Estimate improper integral: $\displaystyle{\int_{\gamma}\dfrac{e^{i\,t\,z}}{z^2+1}}$

Question

Here $\gamma$ is a semicircle with parametrisation $\gamma(t) = Re^{i\varphi}\quad \varphi\in[0,\pi]$. Also $R \to \infty$. That's why improper. I started my estimation as follows

$$\begin{align} &\left|\int_{\gamma}\dfrac{e^{itz}}{z^2+1}\,\mathrm{dz}\right| \\\\\leq &\int_{\gamma} \dfrac{\vert e^{itz}\vert}{\vert z^2+1\vert}\,\vert\mathrm{dz}\vert \\\\ \leq&\int_{\gamma}\dfrac{\vert e^{itz}\vert}{ R^2+1}\,\vert\mathrm{dz}\vert \end{align}$$

Here I'm not sure how to advance. Taking the absolute value of:

$\vert e^{itz}\vert = \vert \cos(tz)+i\sin(tz)\vert = 1?$

However Wiki seems to move in a different direction. It seems like they plugged in the parametrisation of $\gamma$ first:

$\gamma = R\,e^{i\varphi t} = \vert z \vert \,e^{i\varphi t} = \vert z \vert\, (\cos(\varphi)+i\sin(\varphi))$ $\quad\Rightarrow \quad$ $\vert e^{itz}\vert =\vert e^{it\vert z\vert(\cos(\varphi z)+i\sin(\varphi z))}\vert = e^{-t\vert z \vert \sin(\varphi)} < 1$

I'm really not not sure how they came up with the last equality ($ = e^{-t\vert z \vert \,\sin(\varphi)}$) even though it seems right because it explains why $t$ has to be only positive for the whole integral to converge. Once you accepted that move the rest should be clear:

$$\begin{align} \leq&\int_{\gamma}\dfrac{1}{ R^2+1}\,\vert\mathrm{dz}\vert \\\\ =&\dfrac{1}{ R^2+1}\,\int_{\gamma}\vert\mathrm{dz}\vert \\\\ =&\dfrac{1}{ R^2+1}\,\pi\,R \end{align}$$

Taking the limit now yields: $\displaystyle{\lim_{R\to\infty}\dfrac{1}{ R^2+1}\,\pi\,R} = 0$.I just don't understand the middle part in between (and weather it indeed explains why $t$ should be positive)

Answer 1

After parametrization, and using that:

$Re^{i\theta}=R(cos\theta+isin\theta)$ ,

( this identity should also solve one of your doubts), one has:

$I=\int_o^{\pi} iRe^{i\theta}e^{itRcos\theta-Rtsin\theta}/(R^2e^{i2\theta}+1)d\theta$

This means that :

$|I|\le\int_0^{\pi} |iRe^{i\theta}||e^{itRcos\theta-Rsin\theta}|/|(R^2e^{i2\theta}+1)|$

Now:

$|iRe^{i\theta}|=R$

$|e^{itRcos\theta-Rtsin\theta}|=|e^{itRcos\theta}e^{-Rtsin\theta}|=|e^{itRcos\theta}||e^{-Rtsin\theta}|=e^{-Rtsin\theta} \le 1$

(note e.g. that the sine is positive in $(0,\pi)$ and $e^{-x}\le 1$ for positive $x$), and

$1/|(R^2e^{i2\theta}+1)|\le 1/(R^2-1)$ (comes from triangular inequality note that we have equality here with $\theta=\pi/2$)

This leads to the estimate (at this point, valid also for finite $R$):

$0\le|I|\le \pi \frac{R}{R^2-1}$

Now for large $R$ this implies that $|I| \rightarrow 0$, i.e. the integral on this semicircle goes to 0.

Answer 2

This is a special case of Jordans lemma:

Consider a complex-valued, continuous function $f$, defined on a semicircular contour $C_{R}=\{Re^{i\theta }\mid \theta \in [0,\pi ]\}$ of positive radius $R$ lying in the upper half-plane, centered at the origin. If the function $f$ is of the form $f(z)=e^{iaz}g(z),\: z\in C_{R}$ with a positive parameter $a$, then Jordan's lemma states the following upper bound for the contour integral $$\left|\int _{C_{R}}f(z)\,\mathrm dz\right|\leq {\frac {\pi }{a}}\max _{\theta \in [0,\pi ]}\left|g\left(Re^{i\theta }\right)\right|$$ with equality when $g$ vanishes everywhere, in which case both sides are identically zero.

Note that in your problem $f(z)=\frac{e^{itz}}{z^2+1}$ and therefore $g(z)=\frac{1}{z^2+1}$. Here $g$ vanishes everywhere i.e. $\lim\limits_{R\to\infty}\max\limits_{\theta \in [0,\pi ]}\left|g\left(Re^{i\theta }\right)\right|=0$ so the integral is also equal to zero.

A proof for Jordans lemma can be found here.