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This problem comes from a preliminary exam from 2009

Find the Principal Value of the integral $$\int_{-\infty}^\infty \frac{x \sin(x)}{x^2+2x+2}\,dx$$

My attempt at solution: Letting $f(z)=\frac{z \sin(z)}{z^2+2z+2}$, the residue theorem implies that $ \int_{-\infty}^\infty f(x)dx=2\pi i \text{Res}(f, -1+i)$, but I get that Res$(f, -1+i)=(\frac{1}{2}+\frac{i}{2})\sin(-1+i)$, and I don't know how to compute $\sin(-1+i)$.

Also, I was hoping someone could help me to understand why the problem asks for the "Principal Value" rather than the value because the only singularities of the function $f(z)=\frac{z \sin(z)}{z^2+2z+2}$ are at $-1\pm i$, neither of which are on the real axis.

Mark Viola
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user140776
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  • $\sin(-1+i)=\sin(1)\cos(i)-\sin(i)\cos(1)$, $\sin(i)=i\sinh(1)$ and $\cos(i)=\cosh(1)$, so $\sin(i-1)=\sin(1)\cosh(1)-i\sinh(1)\cos(1).$ It's not pretty, but there it is. – Plutoro Jul 13 '15 at 02:45
  • @AlexS Thanks! Although I think you computed $-\sin(-1+i)$. – user140776 Jul 13 '15 at 02:53
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    Because the integral is not absolutely convergent, it exists as an improper Riemann integral but not as a (proper) Lebesgue integral — see here for further discussion. Probably that's why the principal value is asked for... – Micah Jul 13 '15 at 03:03
  • @user140776 With respect to the Cauchy PV, this is due to the limits at $\pm \infty$. Second, to use contour integration, we use a trick in which we write $\sin x=\text{Im}(e^{ix})$, carry out the integration in the complex plane and take imaginary part at the end. Please let me know how I can improve my posted answer. I just want to give you the best answer I can. – Mark Viola Jul 13 '15 at 03:36

1 Answers1

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Note that the integral of interest can be written

$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(\int_{-\infty}^{\infty}\frac{xe^{ix}}{x^2+2x+1}dx\right) \tag 1$$

Let's examine the integral

$$\oint_C\frac{ze^{iz}}{z^2+2z+1}dz$$

where $C$ is the closed contour comprised of

$(i)$ the real-line segment from $(-R,0)$ to $(R,0)$

$(ii)$ the semicircle $C_R$ in the upper-half plane, centered at the origin with radius $R$.

Thus, we can write

$$\begin{align} \oint_C\frac{ze^{iz}}{z^2+2z+1}dz&=\int_{-R}^{R}\frac{xe^{ix}}{x^2+2x+1}dx+ \int_{C_R}\frac{ze^{iz}}{z^2+2z+1}dz \tag 2\\\\ \end{align}$$

Note that in the limit as $R\to \infty$, the imaginary part of first integral on the right-hand side of $(2)$ equals the Cauchy Principal Value of the integral of interest in $(1)$. The second integral on the right-hand side of $(2)$ can be shown using Jordan's Lemma to go to zero in the limit as $R\to \infty$. Thus, from the residue theorem we have

$$\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\text{Im}\left(2\pi i \,\text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)\right)\tag 3$$

The residue in $(3)$ can be evaluated as

$$\begin{align} \text{Res}\left(\frac{ze^{iz}}{z^2+2z+2},z=-1+i\right)&= \lim_{z\to -1+i}\frac{(z+1-i)ze^{iz}}{(z+1-i)(z+1+i)}\\\\ &= \frac{(-1+i)e^{i(-1+i)}}{2i} \\\\ &=\frac{e^{-1}}{2i}\sqrt{2}e^{i(3\pi/4-1)}\tag 4 \end{align}$$

Substituting $(4)$ into $(3)$ reveals that

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+2x+1}dx=\pi e^{-1}(\cos 1 -\sin 1)}$$

Mark Viola
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