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I am to evaluate $\displaystyle\int_0^{\infty} \dfrac{\sin x}{x(x^2+1)}dx$ via contour integration.

Now I used an indented semicircular contour, and the parts lying on the real line and the big arc were no problem, but the small arc is being resistant, and I'm not sure what to do. Usually, on the small arc from $-\varepsilon$ to $\varepsilon$ I can take a laurent expansion of the integrand, and consider integrating its principle part over the arc, letting the rest go to zero in the limit $\varepsilon \to 0$ as the "holomorphic part". My issue is this particular integrand doesn't have a principle part...

The end result is $\dfrac{(e-1)\pi}{2e}$, and so far I have the integral over the whole contour as $\dfrac{-\pi i}{e}$ (I'm not sure why this came out imaginary..) so this part is going to have to contribute something. What should I do to get something out?

FireGarden
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3 Answers3

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My issue is this particular integrand doesn't have a principal part...

Then the integral over the large semicircle won't work out. Remember that

$$\lvert \sin (x+iy)\rvert^2 = \lvert \sin x \cos (iy) + \sin(iy)\cos x\rvert^2 = \sin^2 x+ \sinh^2 y,$$

so if you keep the sine in the integrand, you have no growth control to conclude that the integral over the large semicircle tends to $0$.

Use the symmetry to get

$$\begin{align} \int_0^\infty \frac{\sin x}{x(x^2+1)}\,dx &= \frac{1}{2}\int_{-\infty}^\infty \frac{\sin x}{x(x^2+1)}\,dx\\ &= \frac{1}{2i}\operatorname{v.p.} \int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)}\,dx, \end{align}$$

then you have an integrand with a simple pole on the contour, and the small semicircle gives you $\pi i$ times the residue in $0$, the latter being $1$.

Using both semicircles in the upper half-plane, the integral over the closed contour is

$$2\pi i \operatorname{Res}\left(\frac{e^{iz}}{z(z^2+1)}; i\right) = 2\pi i \frac{e^{-1}}{i\cdot 2i} = - \frac{\pi i}{e}.$$

Hence

$$\operatorname{v.p.}\int_{-\infty}^\infty \frac{e^{ix}}{x(x^2+1)}\,dx = \lim_{\varepsilon \downarrow 0} \int_{\lvert x\rvert > \varepsilon} \frac{e^{ix}}{x(x^2+1)}\,dx = \pi i \frac{e-1}{e}$$

