Week 11 Worksheet - Solutions

Home | Assessment | Notes | Index | Worksheets | Blackboard

Infinite Real Integrals

1. 1. The integrand satisfies \[ |f(t)| = \dfrac{1}{|z^2 + 1|} \le \dfrac{1}{|t^2|} = \dfrac{1}{|t|^2} \] for $|t| > 1$ so the infinite real integral exists by our lemma.
   2. Define $f : \C \setminus \{i,-i\} \to \C$ by $f(z) = \dfrac{1}{z^2 + 1}$. For the semi-circular arc $\eta(t) = Te^{it}$ on $[0,\pi]$ we have \[ \left| \int\limits_\eta f(z) \intd z \right| = \left| \int\limits_0^{\pi} \dfrac{1}{(Te^{it})^2 + 1} iT e^{it} \intd t \right| \le \int\limits_0^\pi \dfrac{T}{|T^2 e^{2it} - (-1)|} \intd t \le \int\limits_0^\pi \dfrac{T}{|T^2 - 1|} \intd t \] by the reverse triangle inequality. When $T > 2$ we have $|T^2 - 1| = T^2 - 1 \ge T^2 / 2$ so \[ \left| \int\limits_0^\pi \dfrac{1}{z^2 + 1} \intd z \right| \le \dfrac{2 \pi}{T} \] and this goes to zero as $T \to \infty$. When $T$ is large our semi-circular contour $\Gamma$ contains $i$ so we must calculate the residue of $f$ there. Since $f$ is a ratio \[ \Res(f,i) = \dfrac{1}{2i} \] and Cauchy's residue theorem gives \[ \int\limits_{-T}^T \dfrac{1}{t^2 + 1} \intd t + \int\limits_0^{\pi} \dfrac{Tie^{it}}{(Te^{it})^2 + 1} \intd t = 2 \pi i \Res(f,i) = \pi \] which, upon taking the limit as $T \to \infty$ gives the answer $\displaystyle\int\limits_{-\infty}^\infty \dfrac{1}{t^2 + 1} \intd t = \pi$
   3. According to the definition we must calculate the following limit. \begin{align*} \lim_{A,B \to \infty} \int\limits_{-A}^B \dfrac{1}{t^2 + 1} \intd t = & \lim_{A,B \to \infty} \Big[ \arctan(t) \Big]_{-A}^B \\ = & \lim_{A,B \to \infty} \arctan(B) - \arctan(-A) = \dfrac{\pi}{2} - \left(-\dfrac{\pi}{2}\right) = \pi \end{align*}
2. 1. Define $f : \C \setminus \{i,-i\} \to \C$ by $f(z) = \dfrac{\exp(2iz)}{z^2 + 1}$ which has absolute value satisfying \[ |f(z)| = \dfrac{e^{-2y}}{|z^2 + 1|} \le \dfrac{1}{z^2 + 1} \] 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 $T \to \infty$ just as in the previous question. There is again only one residue inside $\Gamma$ and it is at $i$. Now \[ \Res(f,i) = \dfrac{\exp(2i \cdot i)}{2i} = \dfrac{1}{2 i e^2} \] so Cauchy's residue theorem gives \[ \int\limits_{-T}^T \dfrac{\exp(2it)}{t^2 + 1} \intd t = 2 \pi i \Res(f,i) = \dfrac{\pi}{e^2} \] after taking the limit as $T \to \infty$.
   2. Writing $\exp(2it) = \cos(2t) + i \sin(2t)$ gives \[ \dfrac{\pi}{e^2} = \int\limits_{-\infty}^\infty \dfrac{\exp(2it)}{t^2 + 1} \intd t = \int\limits_{-\infty}^\infty \dfrac{\cos(2t)}{t^2 + 1} \intd t + i \int\limits_{-\infty}^\infty \dfrac{\sin(2t)}{t^2 + 1} \intd t \] 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 \[ \int\limits_{-\infty}^\infty \dfrac{\cos(2t)}{t^2 + 1} \intd t = \dfrac{\pi}{e^2} \qquad \qquad \int\limits_{-\infty}^\infty \dfrac{\sin(2t)}{t^2 + 1} \intd t = 0 \] by taking real and imaginary parts.
