Week 10 Worksheet - Solutions

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Infinite Real Integrals

1. 1. The integrand satisfies $$|f(t)|={\displaystyle \frac{1}{|z^2+1|}}\le {\displaystyle \frac{1}{|t^2|}}={\displaystyle \frac{1}{|t{|}^2}}$$ for $|t|>1$ so the infinite real integral exists by our lemma.
   2. Define $f:\mathbb{C}\setminus \{i,-i\}\to \mathbb{C}$ by $f(z)={\displaystyle \frac{1}{z^2+1}}$. For the semi-circular arc $\eta (t)=T{e}^{it}$ on $[0,\pi ]$ we have $$|\underset{\eta }{\int }f(z)\mathsf{d}z|=\left|\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{1}{(T{e}^{it}{)}^2+1}}iT{e}^{it}\mathsf{d}t\right|\le \underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{T}{|T^2{e}^{2it}-(-1)|}}\mathsf{d}t\le \underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{T}{|T^2-1|}}\mathsf{d}t$$ by the reverse triangle inequality. When $T>2$ we have $|T^2-1|=T^2-1\ge T^2/2$ so $$\left|\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{1}{z^2+1}}\mathsf{d}z\right|\le {\displaystyle \frac{2\pi }{T}}$$ and this goes to zero as $T\to \mathrm{\infty }$. When $T$ is large our semi-circular contour $\mathrm{\Gamma }$ contains $i$ so we must calculate the residue of $f$ there. Since $f$ is a ratio $$\mathsf{Res}(f,i)={\displaystyle \frac{1}{2i}}$$ and Cauchy's residue theorem gives $$\underset{-T}{\overset{T}{\int }}{\displaystyle \frac{1}{t^2+1}}\mathsf{d}t+\underset{0}{\overset{\pi }{\int }}{\displaystyle \frac{Ti{e}^{it}}{(T{e}^{it}{)}^2+1}}\mathsf{d}t=2\pi i\mathsf{Res}(f,i)=\pi$$ which, upon taking the limit as $T\to \mathrm{\infty }$ gives the answer ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{t^2+1}}\mathsf{d}t=\pi }$
   3. According to the definition we must calculate the following limit. $$\begin{array}{rl}\underset{A,B\to \mathrm{\infty }}{lim}\underset{-A}{\overset{B}{\int }}{\displaystyle \frac{1}{t^2+1}}\mathsf{d}t=& \underset{A,B\to \mathrm{\infty }}{lim}[\arctan (t){]}_{-A}^B\\ =& \underset{A,B\to \mathrm{\infty }}{lim}\arctan (B)-\arctan (-A)={\displaystyle \frac{\pi }{2}}-(-{\displaystyle \frac{\pi }{2}})=\pi \end{array}$$
2. 1. Define $f:\mathbb{C}\setminus \{i,-i\}\to \mathbb{C}$ by $f(z)={\displaystyle \frac{\exp (2iz)}{z^2+1}}$ which has absolute value satisfying $$|f(z)|={\displaystyle \frac{e^{-2y}}{|z^2+1|}}\le {\displaystyle \frac{1}{z^2+1}}$$ when $y>0$ because $|\exp (2iz)|=|\exp (2ix)\exp (-2y)|=e^{-2y}$. It is therefore the case that the infinite real integral exists, and that the semi-circular contour integral converges to zero as $T\to \mathrm{\infty }$ just as in the previous question. There is again only one residue inside $\mathrm{\Gamma }$ and it is at $i$. Now $$\mathsf{Res}(f,i)={\displaystyle \frac{\exp (2i\cdot i)}{2i}}={\displaystyle \frac{1}{2i{e}^2}}$$ so Cauchy's residue theorem gives $$\underset{-T}{\overset{T}{\int }}{\displaystyle \frac{\exp (2it)}{t^2+1}}\mathsf{d}t=2\pi i\mathsf{Res}(f,i)={\displaystyle \frac{\pi }{e^2}}$$ after taking the limit as $T\to \mathrm{\infty }$.
   2. Writing $\exp (2it)=\cos (2t)+i\sin (2t)$ gives $${\displaystyle \frac{\pi }{e^2}}=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\exp (2it)}{t^2+1}}\mathsf{d}t=\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\cos (2t)}{t^2+1}}\mathsf{d}t+i\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\sin (2t)}{t^2+1}}\mathsf{d}t$$ where we note that the two infinite real integrals at right both exist by the same bounding technique as above, which still applies because $|\sin (2t)|\le 1$ and $|\cos (2t)|\le 1$ when $t$ is real. We conclude that $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\cos (2t)}{t^2+1}}\mathsf{d}t={\displaystyle \frac{\pi }{e^2}}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\sin (2t)}{t^2+1}}\mathsf{d}t=0$$ by taking real and imaginary parts.
