# Contour integration problem

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?

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

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

• 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

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.

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 $-3\pi/4$