle \int_0^1 \frac{1}{x}dx$	$\displaystyle =$	$\displaystyle \lim_{\varepsilon \to 0+}\int_{\varepsilon}^{1}\frac{1}{x}dx = \l... ...arepsilon}^{1} = \log{1} - \lim_{\varepsilon \to 0+}\log{\vert\varepsilon\vert}$
	$\displaystyle =$	$\displaystyle 0 - (-\infty) = \infty \ensuremath{ \blacksquare}$

**Figure 3.12:** Exercise3-18-2
$\includegraphics[width=3.5cm]{SOFTFIG-3/logx_gr1.eps}$

$\lim_{x \to 0+}\log{x} = -\infty$ .

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 $t = 0$

$\displaystyle \int_0^1 \frac{1}{t}dt$	$\displaystyle =$	$\displaystyle \lim_{\varepsilon \to 0+}\int_\varepsilon^1 \frac{1}{t}dt = \lim_... ...}\right]_{\varepsilon}^1 = \log{1} - \lim_{\varepsilon \to 0+}\log{\varepsilon}$
	$\displaystyle =$	$\displaystyle 1 - (-\infty) = \infty\ensuremath{ \blacksquare}$

Since the function is not continuous at $x = 1$ , we may write $\lim_{\varepsilon \to 0+}\int_{1+\varepsilon}^e \frac{1}{x\log{x}}\:dx$ . But it is better to take u-substitution first.

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

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

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}{\frac{x}{a}} + c$ . $\displaystyle{\lim_{\varepsilon' \to 0+}\sin^{-1}{(1-\varepsilon')} = \frac{\pi}{2}}$ .

**Figure 3.13:** Discontinuity
$\includegraphics[width=3.5cm]{SOFTFIG-3/piecewise.eps}$

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. $\blacksquare$

**Figure 3.14:** Example3-19
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Exercise 3..19 Evaluate the following improper integral

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

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

**Figure 3.15:** Exercise3-19
$% latex2html id marker 47871 \includegraphics[width=3.5cm]{SOFTFIG-3/enshu3-19_gr3.eps}$

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 $\blacksquare$ 2nd kind Improper integral of the 2nd kind is infinite integral.

**Figure 3.16:** Infinite integral
$\includegraphics[width=3.5cm]{SOFTFIG-3/mugensekibun.eps}$

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.

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

This improper integral is used to find the convergence and divergence of other integral. For example, $\int_1^{\infty}\frac{1}{x^2+1}\:dx$ is less than $\int_1^{\infty}\frac{1}{x^2}\:dx$ and $\int_1^{\infty}\frac{1}{x^2}\:dx$ converges. Thsu, converges. 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

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

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.

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

**Figure 3.17:** Exercise3-20
$% latex2html id marker 47956 \includegraphics[width=3.5cm]{SOFTFIG-3/enshu3-20_gr1.eps}$

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 $\displaystyle{\Gamma(n) = (n-1)!}$

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

Gamma Function Gamma function was created to extend the factorial.

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

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


Maclaurin series expansion of is one of the most important series expansion. As we know the derivative of is . Then in the Maclaurin series expansion, every term is the derivative of the next term. Thus, $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \cdots$ .

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,

$\displaystyle \Gamma(\frac{1}{2})$

$\displaystyle =$

$\displaystyle \int_{0}^{\infty}x^{-\frac{1}{2}}e^{-x}\:dx = \int_{0}^{\infty}e^{-t^2}\frac{1}{t}(2tdt) = 2\int_0^\infty e^{-t^2}dt$

$\int e^{-t^2}\:dt$
It is impossible to find $\int e^{-t^2}\:dt$ . In other words, we can not express the indefinite integral of $e^{-t^2}$ using elementary functions.

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
(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

(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

(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 kindLet $agm(a_{0},b_{0}), a_{0} \geq b_{0} > 0$ be the arithmetic mean. Then the following is known

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

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