   3. We could conclude immediately that the \[ \int\limits_{-\infty}^\infty \dfrac{\sin(2t)}{t^2 + 1} \intd t \] is zero because, we may calculate it as \[ \lim_{T \to \infty} \int\limits_{-T}^T \dfrac{\sin(2t)}{t^2 + 1} \intd t \] 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 $f : \C \setminus \{i,-i\} \to \C$ by $f(z) = \dfrac{\exp(-2iz)}{z^2 + 1}$ and would be stuck with \[ |\exp(-2iz)| = \exp(2y) \] in the numerator. This term grows quickly as $y \to \infty$ 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 $y \l 0$, 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 \[ \int\limits_{-\infty}^\infty \dfrac{\exp(-2it)}{t^2 + 1} \intd t = 2 \pi i \wind(\Gamma,-i) \Res(f,-i) \] because our contour $\Gamma = (\gamma,\eta)$ now contains $-i$ and not $i$, and winds clockwise around $-i$ i.e.\ $\wind(\Gamma,-i) = -1$. So \[ \int\limits_{-\infty}^\infty \dfrac{\exp(-2it)}{t^2 + 1} \intd t = 2 \pi i (-1) \dfrac{\exp(-2i \cdot (-i))}{2(-i)} = \dfrac{\pi}{e^2} \]

4. 1. Define $f : \C \setminus \{ i,-i,\sqrt{3} i, -\sqrt{3} i \} \to \C$ by $f(z) = \dfrac{1}{(z^2+1)(z^2+3)}$ and note that $|z| \ge 6$ implies \[ |f(z)| = \dfrac{1}{|z^2-(-3)| |z^2-(-3)|} \le \dfrac{1}{(|z|^2 - 1)(|z|^2 - 3)} \le \dfrac{12}{|z|^4} \] 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 $T \to \infty$. Indeed, for the latter statement note that \[ \left| \int\limits_0^\pi f(Te^{it}) iTe^{it} \intd t \right| \le 2 \pi \dfrac{12T}{T^4} = \dfrac{24\pi}{T^3} \] which goes to zero as $T \to \infty$. We conclude from Cauchy's residue theorem that \[ \int\limits_{-\infty}^\infty \dfrac{1}{(t^2+1)(t^2+3)} \intd t = 2 \pi i \Res(f,i) + 2 \pi i \Res(f,\sqrt{3} i) \] 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 \[ \left( (t^2 + 1)(t^2 + 3) \right)' \left( t^4 + 4t^2 + 3 \right)' = 4t^3 + 8t \] and plugging in our poles. We get \[ \int\limits_{-\infty}^\infty \dfrac{1}{(t^2+1)(t^2+3)} \intd t = 2 \pi i \dfrac{1}{4(i)^3 + 8i} + 2 \pi i \dfrac{1}{4(\sqrt{3} i)^3 + 8\sqrt{3} i} = \dfrac{\pi}{6}(3 - \sqrt{3}) \]
   2. We can factor the denominator as $28 + 11t^2 + t^4 = (7 + t^2)(4 + t^2)$ and so define $f : \C \setminus \{2i,-2i, \sqrt{7}i,-\sqrt{7}i \} \to \C$ by \[ f(z) = \dfrac{1}{(z^2+7)(z^2+4)} \] which satisfies \[ |f(z)| \le \dfrac{112}{|z|^4} \] giving existence and decay of the semi-circular contour integral just as in the previous part. Cauchy's residue theorem gives \[ \int\limits_{-\infty}^\infty \dfrac{1}{28 + 11t^2 + t^4} \intd t = 2 \pi i \Res(f,2i) + 2 \pi i \Res(f,\sqrt{7} i) \] and the derivative of the denominator is $4t^3 + 22t$ so \begin{align*} \int\limits_{-\infty}^\infty \dfrac{1}{28 + 11t^2 + t^4} \intd t & = 2 \pi i \dfrac{1}{22(2i) + 4(2i)^3} + 2 \pi i \dfrac{1}{22(\sqrt{7} i) + 4(\sqrt{7}i)^3} \\ & = 2 \pi i \dfrac{1}{44i - 32i} + 2 \pi i \dfrac{1}{22\sqrt{7} i - 28\sqrt{7}i} \\ & = \dfrac{\pi}{6} - \dfrac{\pi}{3 \sqrt{7}} = \dfrac{\pi}{6} \left( 1 - \dfrac{2}{\sqrt{7}} \right) \end{align*}
