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I'm trying to evaluate an integral $\int\limits_{-\infty}^\infty \frac{\sin 2x}{x^3}\,dx$ using Cauchy's theorem. Considering an integral from $-R$ to $-\epsilon$, then a semicircular indentation around $x=0$, then $\epsilon$ to $R$, then a semicircular contour from $R$ to $-R$. Around the pole at $x=0$, the semicircular contribution gives $$\int\limits_\pi^0 \, dz\frac{e^{2iz}}{z^3}=\int_\pi^0 (\epsilon e^{i\theta})(i \, d\theta) \frac{e^{\epsilon e^{i\theta}}}{(\epsilon e^{i\theta})^3}$$ What I need is the limiting value of this integral as $\epsilon\rightarrow 0$. But it seems to diverge.

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  • $\begingroup$ You probably mean to have lower limit $-\infty$. In any case, the integral is divergent. The singularity at $0$ is too strong. $\endgroup$ – mickep May 27 '16 at 19:13
  • $\begingroup$ Do you mean that the principal value of this integral doesn't exist? $\endgroup$ – SRS May 27 '16 at 19:15
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    $\begingroup$ Yes, your function is even so you get no cancellation. $\endgroup$ – mickep May 27 '16 at 19:19
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We have that $\;z=0\;$ is clearly a triple pole of

$$f(z):=\frac{e^{i2z}}{z^3} ,\;\;\text{and in this case it is probable easier to use power series for the residue:}$$

$$\frac{e^{2iz}}{z^3}=\frac1{z^3}\left(1+2iz-\frac{4z^2}{2!}-\ldots\right)=\frac1{z^3}+\frac{2i}{z^2}-\frac2z-\ldots\implies\text{Res}_{z=0}(f)=-2$$

so taking the usual contour with a "bump" around zero, we get

$$0=\lim_{R\to\infty,\,\epsilon\to0}\oint_{\Gamma_R}f(z)=\int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx-\int_{\gamma_\epsilon}f(z)dz= \int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx+2\pi i\implies$$

$$-2\pi i=\int_{-\infty}^\infty\frac{e^{2ix}}{x^3}dx=\int_{-\infty}^\infty\frac{\cos2x+i\sin2x}{x^3}dx\implies \int_{-\infty}^\infty\frac{\sin2x}{x^3}dx=-2\pi$$

the last equality following from comparing real and imaginary parts in both sides. The above though is just CPV (Cauchy's Principal Value) of the integral, since it doesn't converge in the usual sense of the word.

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  • $\begingroup$ How did you replace $-2\pi i$ for $\int_{\gamma_{\epsilon}} f(z)dz$? $\endgroup$ – SRS May 27 '16 at 20:14
  • $\begingroup$ @RoopamSinha Read the lemma, and in particular its corollary, in the most upvoted answer here: math.stackexchange.com/questions/83828/… It explains what happens to that integral when $\;\epsilon\to0\;$ $\endgroup$ – DonAntonio May 27 '16 at 21:21
  • $\begingroup$ @Joanpemo- As I understand, the lemma you referred to works for simple pole. Isn't it? But here we have a pole of order 3. Please rectify if I'm wrong. $\endgroup$ – SRS May 28 '16 at 6:08
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It is not difficult to show (through the Laplace transform, for instance) that $$ \int_{-\infty}^{+\infty}\frac{\sin(2z)-2z}{z^3}\,dz = -2\pi \tag{1}$$ so the principal value of your integral is given by $-2\pi$ plus twice the principal value of $$ \int_{-\infty}^{+\infty}\frac{dz}{z^2} \tag{2} $$ that is just $+\infty$, since the integrand function is even and it has a double pole at the origin.

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$$ \frac{\sin (2x)}{x^3} = \left( \vphantom{\frac2{x^2}} \right.\underbrace{\frac{\sin(2x)}{2x}}_{\begin{smallmatrix} \text{This} \\ \text{approaches} \\[2pt] \text{$1$ as $x\to0$} \\[2pt] {} \end{smallmatrix}} \cdot \left. \frac 2 {x^2} \right) $$

For $x$ close enough to $0$ you have $\dfrac{\sin(2x)}{2x}> 0.9$ and then $$ \int_{-a}^a \frac{\sin(2x)}{2x} \cdot \frac 2 {x^2} \, dx \ge 0.9 \int_{-a}^a \frac{2\, dx}{x^2} = +\infty. $$

And $$ \int_a^\infty \left|\frac{\sin(2x)}{x^3}\right| \, dx \le \int_a^\infty \frac{dx}{x^3} <\infty $$ so here we have absolute convergence. (And $x\mapsto \dfrac{\sin(2x)}{x^3}$ is an even function, so that $\text{takes care of } \displaystyle \int_{-\infty}^{-a} \frac{\sin(2x)}{x^3} \, dx$.)

So you have $+\infty$ plus something that converges absolutely, and you conclude that the whole thing diverges to $+\infty$.

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ We can 'jump' directly into the $\mathrm{P.V.}$ definition:

\begin{align} &\color{#f00}{% \mathrm{P.V.}\int_{-\infty}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x}\ \stackrel{\mbox{def.}}{=}\ \lim_{\epsilon \to 0^{+}}\bracks{% \int_{-\infty}^{-\epsilon}{\sin\pars{2x} \over x^{3}}\,\dd x + \int_{\epsilon}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x} \\[3mm] = &\ 2\lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{\infty}{\sin\pars{2x} \over x^{3}}\,\dd x = 2\lim_{\epsilon \to 0^{+}}\bracks{% {\sin\pars{2\epsilon} \over 2\epsilon^{2}} + \int_{\epsilon}^{\infty}{\cos\pars{2x} \over x^{2}}\,\dd x} \\[3mm] & = 2\lim_{\epsilon \to 0^{+}}\bracks{% {\sin\pars{2\epsilon} \over 2\epsilon^{2}} + {\cos\pars{2\epsilon} \over \epsilon} - 2\int_{\epsilon}^{\infty}{\sin\pars{2x} \over x}\,\dd x} = \color{#f00}{+\infty} \end{align}

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