Understanding an example about Cauchy's integral formula

Question

I have two questions about the following example taken from Palka's "An introduction to complex function theory". I highlighted with red the parts that I don't understand.

1. Why does the first equality in 5.11 hold?
2. Since $r \to \infty$, won't we have that eventually the disk $\Delta$ won't encompass $| \gamma |$? (where $|\gamma|$ is the curve parametrized by $\gamma$). I think we cant't make $\Delta$ bigger because it would contain $-i$ and $f$ must be analytic in $\Delta$ to apply the local Cauchy's integral formula.

Answer 1

$$\int_0^r\frac{\cos t}{1+t^2}\,\mathrm{d}t=\frac12\int_{-r}^r\frac{\cos t}{1+t^2}\,\mathrm{d}t$$ because the integrand is an even function. $$\int_0^r\frac{\sin t}{1+t^2}\,\mathrm{d}t=0$$ because the integrand is an odd function.

I'm not sure I understand what your second question is. Are you thinking that the center of the disk remains fixed as $r$ increases? This isn't so. No matter how big $r$ gets we can construct a disk that includes the segment $[-r, r]$ and doesn't contain $-i$.

EDIT

More specifically, we can construct a circle passing through the points $-r, r, \frac{-i}2$. (The center will lie on the positive imaginary axis.) Now if we increase the radius by some $0<\varepsilon<\frac12$ the circle will look like the one in Figure $14$. As $r$ increases, the center of the circle we construct will travel "north" on the imaginary axis.

Answer 2

Regarding your first question:

$$\int_0^r \frac{\cos t \, dt}{t^2 + 1} = \int_{-r}^0 \frac{\cos t \, dt}{t^2 + 1}$$ because the integrand is even. Adding the two sides together and dividing by $2$ will equal the left-hand side.

Meanwhile, $$\int_{-r}^r \frac{\sin t \, dt}{t^2 + 1} = 0$$ because the integrand is odd.

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Regarding your second question: the application of Cauchy's integral formula holds as long as $\Delta$ is an open set containing $\gamma$ but not $-i$. You are right that as $r \to \infty$, the circle described in the proof will not work, but you can replace it with something else, like $\{z : \Im(z) > -1/2\}$.