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


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}


\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}


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Create a free website or blog at

%d bloggers like this: