\[
\newcommand{\Ann}{\mathsf{Ann}}
\newcommand{\Arg}{\mathsf{Arg}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Im}{\mathsf{Im}}
\newcommand{\intd}{\,\mathsf{d}}
\newcommand{\Re}{\mathsf{Re}}
\newcommand{\Res}{\mathsf{Res}}
\newcommand{\ball}{\mathsf{B}}
\newcommand{\wind}{\mathsf{wind}}
\newcommand{\Log}{\mathsf{Log}}
\newcommand{\l}{<}
\]
Week 10 Worksheet
Home | Assessment | Notes | Worksheets | Blackboard
Infinite Real Integrals
- Explain why $\displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{t^2+1}\intd t$ exists.
- Use Cauchy's Residue Theorem to evaluate $\displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{t^2+1}\intd t$
- Evaluate $\displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{t^2+1}\intd x$ using only calculus and limits.
- Use Cauchy's residue theorem to evaluate, the integral $\displaystyle\int\limits_{-\infty}^{\infty} \frac{\exp(2 it)}{t^2+1} \intd t$
- By taking real and imaginary parts, calculate $\displaystyle\int\limits_{-\infty}^{\infty} \frac{ \cos(2t)}{t^2+1} \intd t$ and $\displaystyle\int\limits_{-\infty}^{\infty} \frac{ \sin (2t)}{t^2+1} \intd t$
- How can you tell, without evaluating the integrals, that one of them integrals is zero?
- Why does the contour from the ``Infinite Real Integrals'' video fail when we try to integrate
\[
\int\limits_{-\infty}^{\infty} \frac{\exp(-2 it)}{t^2+1} \intd t
\]
using our approach?
By choosing a different contour, explain how one could evaluate this integral using Cauchy's Residue Theorem.
- Use Cauchy's Residue Theorem to evaluate the following real integrals.
- $\displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{(t^2+1)(t^2+3)} \intd t$
- $\displaystyle\int\limits_{-\infty}^{\infty} \frac{1}{28+11t^2+t^4} \intd t$
- By considering the function
\[
f(z) = \frac{\exp(iz)}{z^2+4z+5}
\]
integrated around a suitable contour, show that
\[
\int\limits_{-\infty}^{\infty} \frac{\sin t}{t^2+4t+5} \intd t = \frac{-\pi\sin 2}{e}.
\]
- This question is about the integral $\displaystyle\int\limits_0^{2\pi} \frac{1}{13+5\cos t} \intd t$
- Use the substitution $z=e^{it}$ to show that
\[
\int\limits_0^{2\pi} \frac{1}{13+5\cos t} \intd t
=
\frac{2}{i} \int_{\gamma} \frac{1}{5z^2+26z+5} \intd z
\]
where $\gamma(t) = e^{it}$ on $[0,2\pi]$.
- Show that
\[
f(z)=\dfrac{1}{5z^2+26z+5}
\]
has simple poles at $z=-5$ and $z=-1/5$. Show that $\Res(f,-1/5) = 1/24$.
- Use Cauchy's Residue Theorem to show that
\[
\int\limits_0^{2\pi} \frac{1}{13+5\cos t} \intd t = \frac{\pi}{6}
\]
- Use
\[
\cos t = \dfrac{\exp(it) + \exp(-it)}{2}
\]
to convert the following real integrals into complex integrals around the unit circle in the complex plane. Then apply Cauchy's Residue Theorem to evaluate them.
- $\displaystyle\int\limits_0^{2\pi} 2(\cos t)^3 + 3 (\cos t)^2 \intd t$
- $\displaystyle\int\limits_0^{2\pi} \frac{1}{1+\cos^2 t} \intd t$