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)

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    $\begingroup$ $|e^{ix}|=1$ only if $x$ is real. In this case $x=tz$ is clearly not real. $\endgroup$ Jul 3, 2021 at 23:51
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    $\begingroup$ @Leon I wrote an answer, which looks at the general case you will encounter far more often. I hope it has helped you :) $\endgroup$
    – vitamin d
    Jul 5, 2021 at 19:16
  • $\begingroup$ Yea it has. I do find it more intuitive to estimate step by step, like we have been doing here, but in future that lemma could shorten the whole process. I'll take it into consideration, thank you! $\endgroup$
    – Leon
    Jul 5, 2021 at 19:34
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    $\begingroup$ @Leon Note that I have also linked a proof to Jordans lemma, which does it step by step, exactly like desired - but more general. $\endgroup$
    – vitamin d
    Jul 5, 2021 at 20:30

2 Answers 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.


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 :

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



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

(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.

  • $\begingroup$ It almost comes together. The only thing I'm missing is why $|e^{itRcos\theta-Rtsin\theta}|$ equals to $e^{-Rtsin\theta}$ ? $\endgroup$
    – Leon
    Jul 4, 2021 at 10:51
  • $\begingroup$ because $|zw|=|z||w|$ for $z,w$ complex, $|e^{ia}|=1$ for $a$ real, and $|a|=a$ if $a$ real and positive. Does this answer your doubt? $\endgroup$
    – Thomas
    Jul 4, 2021 at 10:54
  • $\begingroup$ Sorry I'm still not getting it: abosolute value of $|e^{i\,t\,R\,\cos(\theta)-t\,R\,\sin(\theta)}|$ means to me: $\sqrt{\left(\cos(-t\,R\,\sin(\theta))\right)^2+\left(\sin(t\,R\,\cos(\theta))\right)^2}$ $\endgroup$
    – Leon
    Jul 4, 2021 at 11:00
  • $\begingroup$ Oh hang on now I see, the $i$ is not multiplied on both exponents so you just split the exponential and take absolute value! Do you? $\endgroup$
    – Leon
    Jul 4, 2021 at 11:04
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    $\begingroup$ Yes, thank you! My doubts vanished $\endgroup$
    – Leon
    Jul 4, 2021 at 11:10

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