I'm on the last question of my homework and it's involving using the residue theory, which I dont really understand, so could somebody lend me a hand?

I have to evaluate the real convergent improper integral below using residue theory:

$$ \int_0^\infty \frac{ \sin \pi x}{x(1-x^2)} \; \textrm{d}x$$

  • 5
    $\begingroup$ There are a few observations that will help here. First, the "singularity" at 0 is removable, and so will not contribute a residue. Second, the function is even, and so it is enough to evaluate the integral from $-\infty$ to $\infty$ and divide by 2. The last observation is that if you take a large semicircle with base $[-n,n]$ on the $x$ axis, the integral along the circular part (whether you take the top or the bottom) will approach zero as $n$ grows (I believe). From here, the problem becomes a standard residue calculus problem. $\endgroup$ – Aaron Feb 27 '12 at 19:18

Note that the integrand is an even function of $x$, so we will compute the integral of half the integrand over the whole real line.

Using partial fractions, we get $$ \frac{1/2}{x(1-x^2)}=\frac{1/2}{x}+\frac{1/4}{1-x}-\frac{1/4}{1+x} $$ Since the singularities are removable, we can use the contour $\gamma$ from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ instead of $\mathbb{R}$. The key step is to break up the integral into two along two closed contours

  1. $\gamma_+$ which goes from $-N-\frac iN$ to $+N-\frac iN$ then counterclockwise around the semicircle centered at $-\frac iN$ from $+N-\frac iN$ back to $-N-\frac iN$

  2. $\gamma_-$ which goes from $-N-\frac iN$ to $+N-\frac iN$ then clockwise around the semicircle centered at $-\frac iN$ from $+N-\frac iN$ back to $-N-\frac iN$

$$ \frac{1}{2i}\oint_{\gamma_+}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{i\pi z}\mathrm{d}z-\frac{1}{2i}\oint_{\gamma_-}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{-i\pi z}\mathrm{d}z $$ As $N\to\infty$, the contribution from the semi-circular parts vanishes and we are left with the integral along $\gamma$ of $\frac{1/2}{x(1-x^2)}\sin(\pi x)=\frac{1/2}{x(1-x^2)}\dfrac{e^{i\pi x}-e^{-i\pi x}}{2i}$.

There are no singularities inside $\gamma_-$, so that integral is $0$. Thus, the whole integral boils down to $$ \frac{1}{2i}\oint_{\gamma_+}\left(\frac{1/2}{z}+\frac{1/4}{1-z}-\frac{1/4}{1+z}\right)e^{i\pi z}\mathrm{d}z $$ Summing up the residues at $-1,0,\text{and }1$ yields $\dfrac{1}{4i}2\pi i+\dfrac{1}{8i}2\pi i+\dfrac{1}{8i}2\pi i=\pi$.

Thus, $$ \int_0^\infty\frac{\sin(\pi x)}{x(1-x^2)}\mathrm{d}x=\pi $$


This approach doesn't uses residue theory, but may be it will fit your needs. Since $f(x)=\frac{\sin\pi x}{x(1-x^2)}$ is even then $$ \int\limits_{\mathbb{R}_+}f(x)dx= \frac{1}{2}\int\limits_{\mathbb{R}}f(x)dx= \frac{1}{2}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x(1-x^2)}dx= $$ $$ \frac{1}{2}\int\limits_{\mathbb{R}}\sin\pi x\left(\frac{1}{x}-\frac{1}{2(x-1)}-\frac{1}{2(x+1)}\right)dx= $$ $$ \frac{1}{2}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx- \frac{1}{4}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x-1}dx- \frac{1}{4}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x+1}dx= $$ Note that $$ \int\limits_{\mathbb{R}}\frac{\sin\pi x}{x-1}dx=\{t=x-1\}= \int\limits_{\mathbb{R}}\frac{\sin\pi (t+1)}{t}dt= -\int\limits_{\mathbb{R}}\frac{\sin\pi t}{t}dt= -\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx $$ Similarly, $$ \int\limits_{\mathbb{R}}\frac{\sin\pi x}{x+1}dx= \{t=x+1\}= \int\limits_{\mathbb{R}}\frac{\sin\pi (t-1)}{t}dt= -\int\limits_{\mathbb{R}}\frac{\sin\pi t}{t}dt= -\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx $$ Thus, we have $$ \int\limits_{\mathbb{R}_+}f(x)dx= \frac{1}{2}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx- \frac{1}{4}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x-1}dx- \frac{1}{4}\int\limits_{\mathbb{R}}\frac{\sin\pi x}{x+1}dx= \int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx $$ The last integral reduces to so called Dirichlet integral $$ \int\limits_{\mathbb{R}}\frac{\sin\pi x}{x}dx= \{t=\pi x\}= \int\limits_{\mathbb{R}}\frac{\sin t}{t}dt=\pi $$

  • $\begingroup$ You might want to discuss the removable singularities at $-1,0,\text{and }1$ and use residues to show why $$\int_0^\infty\frac{\sin(\pi x)}{x}\mathrm{d}x=\pi$$ $\endgroup$ – robjohn Feb 27 '12 at 20:08
  • $\begingroup$ This argument just repeats yours, so I'm lazy to do it. If you want you can edit my answer, but I don't insist. $\endgroup$ – Norbert Feb 27 '12 at 20:11
  • $\begingroup$ Your argument is very nice once you have $\int_0^\infty\frac{\sin(\pi x)}{x}\mathrm{d}x=\pi$, but since the OP asked about residues, they probably don't want to start with that integral. Just my feeling. :-) $\endgroup$ – robjohn Feb 27 '12 at 20:24
  • $\begingroup$ Approach of complex analysis is useless here because I can't apply Jordan lemma to the integral $$\oint\limits_{\gamma}\frac{\sin z}{z}dz.$$ But using sledgehammer of Sokhatsky theorem this problem can be avoided. $\endgroup$ – Norbert Feb 27 '12 at 20:32
  • $\begingroup$ Do you need to use Sokhatsky since the singularities are removable? Just offset the contour a bit so it doesn't pass through the singularities. However, since you can quote the Dirichlet Integral, you can avoid all of this :-) $\endgroup$ – robjohn Feb 27 '12 at 20:42

$$f(z) = \frac{e^{\pi iz}}{z(1-z^2)}$$

Integrating along $C$, a large semicicle in the half plane, indented around $0$ and $\pm 1$, we have

$$0=\oint_C f(z)\, dz = P.V.\int_{-\infty}^\infty f(z)\, dz-i \pi\operatorname*{Res}_{z=0}f-i \pi\operatorname*{Res}_{z=1}f-i \pi\operatorname*{Res}_{z=-1}f\tag{1}$$ Trivially, $$\operatorname*{Res}_{z=0}f = 1$$

$$\operatorname*{Res}_{z=-1}f = \frac 12$$

$$\operatorname*{Res}_{z=1}f = \frac 1 2$$

thus, taking the real part of $(1)$ and dividing by 2:

$$\int_0^\infty \frac{\sin (\pi x)}{x(1-x^2)}\,dx=\pi$$


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