# Ngô Quốc Anh

## July 30, 2009

### A couple of complex integrals involving exp(itx) for a real parameter t

In this turn, I will consider a couple of examples of complex contour integrals with respect to variable $x$ involving the following factor $e^{itx}$ where $t$ a real parameter.

Problem 1. Evaluate the integral

$\displaystyle I\left( t \right) = \int\limits_{ - \infty }^\infty {\frac{{{e^{itx}}}} {{{{\left( {x + i} \right)}^2}}}dx}$

where $-\infty < t<\infty$.

Solution. Let

$\displaystyle {f_t}(z) = \frac{{{e^{itz}}}}{{{{(z + i)}^2}}}$

and consider first the case $t>0$. Then $|f_t(z)|$ is bounded in the upper half-plane by

$\displaystyle\frac{1}{|z+i|^2}$.

For $R>1$ let

$\displaystyle C_R=\Gamma_R \cup [-R, R]$,

where $\Gamma_R$ is the semicircle centered at the origin joining $R$ and $-R$, oriented counterclockwise.

Then function $f_t$ is holomorphic on $C_R$ and its interior, so, by Cauchy’s Theorem, we have

$\displaystyle 0 = \int\limits_{{C_R}} {{f_t}(z)dz} = \int_{ - R}^R {{f_t}(z)dz} + \int_{{\Gamma _R}} {{f_t}(z)dz}$.

The absolute value of the second summation on the right is at most

$\displaystyle\frac{\pi R}{R^2}$

since

$\displaystyle |f_t(z)| \leq \frac{1}{R^2}$

on $\Gamma_R$. Taking the limit as $R \to \infty$ we obtain $I(t)=0$.

Suppose now $t<0$. Then $|f_t|$ is bounded in the lower half-plane by $\frac{1}{|z+i|^2}$. Let $C'_R$ be the reflection of $C_R$ with respect to the real axis, oriented clockwise.

By Residue Theorem, we have

$\displaystyle\int\limits_{{C_R'}} {{f_t}\left( z \right)dz} = - 2\pi i\mathop {\rm Res}\limits_{z = - i} {f_t}\left( z \right)$.

A calculation shows that the residue equals to $ite^t$, so

$\displaystyle\int\limits_{C'_R} {{f_t}(z)dz} = 2\pi t{e^t}$.

As $R \to \infty$, the contribution to the last integral from the semicircle tends to 0 since $|f_t| \leq (R-1)^2$ on the semicircle, giving

$\displaystyle I\left( t \right) = 2\pi t{e^t}$.

For $t=0$, the integral is elementary as shown below

$\displaystyle\int_{ - R}^R {{f_t}\left( x \right)dx} = \int_{ - R}^R {\frac{1}{{{{\left( {x + i} \right)}^2}}}dx} = \frac{{ - 1}}{{x + i}}\bigg|_{ - R}^R \to 0$

as $R \to \infty$. Thus $I(0)=0$.

Problem 2. Evaluate the integral

$\displaystyle F(t)=\int_{ - \infty }^\infty {\frac{{{e^{ - itx}}}}{{{{(x + i)}^3}}}dx}$

where $-\infty < t<\infty$.

Solution. For $t=0$, the integral is elementary and finally we get $F(0)=0$. For $t<0$, the function

$\displaystyle {f_t}(z) = \frac{{{e^{ - itz}}}}{{{{(z + i)}^3}}}$

is bounded in the upper half-plan and in fact is $O(|z|^{-3})$ there. In this case, one has $F(t)=0$ for all $t<0$. For $t>0$, by Residue Theorem,

$\displaystyle\int\limits_{{{C'}_R}} {{f_t}(z)dz} = - 2\pi i\mathop {\rm Res}\limits_{z = - i} {f_t}(z) = \underbrace { - 2\pi i\frac{1}{{2!}}\frac{{{d^2}}}{{d{z^2}}}({e^{ - itz}}){\bigg|_{z = - i}}}_{\pi i{t^2}{e^{ - t}}}$

Finally, we obtain

$\displaystyle F\left( t \right) = \pi i{t^2}{e^{ - t}}$

for $t>0$.

Similar question. Estimate the limit

$\displaystyle G(t)=\mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {{{\left( {\frac{{\sin x}} {x}} \right)}^2}{e^{itx}}dx}$

for a real number $t$.

Hint. For $t=0$, the limit equals $\pi$ which I had considered here. For $t \ne 0$, we need to calculate the following residue

$\displaystyle\mathop {\rm Res}\limits_{z = 0} \left[ {\frac{{1 - {e^{2iz}}}} {{{z^2}}}{e^{itz}}} \right]$

which equals $2t+2$ by using the following helpful formula

$\displaystyle\mathop {\rm Res}\limits_{z = {z_0}} \frac{{\varphi \left( z \right)}}{{\phi \left( z \right)}} = {\text{ }}\frac{{\varphi ({z_0})}}{{\phi '\left( {{z_0}} \right)}}$

where $\varphi$ and $\phi$ are holomorphic and $z_0$ is a simple pole.

The final answer is

$\displaystyle G(t) = \begin{cases} 0,&t \leqslant - 2, \hfill \\ \pi + \frac{\pi }{2}t,& - 2 \leqslant t \leqslant 0, \hfill \\ \pi - \frac{\pi }{2}t,&0 \leqslant t \leqslant 2, \hfill \\ 0,&2 \leqslant t. \hfill \\ \end{cases}$