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How should I calculate this integral

$$\int\limits_{-\infty}^\infty\frac{\sin x}{x(1+x^2)}\,dx\quad?$$

I have tried forming an indented semicircle in the upper half complex plane using the residue theorem and I tried to integrate along a curve that went around the complex plane and circled the positive real axis (since the integrand is even). Nothing has worked out for me. Please help!

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  • $\begingroup$ In the upper half plane, I think I just figured out an appropriate bound to make that integral go to zero; I had left off a factor of R^4 which made things cancel nicely. Any suggestions for the indented part? $\endgroup$ – Forgiven Mar 27 '13 at 15:34
  • $\begingroup$ The indented part is a half-circle going clockwise. Since the residue at $0$ is $1$, the integral around the half-circle is $-\pi i$. $\endgroup$ – robjohn Mar 27 '13 at 22:59
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$$ \int_{-\infty}^\infty\frac{\sin x}{x(1+x^2)}\mathrm dx=\int_{-\infty}^\infty\frac{\Im\mathrm e^{\mathrm ix}}{x(1+x^2)}\mathrm dx=\int_{-\infty}^\infty\frac{\Im\left(\mathrm e^{\mathrm ix}-1\right)}{x(1+x^2)}\mathrm dx=\Im\int_{-\infty}^\infty\frac{\mathrm e^{\mathrm ix}-1}{x(1+x^2)}\mathrm dx\;. $$

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  • $\begingroup$ If I may, how is the particularly helpful? I simply don't see it, and I'm curious. $\endgroup$ – Clayton Mar 27 '13 at 16:04
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    $\begingroup$ @Clayton: I understood the OP's problem to lie in dealing with an indented semicircle. This obviates the need for indenting the semicircle, because the integrand on the right-hand side is analytic at the origin and yet decays sufficiently in the upper half plane to allow the contour to be completed by a semicircle. All that remains is to determine the residue at $\mathrm i$ and take the imaginary part of the integral. $\endgroup$ – joriki Mar 27 '13 at 16:16
  • $\begingroup$ Subtracting $1$ is a nice alternative to bumping the contour to miss $0$. (+1) $\endgroup$ – robjohn Mar 27 '13 at 22:48
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Defining

$$\gamma_M:=\{z\in\Bbb C\;;\;z=Me^{it}\;,\;0\le t\le\pi\}\;,\;\;M\in (0,\infty),$$

and

$$f(z):=\frac{e^{iz}}{z(z^2+1)}\;,\;\;C_R:=[-R,-\epsilon]\cup\gamma_\epsilon\cup[\epsilon,R]\cup\gamma_R\;,\;\;0<\epsilon<<R$$

We get that, as the only pole of $\,f\,$ within the region limited by $\,C_R\,$ is $\,z=i\,$ , that

$$\oint\limits_{C_R}f(z)\,dz=2\pi i\,Res_{z=i}(f)$$

Now:

$$Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{e^{i^2}}{i(2i)}=-\frac{e^{-1}}{2}$$

$$Res_{z=0}(f)=\lim_{z\to 0}zf(z)=\frac{e^0}{1}=1$$

So, using the corollary to the lemma in the first answer here, and taking into account that we integrate on the negative direction on $\,\gamma_\epsilon\,$ , we get:

$$\frac{\pi i}{e}=\int\limits_{C_R}f(z)\,dz=\int\limits_{-R}^{-\epsilon}f(x)dx+\int\limits_{\gamma_\epsilon}f(z)\,dz+\int\limits_\epsilon^Rf(x)\,dz+\int\limits_{\gamma_R}f(z)\,dz\xrightarrow[\stackrel{\epsilon\to\ 0}{R\to\infty}]{}$$

$$\xrightarrow[\stackrel{\epsilon\to\ 0}{R\to\infty}]{}\int\limits_{-\infty}^\infty f(x)\,dx-\pi i\Longrightarrow$$

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

and comparing real and imaginary parts in both sides we get

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

Note: You can either use Jordan's Lemma or directly evaluate by Cauchy's to get

$$\int\limits_{C_R}f(z)\,dz\xrightarrow[R\to\infty]{}0$$

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  • $\begingroup$ Same contour, more detail (+1) I had originally thought as joriki did that the OP was interested in the contour. I posted and then filled in the details when I saw that you had done what looked like a complete answer. Upon close inspection, I see that ours are very close. If it's okay with you, I'll leave mine as it is a bit more concise and deals more with why I chose this contour. $\endgroup$ – robjohn Mar 27 '13 at 22:57
  • $\begingroup$ Thanks @robjohn, I insist you leave yours by all means. $\endgroup$ – DonAntonio Mar 27 '13 at 23:11
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$$ \begin{align} \int_{-\infty}^\infty\frac{\sin(x)}{x(1+x^2)}\,\mathrm{d}x &=\mathrm{Im}\left(\int_{-\infty}^\infty\frac{e^{ix}}{x(1+x^2)}\,\mathrm{d}x\right)\\ &=\mathrm{Im}\left(\int_\gamma\frac{e^{iz}}{z(1+z^2)}\,\mathrm{d}z\right)\\ \end{align} $$ where $\gamma$ goes from $(1/R,0)$ to $(R,0)$ circles counterclockwise from $(R,0)$ to $(-R,0)$ from $(-R,0)$ to $(-1/R,0)$, then circles clockwise from $(-1/R,0)$ to $(1/R,0)$.

The integral on the flat pieces gives the integral you are looking for.

We use the contour through the upper half-plane since the integrand vanishes quickly there. That is, the integral along the large half-circle is $0$.

The residue at $z=0$ is $1$ so the integral along the small half-circle is $-\pi i$.

The residue at $z=i$ is $\frac{e^{-1}}{i(i+i)}=-\frac1{2e}$, Thus, the integral along the entire contour is $-\frac{\pi i}{e}$.

Thus, the integral along the two flat pieces is $\pi i-\frac{\pi i}{e}$. Therefore, $$ \int_{-\infty}^\infty\frac{\sin(x)}{x(1+x^2)}\,\mathrm{d}x=\pi\frac{e-1}{e} $$

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