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 \[ |f(z)| = \dfrac{|\exp(iz)|}{|(z+2)^2 + 1|} \le \dfrac{e^{-y}}{|z+2|^2 - 1} \le \dfrac{4}{|z+2|^2} \le \dfrac{4}{(|z| - 2)^2} \le \dfrac{16}{|z|^2} \] 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 $T \to \infty$ from this bound. Next we apply Cauchy's residue theorem. \begin{align*} \int\limits_{-\infty}^\infty \dfrac{\exp(it)}{t^2 + 4t + 5} \intd t & = 2 \pi i \Res(f,i-2) \\ & = 2 \pi i \dfrac{\exp(i(i-2))}{2(i-2) + 4} = \pi \exp(-1 - 2i) = \dfrac{\pi}{e} \left( \cos 2 - i \sin 2 \right) \end{align*} 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 $\intd z = ie^{it} \intd t$ and \begin{align*} \frac{2}{i} \int\limits_\gamma \dfrac{1}{5z^2 + 26z + 5} \intd z & = \int\limits_0^{2\pi} \dfrac{1}{5(e^{it})^2 + 26 e^{it} + 5} ie^{it} \intd t \\ & = \int\limits_0^{2\pi} \dfrac{i}{5e^{it} + 5e^{-it} + 26} \intd t \\ & = \int\limits_0^{2\pi} \dfrac{i}{10 \cos(t) + 26} \intd t = \dfrac{i}{2} \int\limits_0^{2\pi} \dfrac{i}{5 \cos(t) + 13} \intd t \end{align*} as claimed.
   2. If we multiply out \[ (z-(-5))(z-(-1/5)) = (z^2 + \dfrac{26}{5} + 1) = \dfrac{1}{5} (5z^2 + 26z + 5) \] 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 \[ \Res(f,-1/5) = \dfrac{1}{10(-1/5) + 26} = \dfrac{1}{24} \]
   3. The only pole of $f$ within $\gamma$ is $-1/5$. Cauchy's residue theorem tells us that \[ \int\limits_0^{2\pi} \dfrac{1}{13 + 5 \cos(t)} \intd t = \dfrac{2}{i} \int\limits_\gamma f = \dfrac{2}{i} 2 \pi i \Res(f,-1/5) = 4 \pi \dfrac{1}{24} = \dfrac{\pi}{6} \]
7. 1. We have \begin{align*} (\cos t)^3 &= \dfrac{1}{8} (e^{it} + e^{-it})^3 = \dfrac{e^{3it} + 3e^{it} + 3e^{-it} + e^{-3it}}{8} \\ (\cos t)^2 &= \dfrac{1}{4} (e^{it} + e^{-it})^2 = \dfrac{e^{2it} + 2 + e^{-2it}}{4} \end{align*} so our integral becomes \begin{align*} & \dfrac{1}{4} \int\limits_0^{2\pi} e^{3it} + 3e^{2it} + 3e^{it} + 6 + 3e^{-it} + 3e^{-2it} + e^{-3it} \intd t \\ = & \dfrac{1}{4i} \int\limits_0^{2\pi} (e^{2it} + 3e^{it} + 3 + 6e^{-it} + 3e^{-2it} + 3e^{-3it} + e^{-4it}) ie^{it} \intd t \\ = & \dfrac{1}{4i} \int\limits_\gamma z^2 + 3z + 3 + \dfrac{6}{z} + \dfrac{3}{z^2} + \dfrac{3}{z^3} + \dfrac{1}{z^4} \intd z \end{align*} 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 \[ \int\limits_0^{2\pi} 2(\cos t)^3 + 3 (\cos t)^2 \intd t = \dfrac{1}{4i} \cdot 2 \pi i \cdot 6 = 3\pi \]
