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I have the following integral $$ I = \int_{-\infty}^{+\infty}\frac{\sin x}{(x-\pi)^2(x+\pi/2)}dx\text{,} $$ that I am supposed to solve with the theorem of residues.

I see that there are two simple poles in $\pi$ (given that $\pi$ is a simple zero of the numerator and a double zero of the denominator) and in $-\pi/2$.

Having only simple poles, this should mean that the integral exists in the sense of Cauchy principal value.

When I take the usual step of considering the analytic function $$ f(z)=\frac{e^{iz}}{(z-\pi)^2(z+\pi/2)}\text{.} $$ I see that the simple pole $\pi$ becomes a double pole, because it is not anymore a zero of the numerator.

Can I still proceed with $$ I = \operatorname{Im}\big[\pi i\big(R[\pi]+R[-\pi/2]\big)\big] \text{,} $$ where $R[z_0]$ is the residue of $f$ in $z_0$, or there is some other consideration to take into account?

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  • $\begingroup$ To implement a contour integration program, you should make a closed contour - for example, adding a big half-circle in the upper half-plane and two small half-circles above $z=\pi$ and $z=-\pi/2$. Together with the initial integral you get a closed contour $C$ with no poles inside, so $\oint_C=0$. Integral along a big half-circle $\to0$ as the radius $\to\infty$ (Jordan's lemma), so you will be left with the integrals along small half-circles. $\endgroup$
    – Svyatoslav
    Jun 25 '21 at 10:08
  • $\begingroup$ @Svyatoslav, yes, I know, all these considerations are left implicit. Also, there should be theorems to justify that on the big semicircle, when $R\to\infty$ the integral is $0$. The same on the small semicircle around $-\pi/2$, when $\varepsilon\to0$. My problem is the double pole. $\endgroup$
    – enzotib
    Jun 25 '21 at 10:13
  • $\begingroup$ You may use the approach implemented here: math.stackexchange.com/questions/4062304/… - to present $\exp(re^{it})$ as the series for small $r$, then lead $r\to0$. I may suppose that in your case the imaginary part of integral ($\sin z=\Im e^{iz}$) will give a finite answer (so the initial integral in the PV sense is finite), but the real part (with $\cos z$) - diverges. If you face difficulties to do the evaluation - I could help $\endgroup$
    – Svyatoslav
    Jun 25 '21 at 10:26
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Can I still proceed with $$I = \operatorname{Im}\big[\pi i\big(R[\pi]+R[-\pi/2]\big)\big] \text{,}$$ where $R[z_0]$ is the residue of $f$ in $z_0$, or there is some other consideration to take into account?

"No" for the pole at $z_0=\pi$, because $\displaystyle\lim_{r\to0^+}\int_{C_r}f(z)\,\mathrm{d}z$ diverges for the semi-circular arc $C_r$ centered at $z_0=\pi$ and radius $r$.


For the integrand $$f(z)=\frac{e^{iz}}{(z-\pi)^2(z+\pi/2)}$$ The Laurent series of $f(z)$ near $z_0=\pi$ is $$f(z)=\frac{b_2}{(z-z_0)^2}+\frac{b_1}{z-z_0}+\mathcal{O}(1)$$ For the upper semi-circular arc $C_r$, let's use parameterization $z=z_0+re^{i\theta}$ with $0\leq\theta\leq\pi$, $$\begin{aligned} \lim_{r\to0^+}\int_{C_r}f(z)\,\mathrm{d}z &=\lim_{r\to0^+}\int_0^\pi f(z_0+re^{i\theta})rie^{i\theta}\,\mathrm{d}\theta \\ &=\lim_{r\to0^+}\int_0^\pi\left[\frac{b_2}{r^2e^{2i\theta}}+\frac{b_1}{re^{i\theta}}+\mathcal{O}(1)\right]rie^{i\theta}\,\mathrm{d}\theta \\ &=\lim_{r\to0^+}\int_0^\pi\left[\frac{b_2}{re^{i\theta}}+ib_1+\mathcal{O}(r)\right]\,\mathrm{d}\theta \\ &=\lim_{r\to0^+}\frac{-2ib_2}{r}+\pi i b_1 \end{aligned}$$

As $r\to0^+$, $\mathcal{O}(r)\to0$ but $1/r\to\infty$, hence it diverges. The same can be shown using the lower semi-circular arc with $\pi\leq\theta\leq2\pi$.

Generally, the even power terms with $n\geq2$ (such as $z^{-2}$, $z^{-4}$, $\cdots$, etc.) in the principal part of the Laurent series diverges. On the other hand, the odd power terms with $n\geq3$ (such as $z^{-3}$, $z^{-5}$, $\cdots$, etc.) all vanishes. The coefficient of $z^{-1}$ term is the residue.


@Svyatoslav's method in the comments can work because the Laurent series of the complex integrand in that post has only $z^{-3}$ and $z^{-1}$ terms.

We may try the regularization method. One possible way is to "shift" the poles up from real line. The integrand is rewritten as $$f(z)=\frac{e^{iz}}{(z-\pi-i\epsilon)^2(z+\pi/2-i\epsilon)}$$ Then proceed with residue calculus as usual, and take the limit $\epsilon\to0^+$ at the end. The problem of regularization is that there is no unique way to "shift" the poles, and each gives different results.

Example

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  • $\begingroup$ Thank you for the detailed answer, I will look into it carefully $\endgroup$
    – enzotib
    Jun 26 '21 at 14:10
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If we decompose in simple fractions the rational part of the function under the integral sign $$ \frac{1}{(x-\pi)^2(x+\pi/2)}=\frac{4}{9\pi^2}\cdot\frac{1}{x+\pi/2}-\frac{4}{9\pi^2}\cdot\frac{1}{x-\pi}+\frac{2}{3\pi}\cdot\frac{1}{(x-\pi)^2} $$ and taking into account that we have $$ \text{p.v.}\int_{-\infty}^{+\infty}\frac{\sin x}{(x-\pi)^2}dx= {}-\text{p.v.}\int_{-\infty}^{+\infty}\frac{\sin t}{t^2}dt=0 $$ we see that we can only consider the first two terms of the decomposition $$ \frac{4}{9\pi^2}\cdot\frac{1}{x+\pi/2}-\frac{4}{9\pi^2}\cdot\frac{1}{x-\pi}= -\frac{2}{3\pi}\cdot\frac{1}{(x-\pi)(x+\pi/2)} $$ so the integral becomes $$ -\frac{2}{3\pi}\int_{-\infty}^{+\infty}\frac{\sin x}{(x-\pi)(x+\pi/2)}dx $$ that can be evaluated with standard methods.

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