Week 10 Worksheet

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Infinite Real Integrals

1. 1. Explain why ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{t^2+1}\mathsf{d}t}$ exists.
   2. Use Cauchy's Residue Theorem to evaluate ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{t^2+1}\mathsf{d}t}$
   3. Evaluate ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{t^2+1}\mathsf{d}x}$ using only calculus and limits.
2. 1. Use Cauchy's residue theorem to evaluate, the integral ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\exp (2it)}{t^2+1}\mathsf{d}t}$
   2. By taking real and imaginary parts, calculate ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\cos (2t)}{t^2+1}\mathsf{d}t}$ and ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (2t)}{t^2+1}\mathsf{d}t}$
   3. How can you tell, without evaluating the integrals, that one of them integrals is zero?
3. Why does the contour from the ``Infinite Real Integrals'' video fail when we try to integrate $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\exp (-2it)}{t^2+1}\mathsf{d}t$$ using our approach? By choosing a different contour, explain how one could evaluate this integral using Cauchy's Residue Theorem.
4. Use Cauchy's Residue Theorem to evaluate the following real integrals.
   1. ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{(t^2+1)(t^2+3)}\mathsf{d}t}$
   2. ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{1}{28+11t^2+t^4}\mathsf{d}t}$
5. By considering the function $$f(z)=\frac{\exp (iz)}{z^2+4z+5}$$ integrated around a suitable contour, show that $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{\sin t}{t^2+4t+5}\mathsf{d}t=\frac{-\pi \sin 2}{e}.$$
6. This question is about the integral ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{13+5\cos t}\mathsf{d}t}$
   1. Use the substitution $z=e^{it}$ to show that $$\underset{0}{\overset{2\pi }{\int }}\frac{1}{13+5\cos t}\mathsf{d}t=\frac{2}{i}{\int }_{\gamma }\frac{1}{5z^2+26z+5}\mathsf{d}z$$ where $\gamma (t)=e^{it}$ on $[0,2\pi ]$.
   2. Show that $$f(z)={\displaystyle \frac{1}{5z^2+26z+5}}$$ has simple poles at $z=-5$ and $z=-1/5$. Show that $\mathsf{Res}(f,-1/5)=1/24$.
   3. Use Cauchy's Residue Theorem to show that $$\underset{0}{\overset{2\pi }{\int }}\frac{1}{13+5\cos t}\mathsf{d}t=\frac{\pi }{6}$$
7. Use $$\cos t={\displaystyle \frac{\exp (it)+\exp (-it)}{2}}$$ 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.
   1. ${\displaystyle \underset{0}{\overset{2\pi }{\int }}2(\cos t{)}^3+3(\cos t{)}^2\mathsf{d}t}$
   2. ${\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{1}{1+{\cos }^{2}t}\mathsf{d}t}$