In class, we saw a surprising way to show that
$$\int_{-\infty}^{\infty} \frac{1}{x^2+1} dx = \pi.$$
In fact, there is a very powerful and general technique at work here that allows us to evaluate quite a lot of definite integrals - often, when there isn't even a closed form expression for the anti-derivative.
Let's see similar solutions for problems 16, 17, and 18 from chapter 5. Note that problem 17 outlines the technique quite specifically.
I solved 5.16
$$\int_{\infty}^{\infty} \frac{dz}{z^4 + 1}$$ can be solved by considering first the integral about a semicircle of radius $R$ centered at $0$ where $R > 1$ and the path is counterclockwise starting at $R$. Thus we have
$$\int_{S_r(0)} \frac{dz}{z^4 + 1} = \int_{\gamma}\frac{dz}{z^4 + 1} + \int_{\infty}^{\infty} \frac{dz}{z^4 + 1} $$
Here $\gamma$ denotes the semicircular arc from $R$ and ending at $-R$. We can attain a bound on it by using the following inequality. We have
$$\int_{\gamma} \frac{dz}{z^4 + 1} \leq max_{z \in \gamma} | {\frac{1}{z^4 + 1}}|\pi R $$ We know that $$max_{z \in \gamma} | {\frac{1}{z^4 + 1}} | \leq \frac{1}{min (|z^4 +1|)} \leq \frac{1}{min(|z|^4 - 1)}= \frac{1}{R^4 - 1}$$ therefore we have $$\int_{\gamma} \frac{dz}{z^4 + 1} \leq \frac{\pi R}{R^4 - 1}$$ which tends to $0$ as $R$ tends to infinity. Thus we know that the contribution of this integral to the integral about the semicircle tends to $0$. Thus, as $R$ gets large we have
$$\int_{S_r(0)} \frac{dz}{z^4 + 1} = \int_{\infty}^{\infty} \frac{dz}{z^4 + 1}$$
The left hand side can be solved as
$$\int_{S_r(0)} \frac{dz}{z^4 + 1}dz = \int_{S_r(0)} \frac{\frac{1}{(z-e^{3/4 \pi i})((z-e^{5/4 \pi i})(z-e^{7/4 \pi i})}}{(z-e^{\pi i / 4})}dz + \int_{S_r(0)} \frac{\frac{1}{(z-e^{ \pi i / 4})((z-e^{5/4 \pi i})(z-e^{7/4 \pi i})}}{(z-e^{ 3 \pi i / 4})}dz$$
Now, applying the Cauchy integral formula
$$ \frac{2 \pi i}{\sqrt{2}(2+2i)}+ \frac{2 \pi i}{\sqrt{2}(-2 + 2i)} = \frac{\pi}{\sqrt{2}}$$
So we have, $$\int_{- \infty}^{\infty} \frac{1}{z^4 + 1} dz = \frac{\pi}{\sqrt{2}}$$
5.17
In this problem $f(z)=e^{iz}*\frac{1}{z^2+1}$ and $R>1$.
a.) We can show that $\int_{\sigma_R}e^{iz}*\frac{1}{z^2+1}=\frac{\pi}{e}$ by integrating over $\gamma_R$ such that
$$\int_{\gamma_R}e^{iz}*\frac{1}{z^2+1}
=\int_{\gamma_R}e^{iz}*\frac{1}{(z+i)(z-i)}
=\int_{\gamma_R}e^{iz}*\frac{1/(z+i)}{z-i}.$$ If we apply Cauchy's Integral Formula and integrate over a circle of radius R centered at i, then
$$=\int_{\gamma_R}e^{iz}*\frac{1/(z+i)}{z-i}=2{\pi}if(i)=2{\pi}i{\frac{e^{i^2}}{i+i}}
=2{\pi}i{\frac{e^{-1}}{2i}}=\frac{\pi}{e}.$$
b.) Now we need to show that $|e^{iz}|\leq1$. If $z=x+iy$ then $$|e^{iz}|=|e^{i*(x+iy)}|=|e^{ix-y}|=|e^{-y}e^{ix}|=|e^{-y}(cos(x)+isin(x))|=e^{-y}.$$ Since $e^{-y}\leq1$ for $y\geq0$ and $z=x+iy$, $|e^{iz}|\leq{1}$ in the upper half plane.
We also need to show that $|f(z)|\leq\frac{2}{|z|^2}.$
Since $|e^{iz}|\leq{1},$ $$|e^{iz}*\frac{1}{z^2+1}|\leq\frac{1}{z^2+1}.$$ It would suffice to show that $\frac{1}{z^2+1}\leq\frac{2}{|z|^2}$ to show that $|f(z)|\leq\frac{2}{|z|^2}.$
Notice that $$\frac{1}{z^2+1}\leq\frac{1}{z^2}\leq\frac{1}{|z|^2},$$ by the triangle inequality.
If we take a look at $\frac{1}{z^2+1}\leq\frac{1}{|z|^2},$ for a sufficiently large $|z|$, $\frac{1}{z^2+1}\leq\frac{2}{|z|^2}.$
Since $|f(z)|\leq\frac{1}{z^2+1}\leq\frac{2}{|z|^2},$
$|f(z)|\leq\frac{2}{|z|^2}.$