# definite integral calculation with 0 pole and

$$\int_0^\infty \frac{\sin(2\pi x)}{x(x^{2}+3)}$$

I look at $\frac{e^{2\pi i z}}{z^{3}+3z}$ , also calculate the residues, but they don't get me the right answer. ( I use that for this case, it holds that $\int_{-\infty}^\infty f(z)dz = 2\pi i (\sum \operatorname{Res} z_{r}) + \pi i Res_{0}$, but my answer turns out wrong when I check with wolframalpha.

Residue for $0$ is $1$, for $z=\sqrt{3}i$ it's $-\frac{e^{-2\pi}}{2}$ . . .

In a wronger attempt I forgot $2\pi z$ and used $z$ only: $\frac{e^{iz}}{z^{3}+3z}$ and the result was little closer, but missing factor 2 and and $i$.

Does anybody see the right way? Please do tell

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How do you get the residue at 0 to be 1? –  Henning Makholm Nov 20 '11 at 4:12
Once you calculated the residue of $f(z) = \frac{e^{2\pi i z}}{z(z^2+3)}$ at $z = 0$ and $z = i\sqrt{3}$, you will get $$\int_{-\infty}^{\infty} = 2\pi i \operatorname{Res}_{z=i\sqrt{3}} f(z) + \pi i \operatorname{Res}_{z=0} f(z).$$ Then the actual answer must be $\frac{1}{2i} \int_{-\infty}^{\infty} f(z) \; dz$. I guess this is what you have missed. –  sos440 Nov 20 '11 at 5:23
Hey sos440, I do get the right answer with $\frac{1}{2i}$, the $\frac{1}{2}$ comes from the integration boundaries, and the $\frac{1}{i}$ because sine is the imaginary part of $e^{iz}$, right? –  VVV Nov 20 '11 at 13:05
@VVV , what contour are you taking for your calculations?? Also, after the limit when $\,R\to\infty\,$ (or whatever) is done, we still remain with the real part of the complex function ($\,cos 2\pi x\,$ instead of $\,\sin 2\pi x\,$, which does not converge on $\,(0,\infty)\,$... Also, you seem to be taking a contour that contains only the positive imaginary pole and not the negative one. –  DonAntonio Jul 7 '12 at 0:48
@sos440 , what contour are you taking for your calculations? and why the integral is multiplied by $\,1/2i\,$ and not by $\,1/2\pi i\,$ ? –  DonAntonio Aug 20 '12 at 2:39

We make the following lemma:

Lemma. Suppose $f(z)$ is holomorphic near $z = z_0$. Fix $\theta_0 \in (0, 2\pi)$. If $\gamma_{\epsilon}$ denotes a counter-clockwise oriented arc of angle $\theta_0$ on the circle of radius $\epsilon$ centered at $z_0$, then $$\lim_{\epsilon\to0} \int_{\gamma_{\epsilon}} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta=i\theta_0 f(z_0).$$

Proof. By the substitution $\zeta = z_0 + \epsilon e^{i\theta}$, we have \begin{align*} \left| \int_{\gamma_{\epsilon}} \frac{f(\zeta)}{\zeta-z_0}\;d\zeta - i\theta_0 f(z_0)\right| &= \left| i \int_{\theta_1}^{\theta_1+\theta_0} f(z_0 + \epsilon e^{i\theta})\;d\theta - i\theta_0 f(z_0)\right| \\ & \leq \int_{\theta_1}^{\theta_1+\theta_0} \left| f(z_0 + \epsilon e^{i\theta}) - f(z_0) \right| \;d\theta, \end{align*} which clearly goes to zero when $\epsilon \to 0$.

As a corollary, if $f(z)$ has a simple pole at $z = z_0$ then with the same notation as in the Lemma, we have

$$\lim_{\epsilon\to0} \int_{\gamma_{\epsilon}} f(\zeta) \;d\zeta=i\theta_0 \mathrm{Res} \{ f(z), z_0 \}.$$

Now let $C$ be the upper-semicircular contour of radius $R \gg 1$ with a small semicircular indent of radius $\epsilon \ll 1$ at the origin. Let us write $C$ as $$C = \Gamma_{R} + L_{\epsilon,R} - \gamma_{\epsilon},$$ where $\Gamma_R$ and $\gamma_{\epsilon}$ denote counter-clockwise oriented arcs corresponding to the outer circle and the inner circle of $C$, respectively, and $L_{\epsilon,R}$ denote the remaining union of two lines on $C \cap \Bbb{R}$. Now for

$$f(z) = \frac{e^{2\pi i z}}{z(z^2 + 3)},$$

the integral in question, which we denote as $I$, is equal to

$$I = \frac{1}{2i} \lim_{{\epsilon \to 0 \atop R \to \infty}} \int_{L_{\epsilon, R}} f(z) \; dz.$$

