Ngô Quốc Anh

July 17, 2009

3 indefinite integral problems involving sinx/x via residue

Problem 1. Compute

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{\sin x}} {x}dx}

via complex variable methods.

Problem 2. Compute

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{\sin^2 x}} {x^2}dx}

via complex variable methods.

Problem 3. Compute

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{\sin^3 x}} {x^3}dx}

via complex variable methods.

Solution to problem 2. To see that the integral exists, notice that

\displaystyle\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}} {{{x^2}}} = 1

and

\displaystyle\frac{{{{\sin }^2}x}} {{{x^2}}} = O\left( {\frac{1} {{{x^2}}}} \right)

as x \to \infty. As the integrand is an even function, we have

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{{{\sin }^2}x}}{{{x^2}}}dx} = 2\int\limits_0^\infty {\frac{{{{\sin }^2}x}}{{{x^2}}}dx}.

Let

\displaystyle f(z) =\frac{1-e^{2iz}}{z^2}.

Then f is analytic except for a simple pole at 0. For 0 < \varepsilon < R, consider the contour

\displaystyle C_{\varepsilon, R} = \Gamma_R \cup \gamma_\varepsilon \cup [-R, -\varepsilon] \cup [\varepsilon , R]

integrationofsinxoverxviaresidueby Cauchy’s Theorem. We have, by the Residue Theorem,

\displaystyle 0 = \int\limits_{{C_{\varepsilon ,R}}} {f\left( z \right)dz} = \int\limits_{{\Gamma _R}} {} + \int\limits_{\left[ { - R, - \varepsilon } \right]} {} + \int\limits_{{\gamma _\varepsilon }} {} + \int\limits_{\left[ {\varepsilon ,R} \right]} {}.

It is easy to see that on \Gamma_R we have

\displaystyle\left| {f\left( z \right)} \right| = \left| {\frac{{1 - {e^{2iz}}}}{{{z^2}}}} \right| \leqslant \frac{2}{{{{\left| z \right|}^2}}}

since

\displaystyle\left| {1 - {e^{2iz}}} \right| \leqslant 1 + \left| {{e^{2iz}}} \right| \leqslant 2.

Therefore,

\displaystyle 0 = \mathop {\lim }\limits_{R \to \infty } \int\limits_{{C_{\varepsilon ,R}}} {f\left( z \right)dz} = \mathop {\lim }\limits_{R \to \infty } \left( {\int\limits_{\left[ { - R, - \varepsilon } \right]} {} + \int\limits_{{\gamma _\varepsilon }} {} + \int\limits_{\left[ {\varepsilon ,R} \right]} {} } \right).

Since 0 is the simple pole, then

\displaystyle\mathop {\lim }\limits_{\varepsilon \to 0} \int\limits_{{\gamma _\varepsilon }} {f\left( z \right)dz} = - \pi i\mathop {\rm Res}\limits_{z = 0} f\left( z \right).

Thus

\displaystyle 0 = \mathop {\lim }\limits_{\varepsilon \to 0} \mathop {\lim }\limits_{R \to \infty } \int\limits_{{C_{\varepsilon ,R}}} {f\left( z \right)dz} = \mathop {\lim }\limits_{\varepsilon \to 0} \left( {\int\limits_{\left( { - \infty , - \varepsilon } \right]} {f(x)dx} + \int\limits_{\left[ {\varepsilon , + \infty } \right)} {f(x)dx} } \right) - \pi i\mathop {\rm Res}\limits_{z = 0} f\left( z \right).

In other word,

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{1 - {e^{2ix}}}}{{{x^2}}}dx} = \pi i\mathop {\rm Res}\limits_{z = 0} f\left( z \right) = \pi i( - 2i) = 2\pi .

Finally, since

\displaystyle {\sin ^2}x = \frac{1} {2}\Re \left( {1 - {e^{2ix}}} \right)

we have

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{{{\sin }^2}x}}{{{x^2}}}dx} = \frac{1}{4}\Re\int\limits_{ - \infty }^\infty {\frac{{1 - {e^{2ix}}}}{{{x^2}}}dx} = \frac{\pi }{2}.

