Given $r \gt 0$, consider $\gamma_r (t) = re^{it}, 0\le t \le \pi$. Show that $$\lim_{r \to \infty} \int_{\gamma_r} \frac{e^{iz}}{z}dz = 0$$.
For Cauchy's formula we have
$\int_{\gamma_r} \frac{e^{iz}}{z}dz \to f'(0) = \frac{0!}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{(z-0)^1} dz \to 1 = \frac{1}{2 \pi i} \int_{\gamma} \frac{e^{iz}}{z^2} dz \to \int_{\gamma} \frac{e^{iz}}{z^2} dz = 2 \pi i$
So,
$\lim_{r\to \infty} 2 \pi i = 0$? how ?
can I use Lagrange in this case?
Thanks.
There are maybe two ways to see the error in the computation of the integral and/or in its limit. As the comments of Martin R are showing, the limit was calculated in the linked MSE question 4489103 - it is important to note that $\gamma(r):=\gamma_r$ is not a closed path, instead, it is a path starting in $r=re^{i\cdot 0}\in\Bbb R_{>0}$, following then a half-circle in trigonometric direction to $-r=re^{i\cdot \pi}\in\Bbb R_{<0}$.
$(1)$ There is no way to apply Cauchy's formula as in the question, instead, one can estimate the integral $J(r)$ under the limit for big values of $r$ as follows: $$ \begin{aligned} J(r)&:= \oint_{\gamma(r)}\frac{e^{iz}}z\; dz \\ &=\int_0^\pi\frac{e^{ir\exp(it)}}{r\exp(it)}\; ir\exp(it)\; dt =i\int_0^\pi e^{ir\;e^{it}}\; dt =i\int_0^\pi e^{ir\;(\cos t+i\sin t)}\; dt \\ |J(r)| &=\left|i\int_0^\pi e^{ir\;(\cos t+i\sin t)}\; dt\right| \\ &\le \int_0^\pi \left|e^{ir\;(\cos t+i\sin t)}\; dt\right| = \int_0^\pi e^{-r\sin t}\; dt = 2\int_0^{\pi/2} e^{-r\sin t}\; dt \\ &\to 0\text{ for $r\to\infty$ by dominated convergence.} \end{aligned} $$
$(2)$ Alternatively, let $\epsilon>0$ be some "small number", and let us close the contour by adding the path $$\nu=\nu(\epsilon,r)$$ from $-r$ to $r$ with image to the two segments from $-r$ to $-\epsilon$, and from $+\epsilon$ to $+r$, together with the half-circle from $-\epsilon$ to $\epsilon$, drawn in anti-trigonometric sense in the upper half plane. Then let us compute alternatively the integral $K(\epsilon,r)$ of the same function on this alternative contour. We split it in three parts, $K_1$, $K_2$, $K_3$, on corresponding pieces $\nu_1$, $\nu_2,\nu_3$, the first two on the segments, the third on the small arc around zero. $$ \begin{aligned} K_1(\epsilon,r)&:= \oint_{\nu_1(\epsilon,r)}\frac{e^{iz}}z\; dz =\int_{-r}^{-\epsilon} \frac{e^{iu}}u\; du =-\int_\epsilon^r \frac{e^{it}}t\; dt\ ,\qquad\text{ with }t:=-u \\ K_2(\epsilon,r) &:= \oint_{\nu_2(\epsilon,r)}\frac{e^{iz}}z\; dz :=\int_\epsilon^r \frac{e^{it}}t\; dt\ . \\ K_3(\epsilon) &:= \oint_{\nu_3(\epsilon)}\frac{e^{iz}}z\; dz :=i\int_{\pi}^0e^{i\epsilon(\cos t + i\sin t)}\; dt\ , \qquad\text{ with }z:=\epsilon(\cos t+i\sin t)\ . \\ &\qquad\text{ Then:} \\ K_1(\epsilon,r) + K_2(\epsilon,r) & =\int_\epsilon^r \frac{e^{it}-e^{-it}}t\; dt =2i\int_\epsilon^r \frac{\sin t}t\; dt \\ &\to 2i\int_0^\infty\frac{\sin t}t\; dt=2i\cdot\frac \pi2=\color{blue}{i\pi}\ . \\ K_3(\epsilon) &=-i\int_0^\pi e^{i\epsilon(\cos t + i\sin t)}\; dt \\ &\to -i\int_0^\pi e^{i0\cdot(\cos t + i\sin t)}\; dt =\color{blue}{-i\pi} \text{ for $\epsilon\to0$ by dominated convergence.} \end{aligned} $$ So the sum $K_1+K_2+K_3$ converges to zero.
Note: This second way to go is the possible computation of the integral $\int_0^\infty\frac{\sin t}{t}\; dt=\frac\pi2$ using complex analysis. Above we have used this value, so to get the alternative argument we may need to give an other computational idea. One other way is to consider the deformation $$ I(s):=\int_0^\infty e^{-st}\sin t\;\frac{dt}t\ , $$ which is a continuous function in $s\in[0,\infty)$, differentiable in $(0,\infty)$, with limit zero for $s\to\infty$, and with differential $I'(s)=-\int_0^\infty e^{-st}\sin t\;dt=-\frac 1{1+s^2}$.
Let $\epsilon \gt 0$ then $\vert \int_{\gamma} \frac{e^{iz}}{z} – 0 \vert$ = $\vert \int_{\gamma} \frac{e^{iz}}{z} \vert = \vert \int_{0}^{\pi} \frac{e^{ire^{it}}}{re^{it}} rie^{it} dt\vert$ = $\vert i \int_0^{\pi} e^{ir^{it}}dt$ $\vert \le \int_0^{\pi} \vert e^{ir^{it}} \vert dt$.
but
$\vert e^z \vert = e^{Re(ire^it)}$
so
$\vert e^{ire^{it}}\vert = e^{Re(ire^it)}=e^{Re(ircos(t)+irisin(t))}= e^{Re(ircos(t)-rsin(t))}=e^{-rsin(t)}$.
Hence.
$\int_0^{\pi} \vert e^{ire^{it}}\vert dt = \int_0^{\pi} e^{-rsin(t)} dt = \int_0^{\delta_1} e^{-rsin(t)} dt + \int_{\delta_1}^{\pi - \delta_2} e^{-rsin(t)} dt +\int_{\pi - \delta_2}^{\pi} e^{-rsin(t)} dt. $
For $\delta_1 \gt 0$ and $\delta_2 \gt 0$, and since $e^{-rsin(t)}$ is continuous, só $\delta_1 \gt 0$ can be chosen such that. $\int_0^{\delta_1}e^{-rsin(t)} dt \lt \frac{\epsilon}{3}, \forall r \gt 0$.
and $\delta_2 \gt 0$ can be chosen such that $\int_{\pi -\delta_2}^{\pi}e^{-rsin(t)} dt \lt \frac{\epsilon}{3}, \forall r \gt 0$.
and because of the Lebesgue’s dominated convergence Theorem we have.
$lim_{r \to \infty} \int_{\delta_1}^{\pi-\delta_2} e^{-rsin(t)}dt = \int_{\delta_1}^{\pi-\delta_2} lim_{r \to \infty} e^{-rsin(t)}dt = lim_{r \to \infty} \int_{\delta_1}^{\pi-\delta_2} e^{-rsin(t)}dt \lt \frac{\epsilon}{3} $.
Therefore.
$$lim_{r \to \infty} \int_{\gamma_r} \frac{e^{iz}}{z} dz = 0.$$