   3. We could conclude immediately that the $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\sin (2t)}{t^2+1}}\mathsf{d}t$$ is zero because, we may calculate it as $$\underset{T\to \mathrm{\infty }}{lim}\underset{-T}{\overset{T}{\int }}{\displaystyle \frac{\sin (2t)}{t^2+1}}\mathsf{d}t$$ and the integrand is an odd function of $T$.

3. The infinite real integral exists for the same reason as before: we still have $|\exp (-2it)|=1$ for all real $t$. However, in this case we would define $f:\mathbb{C}\setminus \{i,-i\}\to \mathbb{C}$ by $f(z)={\displaystyle \frac{\exp (-2iz)}{z^2+1}}$ and would be stuck with $$|\exp (-2iz)|=\exp (2y)$$ in the numerator. This term grows quickly as $y\to \mathrm{\infty }$ so we would not be able to control the semi-circular contour integral.

   We can instead use $\nu (t)=T{e}^{-it}$ on $[0,\pi ]$ to get from $T$ to $-T$. Indeed, since $$|\exp (-2iz)|=|\exp (-2ix)\exp (2y)|=\exp (2y)=e^{-2|y|}$$ when $y<0$, going below the horizontal axis lets us regain control of our semi-circular contour integral. We need to be a bit careful with Cauchy's residue theorem as well. It says $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\exp (-2it)}{t^2+1}}\mathsf{d}t=2\pi i\mathsf{wind}(\mathrm{\Gamma },-i)\mathsf{Res}(f,-i)$$ because our contour $\mathrm{\Gamma }=(\gamma ,\eta )$ now contains $-i$ and not $i$, and winds clockwise around $-i$ i.e.\ $\mathsf{wind}(\mathrm{\Gamma },-i)=-1$. So $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\exp (-2it)}{t^2+1}}\mathsf{d}t=2\pi i(-1){\displaystyle \frac{\exp (-2i\cdot (-i))}{2(-i)}}={\displaystyle \frac{\pi }{e^2}}$$

4. 1. Define $f:\mathbb{C}\setminus \{i,-i,\sqrt{3}i,-\sqrt{3}i\}\to \mathbb{C}$ by $f(z)={\displaystyle \frac{1}{(z^2+1)(z^2+3)}}$ and note that $|z|\ge 6$ implies $$|f(z)|={\displaystyle \frac{1}{|z^2-(-3)||z^2-(-3)|}}\le {\displaystyle \frac{1}{(|z{|}^2-1)(|z{|}^2-3)}}\le {\displaystyle \frac{12}{|z{|}^4}}$$ using the reverse triangle inequality and the fact that $|z{|}^2-K\ge |z{|}^2/2$ whenever $|z|\ge 2K$ and $K>1$. This estimate - which in particular is true for $z=t$ with $|t|\ge 6$ - implies that the infinite real integral exists and that the semi-circular contour integral vanishes as $T\to \mathrm{\infty }$. Indeed, for the latter statement note that $$|\underset{0}{\overset{\pi }{\int }}f(T{e}^{it})iT{e}^{it}\mathsf{d}t|\le 2\pi {\displaystyle \frac{12T}{T^4}}={\displaystyle \frac{24\pi }{T^3}}$$ which goes to zero as $T\to \mathrm{\infty }$. We conclude from Cauchy's residue theorem that $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{(t^2+1)(t^2+3)}}\mathsf{d}t=2\pi i\mathsf{Res}(f,i)+2\pi i\mathsf{Res}(f,\sqrt{3}i)$$ as the denominator has simple zeroes - and therefore $f$ has simple poles - at each of $i,-i,\sqrt{3}i,-\sqrt{3}i$. As we have a ratio with simple poles, we can calculate the residues by differentiating the denominator $${((t^2+1)(t^2+3))}^{\prime }{(t^4+4t^2+3)}^{\prime }=4t^3+8t$$ and plugging in our poles. We get $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{(t^2+1)(t^2+3)}}\mathsf{d}t=2\pi i{\displaystyle \frac{1}{4(i{)}^3+8i}}+2\pi i{\displaystyle \frac{1}{4(\sqrt{3}i{)}^3+8\sqrt{3}i}}={\displaystyle \frac{\pi }{6}}(3-\sqrt{3})$$