Now, by Cauchy integration formula,

$$\oint_{C} f(z) \; dz = 2\pi i \mathrm{Res} \{ f(z), i\sqrt{3} \}.$$

This means that

$$\int_{L_{\epsilon, R}} f(z) \; dz = 2\pi i \mathrm{Res} \{ f(z), i\sqrt{3} \} + \int_{\gamma_{\epsilon}} f(z) \; dz - \int_{\Gamma_R} f(z) \; dz.$$

Taking limit $\epsilon \to 0$ and $R \to \infty$, the integral $\int_{\Gamma_R} f(z) \; dz$ vanishes by Jordan's lemma. Thus by our lemma,

$$\lim_{{\epsilon \to 0 \atop R \to \infty}} \int_{L_{\epsilon, R}} f(z) \; dz = 2\pi i \mathrm{Res} \{ f(z), i\sqrt{3} \} + \pi i \mathrm{Res} \{ f(z), 0 \}.$$

But since

$$\mathrm{Res} \{ f(z), i\sqrt{3} \} = \left. (z-i\sqrt{3})f(z) \right|_{z=i\sqrt{3}} = -\frac{1}{6}e^{-2\pi \sqrt{3}}$$

and

$$\mathrm{Res} \{ f(z), 0 \} = \left. z f(z) \right|_{z=0} = \frac{1}{3},$$

we have

$$I = \frac{1}{2i} \left[ 2\pi i \left(-\frac{1}{6}e^{-2\pi \sqrt{3}}\right) + \pi i \left(\frac{1}{3} \right) \right] = \frac{\pi}{6}\left(1 - e^{-2\pi \sqrt{3}} \right).$$

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+1 Very nice answer! Not only was I answered but I also learned about a rather beautiful, and hopefully pretty useful, generalization of Cauchy's Integral Theorem. Thanks a lot, @sos440 –  DonAntonio Aug 22 '12 at 1:55

\begin{align} \int_0^\infty\frac{\sin(2\pi x)}{x(x^2+3)}\,\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{\sin(2\pi x)}{x(x^2+3)}\,\mathrm{d}x\tag{1}\\ &=\frac12\int_{i-\infty}^{i+\infty}\frac{\sin(2\pi z)}{z(z^2+3)}\,\mathrm{d}z\tag{2}\\ &=\frac12\int_{i-\infty}^{i+\infty}\frac{e^{i2\pi z}-e^{-i2\pi z}}{2iz(z^2+3)}\,\mathrm{d}z\tag{3}\\ &=\frac1{4i}\int_{\gamma_+}\frac{e^{i2\pi z}}{z(z^2+3)}\mathrm{d}z -\frac1{4i}\int_{\gamma_-}\frac{e^{-i2\pi z}}{z(z^2+3)}\mathrm{d}z\tag{4}\\ &=\frac{2\pi i}{4i}\frac{e^{-2\pi\sqrt3}}{i\sqrt3(i\sqrt3+i\sqrt3)} +\frac{2\pi i}{4i}\left(\frac13+\frac{e^{-2\pi\sqrt3}}{-i\sqrt3(-i\sqrt3-i\sqrt3)}\right)\tag{5}\\ &=\frac\pi6\left(1-e^{-2\pi\sqrt3}\right)\tag{6} \end{align} where $\gamma_+$ passes from $i-R$ to $i+R$ then circles back counterclockwise around the upper half-plane, and where $\gamma_-$ passes from $i-R$ to $i+R$ then circles back clockwise around the lower half-plane.

$\gamma_+$ contains the singularity at $i\sqrt3$.

$\gamma_-$ contains the singularities at $0$ and $-i\sqrt3$.

Explanation of steps

$(1)$ integrand is an even function.

$(2)$ there are no singularities for the integrand in the rectangle with corners $i-R,i+R,R,-R$ and the integral over the ends of the rectangle vanishes as $R\to\infty$.

$(3)$ write $\sin(2\pi z)=\dfrac{e^{i2\pi z}-e^{-i2\pi z}}{2i}$

$(4)$ use contour $\gamma_+$ for $e^{i2\pi z}$ and $\gamma_-$ for $e^{-i2\pi z}$ so that the integrand vanishes over the large circles in the upper and lower half-planes.

$(5)$ use residues to compute the integrals over $\gamma_+$ and $\gamma_-$.

$(6)$ simplification.

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