Solution to problem 3. The integral exists since it is absolutely integrable with the same argument as above. Since

\displaystyle {\sin ^3}x = - \frac{1}{{8i}}{\left( {{e^{ix}} - {e^{ - ix}}} \right)^3} = - \frac{1}{{8i}}\left( {{e^{3ix}} - {e^{ - 3ix}} - 3{e^{ix}} + 3{e^{ - ix}}} \right)

then

\displaystyle {\sin ^3}x = \Re\left( {\frac{3} {4}{e^{ix}} - \frac{1} {4}{e^{3ix}}} \right).

The contour is the same to the proof of the problem 2. By Cauchy’s Theorem, one has

\displaystyle 0 = \int\limits_{{C_{\varepsilon ,R}}} {\frac{{3{e^{iz}} - {e^{3iz}}}}{{{z^3}}}dz} = \int\limits_{{\Gamma _R}} {} + \int\limits_{\left[ { - R, - \varepsilon } \right]} {} + \int\limits_{{\gamma _\varepsilon }} {} + \int\limits_{\left[ {\varepsilon ,R} \right]} {} .

The integral over \Gamma_R tends to 0 as R \to \infty. To estimate the integral over \gamma_\varepsilon, we note that

\displaystyle\frac{{3{e^{iz}} - {e^{3iz}}}}{{{z^3}}} = \frac{1}{{{z^3}}}\left( {2 + \frac{{3{{(iz)}^2}}}{2} - \frac{{{{(3iz)}^2}}}{2} + O({z^3})} \right)

which yields

\displaystyle \frac{{3{e^{iz}} - {e^{3iz}}}} {{{z^3}}} = \frac{2} {{{z^3}}} + \frac{3} {z} + O\left( 1 \right), \quad z \to 0.

Hence we cannot apply the above argument as in the proof of Problem 2 to estimate the integral over \gamma_\varepsilon. However, this integral can be computed directly as follows

\displaystyle\int\limits_{{\gamma _\varepsilon }} {\frac{{3{e^{iz}} - {e^{3iz}}}}{{{z^3}}}dz} = \int\limits_\pi ^0 {\left( {\frac{2}{{{\varepsilon ^3}{e^{3i\theta }}}} - \frac{3}{{\varepsilon {e^{i\theta }}}} + O(1)} \right)i\varepsilon {e^{i\theta }}d\theta } = \frac{{2i}}{{{\varepsilon ^2}}}\int\limits_\pi ^0 {{e^{ - 2i\theta }}d\theta } + 3i\int\limits_\pi ^0 {d\theta } + O(\varepsilon ).

Therefore

\displaystyle\int\limits_\pi ^0 {\left( {\frac{2}{{{\varepsilon ^3}{e^{3i\theta }}}} - \frac{3}{{\varepsilon {e^{i\theta }}}} + O(1)} \right)i\varepsilon {e^{i\theta }}d\theta } = 0 - 3\pi i + O(\varepsilon ) \to - 3\pi

as \varepsilon \to 0. And thus,

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{{{\sin }^3}x}} {{{x^3}}}dx} = \frac{{3\pi }} {4}.

Solution to problem 1. The integral clearly exists. The proof is similar to the proof of Problem 2. The answer is

\displaystyle\int\limits_{ - \infty }^\infty {\frac{{\sin x}}{x}dx} = \pi .

3 Comments »

  1. The solution for Problem 2 is wrong. This integral from 0 to infinity is supposed to be pi/2. As an even function, your answer should be pi.

    Comment by Priscila — February 26, 2011 @ 23:58

    • Thanks for commenting. The proof is correct but there is a coefficient-problem. It shoule be

      \displaystyle\int\limits_{ - \infty }^\infty {\frac{{{{\sin }^2}x}}{{{x^2}}}dx} = \frac{1}{2}\int\limits_{ - \infty }^\infty {\frac{{\Re (1 - {e^{2ix}})}}{{{x^2}}}dx} = \frac{1}{2}\Re \int\limits_{ - \infty }^\infty {\frac{{1 - {e^{2ix}}}}{{{x^2}}}dx} = \pi .

      Comment by Ngô Quốc Anh — February 27, 2011 @ 0:07

  2. Hay!

    Comment by TriPhuong — January 25, 2013 @ 18:53


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