   2. Here we use \[ (\cos t)^2 = \dfrac{1}{4} (e^{it} + e^{-it})^2 = \dfrac{e^{2it} + 2 + e^{-2it}}{4} \] to write the integrand as \[ \dfrac{4}{e^{2it} + 6 + e^{-2it}} = \dfrac{4e^{2it}}{e^{4it} + 6e^{2it} + 1} = \dfrac{1}{i} \dfrac{4e^{it}}{e^{4it} + 6 e^{2it} + 1} ie^{it} \] so that \[ \int\limits_0^{2\pi} \dfrac{1}{1 + (\cos t)^2} \intd t = \dfrac{1}{i} \int\limits_\gamma \dfrac{4z}{z^4 + 6z^2 + 1} \intd z \] 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 \[ i \sqrt{3 - \sqrt{8}} \quad -i \sqrt{3 - \sqrt{8}} \quad i\sqrt{\sqrt{8} + 3} \quad -i\sqrt{\sqrt{8} + 3} \] and that only the first two are inside $\gamma$. Thus \[ \dfrac{1}{i} \int\limits_\gamma \dfrac{4z}{z^4 + 6z^2 + 1} \intd z = 2 \pi \Res \left( f,i\sqrt{3 - \sqrt{8}} \right) + 2 \pi \Res \left( f,-i\sqrt{3 - \sqrt{8}} \right) \] 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 \[ \Res \left( f,i\sqrt{3 - \sqrt{8}} \right) = \dfrac{1}{\sqrt{8}} \qquad \Res \left( f,-i\sqrt{3 - \sqrt{8}} \right) = \dfrac{1}{\sqrt{8}} \] so \[ \int\limits_0^{2\pi} \dfrac{1}{1 + (\cos t)^2} \intd t = \dfrac{4 \pi}{\sqrt{8}} = \sqrt{2} \pi \]

Summing Series

8. We take $f : \C \setminus \{0\} \to \C$ to be $f(z) = \dfrac{1}{z^4}$. Recall that $g(z) = \dfrac{\sin(\pi z)}{\cos(\pi z)}$ is uniformly bounded on $\Gamma_N$. We can estimate that \[ \left| \int\limits_{\Gamma_N} fg \right| \le \ell(\Gamma_N) \dfrac{M}{N^4} \le \dfrac{16M}{N^3} \] because $|z| \ge N$ for all $z \in \Gamma_N$. Thus the contour integral converges to zero as $N \to \infty$. Cauchy's residue theorem gives \[ \int\limits_{\Gamma_N} fg = 2 \pi i \sum_{n=-N}^N \Res(fg,n) \wind(\Gamma_N,n) \] because the only poles of $fg$ are at $n \in \{-N,\dots,N\}$. As usual \[ \Res(fg,n) = \dfrac{1}{\pi n^4} \] for all non-zero $n$ and it reamins to calculate the residue at zero. The order of the pole at zero is five. From our lemma \[ \Res(fg,0) = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{z^5 \cos(\pi z)}{z^4 \sin(\pi z)} \right)^{(4)} = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{z}{\sin(\pi z)} \cos(\pi z) \right)^{(4)} \] and the power series \[ \dfrac{w}{\sin w} = 1 + \dfrac{w^2}{6} + \dfrac{w^4}{240} + \cdots \] we get \begin{align*} \Res(fg,0) & = \lim_{z \to 0} \dfrac{1}{4!} \left( \left( \dfrac{1}{\pi} + \dfrac{\pi}{6} z^2 + \dfrac{7\pi^3}{360} z^4 - \cdots \right) \left( 1 - \dfrac{\pi^2}{2} z^2 + \dfrac{\pi^4}{24} z^4 - \cdots \right) \right)^{(4)} \\ & = \lim_{z \to 0} \dfrac{1}{4!} \left( \dfrac{1}{\pi} - \dfrac{\pi}{3} z^2 - \dfrac{\pi^3}{45} z^4 + \cdots \right)^{(4)} \\ & = -\dfrac{\pi^3}{45} \end{align*} and we can put everything together. Cauchy's residue theorem gives \[ 2 \pi i \left( 2 \sum_{n=1}^N \dfrac{1}{\pi n^4} - \dfrac{\pi^3}{45} \right) = \int\limits_{\Gamma_N} fg \] and the limit as $N \to \infty$ gives $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$
9. We would want to use $f(z) = 1/z^3$ but with that choice \[ \Res(fg,-n) = -\dfrac{1}{\pi n^3} = - \Res(fg,n) \] for all $n \in \N$ and Cauchy's residue theorem only gives \[ \int\limits_{\Gamma_N} fg = 2 \pi i \Res(fg,0) \] which does not involve $\displaystyle\sum_{n=1}^N \dfrac{1}{n^3}$