Problem Set: Improper integrals
620-205 Semester I 2010

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 8 April 2010

(1) Improper integrals: Rational functions
(2) Improper integrals: Exponential functions
(3) Improper integrals: Special functions
(4) Improper integrals: Analysis and applications

Improper Integrals: Rational functions

For each of the following integrals:

(a) graph the integrand,

(b) determine if the integral converges, and

(c) evaluate the integral as appropriate.

	${\displaystyle {\int }_0^1\frac{1}{x}dx}$
	${\displaystyle {\int }_0^1\frac{1}{x^3}dx}$
	${\displaystyle {\int }_{-1}^1{(1-x^2)}^n\hspace{0.17em}dx}$
	${\displaystyle \underset{n\to \mathrm{\infty }}{\lim }{\int }_0^1\frac{n{y}^{n-1}}{1+y}\hspace{0.17em}dy}$
	${\displaystyle {\int }_0^3\frac{dx}{{(x-1)}^{\mathrm{2/3}}}}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{x}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{x^2}dx}$
	Show that ${\displaystyle {\int }_1^{\infty }\frac{1}{x^p}dx}$ converges if $p\in ℝ$ and $p>1$.
	Show that ${\displaystyle {\int }_1^{\infty }\frac{1}{x^p}dx}$ diverges if $p\in ℝ$ and $p\le 1$.
	${\displaystyle {\int }_0^{\infty }\frac{1}{x^2+1}dx}$
	${\displaystyle {\int }_0^1\frac{1}{\sqrt{x}}dx}$
	${\displaystyle {\int }_{-1}^1\frac{1}{x^{\mathrm{2/3}}}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{x^{1.001}}dx}$
	${\displaystyle {\int }_0^4\frac{1}{\sqrt{4-x}}dx}$
	${\displaystyle {\int }_0^1\frac{1}{\sqrt{1-x^2}}dx}$
	${\displaystyle {\int }_0^1\frac{1}{x^{0.999}}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{\sqrt{x}}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{x^3}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{x^3+1}dx}$
	${\displaystyle {\int }_0^{\infty }\frac{1}{x^3}dx}$
	${\displaystyle {\int }_0^{\infty }\frac{1}{x^3+1}dx}$
	${\displaystyle {\int }_{-1}^1\frac{1}{x^2}dx}$
	${\displaystyle {\int }_{-1}^1\frac{1}{x^{\mathrm{2/5}}}dx}$
	${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{x}}dx}$
	${\displaystyle {\int }_0^{\infty }\frac{1}{\sqrt{x+x^4}}dx}$
	${\displaystyle {\int }_1^5\frac{4x}{\sqrt{x^2-1}}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{1}{1+x^2}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{x^2}{(x-2){(x^{11}+2)}^{\mathrm{1/4}}}dx}$
	Let $a,b\in ℝ$ with $a<b$. Analyse ${\displaystyle {\int }_a^b{x}^{-\frac{1}{2}}dx}$.
	Let $a,b\in ℝ$ with $a<b$. Analyse ${\displaystyle {\int }_a^b\frac{1}{x}dx}$.
	Let $\alpha \in ℝ$. Analyse ${\displaystyle {\int }_0^1{x}^{\alpha }dx}$.
	Let $a\in ℝ$. Analyse ${\displaystyle {\int }_{-a}^a\frac{1}{x}dx}$.
	${\displaystyle {\int }_0^{\infty }\frac{1}{x+1}dx}$
	Let $m\in {ℝ}_{\le 0}$. Analyse ${\displaystyle {\int }_0^{\infty }{t}^{-m}dt}$.
	${\displaystyle {\int }_{-\infty }^{\infty }\frac{1}{1+t^2}dt}$
	${\displaystyle {\int }_1^{\infty }{t}^{-\mathrm{1/2}}dt}$
	${\displaystyle {\int }_1^{\infty }{t}^{-\mathrm{3/2}}dt}$
	${\displaystyle {\int }_0^1{t}^{-\mathrm{1/2}}dt}$
	${\displaystyle {\int }_0^1{t}^{-\mathrm{3/2}}dt}$
	${\displaystyle {\int }_4^{\infty }\frac{1}{t^2}dt}$
	${\displaystyle {\int }_{2.735}^{\infty }\frac{1}{t^{\mathrm{1/3}}}dt}$
	Let $p\in {ℝ}_{>0}$. Analyse ${\displaystyle {\int }_1^{\infty }\frac{1}{t^p}dt}$.
	${\displaystyle {\int }_0^1\frac{1}{t-\frac{1}{2}}dt}$