and

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

Daniel Fischer
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    v.p.? Is that for the French valeur principale? – Étienne Bézout Mar 27 '14 at 19:44
  • Oui, that's how I learned it, old habits die hard. – Daniel Fischer Mar 27 '14 at 19:44
  • I see :) Cauchy was French after all, so I guess it is nice to honor him by using the French abbreviation. – Étienne Bézout Mar 27 '14 at 19:48
  • I don't understand why you say the simple upper semicircle contour doesn't work: it worked for me... – DonAntonio Mar 27 '14 at 20:18
  • @DonAntonio That's exactly what I do. It doesn't work when one keeps the sine in the integrand and doesn't replace it with $e^{iz}$. – Daniel Fischer Mar 27 '14 at 20:19
  • Oh, I see now @DanielFischer...yes, it is basically hte very same v.p. (=valor principal in spanish) thing, and it is done a lot with this kind of functions. – DonAntonio Mar 27 '14 at 20:22
  • I didn't know there was a French word for principal value. Nice. – ireallydonknow Mar 28 '14 at 13:31
  • @DanielFischer, I dont understand it. How can you replace $\sin(z)$ with $e^{ix}$? $e^{ix} = \cos(x) + i\sin(x)$ it doesnt equal just sine. – Amad27 Dec 22 '14 at 15:31
  • @Amad27 It's the symmetry (or parity). The denominator, $x\mapsto x(x^2+1)$, is and odd function, thus $x\mapsto \frac{\cos x}{x(x^2+1)}$ is also odd, and hence the principal value integral $\operatorname{v.p.}\int_{-\infty}^\infty \frac{\cos x}{x(x^2+1)},dx$ vanishes. Therefore, $$\operatorname{v.p.}\int_{-\infty}^\infty \frac{\cos x + i\sin x}{x(x^2+1)},dx = 0 + i\int_{-\infty}^\infty \frac{\sin x}{x(x^2+1)},dx.$$ – Daniel Fischer Dec 22 '14 at 15:37
  • @DanielFischer, thanks. That is good. So really in this context. $e^{ix}$ really is $i\sin(x)$ because cosine vanishes? How about if in the original statement, it was $\cos(x)$ rather than $\sin(x)$ then why is it allowed to use $e^{ix}$ still? Thanks =) – Amad27 Dec 22 '14 at 15:39
  • @Amad27 Saying "$e^{ix}$ really is $i\sin x$" is wrong, the point is that since we're integrating over $\mathbb{R}$, and the denominator is odd, the change in the integrand cancels out, so replacing $\sin x$ with $\frac{1}{i}e^{ix}$ doesn't change the value of the integral. If we had started with $\cos$ instead of $\sin$, the cancelling would occur when the factor by which $\cos$ is multiplied is an even function, e.g. $\frac{1}{x^2+1}$; then $\frac{\sin x}{x^2+1}$ is an odd function, and integrating over $\mathbb{R}$ annihilates that part. – Daniel Fischer Dec 22 '14 at 15:51
  • @DanielFischer, I asked a question about this here. Take a look, you may find it interesting. http://math.stackexchange.com/questions/1077695/using-eix-instead-of-sine-and-cosine-in-contour-integration/1077701?noredirect=1#comment2192715_1077701 I assume you are were this fact that Jack stated? – Amad27 Dec 22 '14 at 16:06
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Define

$$f(z)=\frac{e^{iz}}{z(z^2+1)}\;,\;\;C_R:=[-R,-\epsilon]\cup\gamma_\epsilon\cup[\epsilon,R]\cup\gamma_R\;\;,\;\;\text{with}$$

$$\gamma_a:=\{z=ae^{it}\;,\;\;a,t\in\Bbb R_+\;,\;\;0<t<\pi\}$$

Then the function has one simple pole within the domain defined by the above contour:

$$\text{Res}_{z=i}(f):=\lim_{z\to i}\,(z-i)f(z)=\lim_{z\to i}\frac{e^{iz}}{z(z+i)}=\frac{e^{-1}}{-2}=-\frac1{2e}$$

Thus, by Cauchy's theorem:

$$-\frac{\pi i}e=-2\pi i \frac1{2e}=\oint\limits_{C_R}f(z)dz=\int\limits_{-R}^{-\epsilon}f(x)dx-\int\limits_{\gamma_\epsilon}f(z)dz+\int\limits_\epsilon^Rf(x)dx+\int\limits_{\gamma_R}f(z)dz$$

By Jordan's Lemma, the last integral above tends to zero when $\;R\to\infty\;$ , while the second equals $\;\pi i\;$ when $\;\epsilon\to 0\;$ by the corollary below the lemma in the answer here:Definite integral calculation with poles at 0 and $\pm i\sqrt{3}$ and because

$$\text{Res}_{z=0}(f):=\lim_{z\to 0}\,zf(z)=\lim_{z\to 0}\frac{e^{iz}}{(z^2+1)}=1$$

Thus, we get passing to the limit $\;R\to\infty\;,\;\;\epsilon\to 0\;$ :

$$-\frac{\pi i}e=\int\limits_{-\infty}^\infty\frac{e^{ix}}{x(x^2+1)}dx-\pi i\implies\ldots$$

Now just take imaginary parts and take into account the real integrand function is even.

Community
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DonAntonio
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My guess is you should have two integrals, one for the residue at $i$ and the other one at $-i$

For the $i $ residue take the angle $3\pi/4$ for the other take $-3\pi/4$

Ellya
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