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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$

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    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.2012-02-27

3 Answers 3

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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 $

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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 $

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    Thanks for your help guys, much appreciated!2012-03-01
1

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

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:

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