Improper Integrals: Exponential functions

For each of the following integrals:

(a) graph the integrand,

(b) determine if the integral converges, and

(c) evaluate the integral as appropriate.

	${\displaystyle {\int }_3^{\mathrm{\infty }}{e}^{-3x}dx}$
	${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx}$
	${\displaystyle {\int }_0^{\mathrm{\infty }}{x}^3{e}^{-x^2}dx}$
	${\displaystyle {\int }_{-\infty }^b{e}^{x-e^x}dx}$
	${\displaystyle {\int }_{-\infty }^{\infty }{e}^{x-e^x}dx}$
	${\displaystyle {\int }_1^{\infty }{e}^{-x^2}dx}$
	${\displaystyle {\int }_0^{\infty }{e}^{-x}\cos xdx}$
	${\displaystyle {\int }_0^{\infty }\frac{1}{1+e^x}dx}$
	${\displaystyle {\int }_0^{\infty }{e}^{-t}dt}$
	${\displaystyle {\int }_0^{\infty }{e}^{-x}dx}$
	${\displaystyle {\int }_{-\infty }^{\infty }{e}^{-t^2}dt}$
	${\displaystyle {\int }_3^{\mathrm{\infty }}\frac{1}{\sqrt{x+e^{2x}}}dx}$
	${\displaystyle {\int }_0^1\frac{e^x}{\sqrt{e^x-1}}dx}$
	${\displaystyle {\int }_1^{\mathrm{\infty }}\frac{3+e^{-x}}{2x^{\mathrm{2/3}}-1}dx}$

Improper Integrals: Special functions

For each of the following integrals:

(a) graph the integrand,

(b) determine if the integral converges, and

(c) evaluate the integral as appropriate.

	${\displaystyle {\int }_0^{\mathrm{\infty }}\cos x\hspace{0.17em}dx}$.
	${\displaystyle {\int }_0^{\pi /2}\tan x\hspace{0.17em}dx}$
	${\displaystyle {\int }_0^1\log x\hspace{0.17em}dx}$
	${\displaystyle {\int }_1^{\infty }\frac{{\left(\log x\right)}^4}{1+x^2}dx}$
	${\displaystyle {\int }_{-\pi }^{\pi }\frac{\sin x}{{\left|x\right|}^{\beta }}\hspace{0.17em}dx}$
	${\displaystyle {\int }_0^1{t}^{\mathrm{1/2}}{e}^{e^t}\hspace{0.17em}dt}$
	${\displaystyle {\int }_0^{100}{e}^{\lfloor t\rfloor }dt}$
	${\displaystyle {\int }_3^{\mathrm{\infty }}\frac{\log x}{\sqrt{x^{\mathrm{3/2}}+1}}dx}$
	${\displaystyle {\int }_0^{\infty }\left|\frac{\sin x}{x^2+1}\right|\hspace{0.17em}dx}$
	${\displaystyle {\int }_0^{\infty }\frac{\sin x}{x}\hspace{0.17em}dx}$
	${\displaystyle {\int }_0^{\infty }{x}^{\alpha -1}\cos x\hspace{0.17em}dx}$
	${\displaystyle {\int }_0^{\infty }{x}^{\alpha -1}\sin x\hspace{0.17em}dx}$
	${\displaystyle B(u,v)={\int }_0^1{t}^{u-1}{(1-t)}^{v-1}\hspace{0.17em}dt}$
	Let ${\displaystyle B(u,v)={\int }_0^1{t}^{u-1}{(1-t)}^{v-1}\hspace{0.17em}dt}$. Show that ${\displaystyle B(u,v)={\int }_0^{\infty }\frac{x^{u-1}}{{(1+x)}^{u+v}}\hspace{0.17em}dx}$ by setting $t={\displaystyle \frac{x}{1+x}}$.