   2. We can factor the denominator as $28+11t^2+t^4=(7+t^2)(4+t^2)$ and so define $f:\mathbb{C}\setminus \{2i,-2i,\sqrt{7}i,-\sqrt{7}i\}\to \mathbb{C}$ by $$f(z)={\displaystyle \frac{1}{(z^2+7)(z^2+4)}}$$ which satisfies $$|f(z)|\le {\displaystyle \frac{112}{|z{|}^4}}$$ giving existence and decay of the semi-circular contour integral just as in the previous part. Cauchy's residue theorem gives $$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{28+11t^2+t^4}}\mathsf{d}t=2\pi i\mathsf{Res}(f,2i)+2\pi i\mathsf{Res}(f,\sqrt{7}i)$$ and the derivative of the denominator is $4t^3+22t$ so $$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{1}{28+11t^2+t^4}}\mathsf{d}t& =2\pi i{\displaystyle \frac{1}{22(2i)+4(2i{)}^3}}+2\pi i{\displaystyle \frac{1}{22(\sqrt{7}i)+4(\sqrt{7}i{)}^3}}\\ & =2\pi i{\displaystyle \frac{1}{44i-32i}}+2\pi i{\displaystyle \frac{1}{22\sqrt{7}i-28\sqrt{7}i}}\\ & ={\displaystyle \frac{\pi }{6}}-{\displaystyle \frac{\pi }{3\sqrt{7}}}={\displaystyle \frac{\pi }{6}}(1-{\displaystyle \frac{2}{\sqrt{7}}})\end{array}$$
5. The denominator can be written as $z^2+4z+5=(z+2{)}^2+1$ so factors as $$(z-(i-2))(z-(-i-2))$$ which shows $f$ has simple poles at $i-2$ and $-i-2$. In absolute value $$|f(z)|={\displaystyle \frac{|\exp (iz)|}{|(z+2{)}^2+1|}}\le {\displaystyle \frac{e^{-y}}{|z+2{|}^2-1}}\le {\displaystyle \frac{4}{|z+2{|}^2}}\le {\displaystyle \frac{4}{(|z|-2{)}^2}}\le {\displaystyle \frac{16}{|z{|}^2}}$$ using $\exp (i(x+iy))=e^{ix}{e}^{-y}$ and the reverse triangle inequality. We again have existence of the infinite real integral and decay of the semi-circular contour integral as $T\to \mathrm{\infty }$ from this bound. Next we apply Cauchy's residue theorem. $$\begin{array}{rl}\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{\displaystyle \frac{\exp (it)}{t^2+4t+5}}\mathsf{d}t& =2\pi i\mathsf{Res}(f,i-2)\\ & =2\pi i{\displaystyle \frac{\exp (i(i-2))}{2(i-2)+4}}=\pi \exp (-1-2i)={\displaystyle \frac{\pi }{e}}(\cos 2-i\sin 2)\end{array}$$ using $\exp (it)=\cos (t)+i\sin (t)$. Doing so again in the integrand and taking the imaginary part of the above equation gives the result.
6. 1. With $z=e^{it}$ we have $\mathsf{d}z=i{e}^{it}\mathsf{d}t$ and $$\begin{array}{rl}\frac{2}{i}\underset{\gamma }{\int }{\displaystyle \frac{1}{5z^2+26z+5}}\mathsf{d}z& =\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{1}{5(e^{it}{)}^2+26e^{it}+5}}i{e}^{it}\mathsf{d}t\\ & =\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{i}{5e^{it}+5e^{-it}+26}}\mathsf{d}t\\ & =\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{i}{10\cos (t)+26}}\mathsf{d}t={\displaystyle \frac{i}{2}}\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{i}{5\cos (t)+13}}\mathsf{d}t\end{array}$$ as claimed.