Improper Integrals: Analysis and applications

	Let $A,r\in ℝ$. Analyse ${\displaystyle {\int }_0^{\mathrm{\infty }}A{e}^{-rt}\hspace{0.17em}dt}$.
	Show that if $z=\sqrt{{\displaystyle \frac{4+x}{1-x}}}$ then ${\displaystyle {\int }_{-1}^1\sqrt{\frac{1+x}{1-x}}}dx={\displaystyle {\int }_0^{\infty }\frac{4z^2}{{(z^2+1)}^2}}dz$. The improper integral on the left is an improper integral of the first kind and the improper integral on the right is an improper integral of the second kind.
	Show that ${\displaystyle {\int }_{-1}^1\sqrt{\frac{1+x}{1-x}}}dx=\pi$.
	Let $n\in {\mathbb{Z}}_{\ge 0}$. Show that $n\text{!}={\displaystyle {\int }_0^{\infty }{e}^{-t}{t}^{n}dt}$.
	Define $\Gamma \left(z\right)={\displaystyle {\int }_0^{\infty }{e}^{-t}{t}^{z-1}dt}$. Show that $\Gamma \left(1\right)=1$ and $\Gamma (z+1)=z\Gamma \left(z\right)$.
	Let $f\left(x\right)=\left\{\begin{array}{cc}{e}^x\text{,}& \text{if }x\in {ℝ}_{\ge 0}\text{ and }x\in \mathbb{Q}\text{,}\\ e^{-x}\text{,}& \text{if }x\in {ℝ}_{\ge 0}\text{ and }x\notin \mathbb{Q}\text{.}\end{array}\right.$ Show that ${\displaystyle {\int }_0^{\mathrm{\infty }}\left|f\right(x\left)\right|\hspace{0.17em}dx}$ exists but ${\displaystyle {\int }_0^{\mathrm{\infty }}f\left(x\right)\hspace{0.17em}dx}$ doesn't.
	Let ${\displaystyle B(m+1)={\int }_0^1{x}^{m-1}{(1-x)}^{-\mathrm{1/2}}\hspace{0.17em}dx}$. (a) Explain why $B\left(m\right)$ is improper for all values of $m$. (b) Find the value of $B\left(1\right)$. (c) Show that $B(m+1)={\displaystyle \frac{2m}{2m+1}}B\left(m\right)$. (d) Find the values of $B\left(2\right)$, $B\left(3\right)$ and $B\left(4\right)$.
	Let ${\displaystyle \Gamma \left(p\right)={\int }_0^{\mathrm{\infty }}{x}^{p-1}{e}^{-x}\hspace{0.17em}dx}$. (a) Explain why $\Gamma \left(p\right)$ is improper for all values of $p$. (b) Evaluate $\Gamma \left(1\right)$ and $\Gamma \left(2\right)$. (c) Show that $\Gamma (p+1)=p\Gamma \left(p\right)$. (d) Find the values of $\Gamma \left(3\right)$, $\Gamma \left(4\right)$ and $\Gamma \left(5\right)$.

References

[Ca] S. Carnie, 620-143 Applied Mathematics, Course materials, 2006 and 2007.

[Hu] B.D. Hughes, 620-158 Accelerated Mathematics 2, Lectures by B.D. Hughes, University of Melbourne, 2009.

[TF] Thomas and Finney, Calculus and Analytic Geometry, Fifth Edition, Addison-Wesley 1979.