   2. If we multiply out $$(z-(-5))(z-(-1/5))=(z^2+{\displaystyle \frac{26}{5}}+1)={\displaystyle \frac{1}{5}}(5z^2+26z+5)$$ we see that the denominator has simple zeroes at $-5$ and $-1/5$. Since the numerator is never zero, the ratio has simple poles at $-5$ and $-1/5$. We calculate $$\mathsf{Res}(f,-1/5)={\displaystyle \frac{1}{10(-1/5)+26}}={\displaystyle \frac{1}{24}}$$
   3. The only pole of $f$ within $\gamma$ is $-1/5$. Cauchy's residue theorem tells us that $$\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{1}{13+5\cos (t)}}\mathsf{d}t={\displaystyle \frac{2}{i}}\underset{\gamma }{\int }f={\displaystyle \frac{2}{i}}2\pi i\mathsf{Res}(f,-1/5)=4\pi {\displaystyle \frac{1}{24}}={\displaystyle \frac{\pi }{6}}$$
7. 1. We have $$\begin{array}{rl}(\cos t{)}^3& ={\displaystyle \frac{1}{8}}(e^{it}+e^{-it}{)}^3={\displaystyle \frac{e^{3it}+3e^{it}+3e^{-it}+e^{-3it}}{8}}\\ (\cos t{)}^2& ={\displaystyle \frac{1}{4}}(e^{it}+e^{-it}{)}^2={\displaystyle \frac{e^{2it}+2+e^{-2it}}{4}}\end{array}$$ so our integral becomes $$\begin{array}{rl}& {\displaystyle \frac{1}{4}}\underset{0}{\overset{2\pi }{\int }}{e}^{3it}+3e^{2it}+3e^{it}+6+3e^{-it}+3e^{-2it}+e^{-3it}\mathsf{d}t\\ =& {\displaystyle \frac{1}{4i}}\underset{0}{\overset{2\pi }{\int }}(e^{2it}+3e^{it}+3+6e^{-it}+3e^{-2it}+3e^{-3it}+e^{-4it})i{e}^{it}\mathsf{d}t\\ =& {\displaystyle \frac{1}{4i}}\underset{\gamma }{\int }{z}^2+3z+3+{\displaystyle \frac{6}{z}}+{\displaystyle \frac{3}{z^2}}+{\displaystyle \frac{3}{z^3}}+{\displaystyle \frac{1}{z^4}}\mathsf{d}z\end{array}$$ with $\gamma (t)=e^{it}$ on $[0,2\pi ]$. The integrand is already a Laurent series so we can read off that its residue at the origin is $6$. Cauchy's residue theorem then gives $$\underset{0}{\overset{2\pi }{\int }}2(\cos t{)}^3+3(\cos t{)}^2\mathsf{d}t={\displaystyle \frac{1}{4i}}\cdot 2\pi i\cdot 6=3\pi$$
   2. Here we use $$(\cos t{)}^2={\displaystyle \frac{1}{4}}(e^{it}+e^{-it}{)}^2={\displaystyle \frac{e^{2it}+2+e^{-2it}}{4}}$$ to write the integrand as $${\displaystyle \frac{4}{e^{2it}+6+e^{-2it}}}={\displaystyle \frac{4e^{2it}}{e^{4it}+6e^{2it}+1}}={\displaystyle \frac{1}{i}}{\displaystyle \frac{4e^{it}}{e^{4it}+6e^{2it}+1}}i{e}^{it}$$ so that $$\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{1}{1+(\cos t{)}^2}}\mathsf{d}t={\displaystyle \frac{1}{i}}\underset{\gamma }{\int }{\displaystyle \frac{4z}{z^4+6z^2+1}}\mathsf{d}z$$ where $\gamma (t)=e^{it}$ on $[0,2\pi ]$. From $$z^4+6z^2+1=(z^2+3{)}^2-8=((z^2+3)-\sqrt{8})((z^2+3)+\sqrt{8})$$ and the fact that $-3-\sqrt{8}$ and $\sqrt{8}-3$ are both negative, we see that the denominator of the latter integrand has zeroes at $$i\sqrt{3-\sqrt{8}}-i\sqrt{3-\sqrt{8}}i\sqrt{\sqrt{8}+3}-i\sqrt{\sqrt{8}+3}$$ and that only the first two are inside $\gamma$. Thus $${\displaystyle \frac{1}{i}}\underset{\gamma }{\int }{\displaystyle \frac{4z}{z^4+6z^2+1}}\mathsf{d}z=2\pi \mathsf{Res}(f,i\sqrt{3-\sqrt{8}})+2\pi \mathsf{Res}(f,-i\sqrt{3-\sqrt{8}})$$ and we need to calculate the residues. The derivative of the denominator is $4z(z^2+3)$ so we can calculate the residues by plugging into $1/(z^2+3)$. We get $$\mathsf{Res}(f,i\sqrt{3-\sqrt{8}})={\displaystyle \frac{1}{\sqrt{8}}}\mathsf{Res}(f,-i\sqrt{3-\sqrt{8}})={\displaystyle \frac{1}{\sqrt{8}}}$$ so $$\underset{0}{\overset{2\pi }{\int }}{\displaystyle \frac{1}{1+(\cos t{)}^2}}\mathsf{d}t={\displaystyle \frac{4\pi }{\sqrt{8}}}=\sqrt{2}\pi$$