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My attempt: We have $$ \int_{\Gamma} f(z) \ dt = \int_{\gamma} f(z) \ dz + \int_{\sigma} f(z) \ dz = 2 \pi i \sum_{z_0 \text{ pole inside } \Gamma} \mathrm{res}_{z_0}(f) $$ where $\Gamma$ is contour of the upper semi-circle of radius $r$, $\gamma$ is the real line segment going from $-r$ to $r$, $\sigma$ is the semi-circle, and $f(z) = \frac{sin(z)}{z^2 + 2z +2}$.

We have $$ \frac{\sin(t)}{t^2+2t+2} = \frac{\sin(t)}{(t+1 -i)(t+1+i)} $$ so the function has simple poles at $z = -1 \pm i$, and therefore

$$ \int_{-r}^r \frac{\sin(t)}{t^2+2t+2} \ dt = - \int_0^{ \pi} \frac{\sin(re^{i \theta} )ri e^{i \theta}}{(re^{i \theta})^2 + 2 re^{i \theta} + 2} \ d\theta + \pi \sin(-1+i). $$

By taking the limit on both sides we get

$$ \int_{-\infty}^{\infty} \frac{\sin(t)}{t^2+2t+2} \ dt = - \lim_{r \to \infty} \int_0^{ \pi} \frac{\sin(re^{i \theta}) ri e^{i \theta}}{(re^{i \theta})^2 + 2 re^{i \theta} + 2} \ d\theta + \pi \sin(-1+i) $$

but I don't know how to simplify the right hand side.

Stephen Brown
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    the problem with this is that the circle integrand is huge in absolute value when $|\theta-\pi/2| < \delta <\pi/2$ and $r$ large so it's hard to estimate; you are better off to take the function $\frac{e^{it}}{t^2+2t+2}$ for which your integral is the imaginary part since now in the upper half plane $|e^{iz}| \le 1$ so you can estimate the circle integral and shows it goes to zero – Conrad Jan 11 '23 at 00:09
  • @Conrad I don't understand why the function $\frac{e^{it}}{t^2+2t+2}$ is useful. Are you suggesting I use $\sin(z) = \frac{e^{iz} - e^{-iz}}{2i}$ and split up the original integral into two integrals? – Stephen Brown Jan 11 '23 at 00:18
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    No just consider $I=\int_{-\infty}^{\infty} \frac{e^{it}}{t^2+2t+2} \ dt$ as your original integral is $\Im I$; repeat your procedure except that now the $r$ circle integral will be $\int_0^{\pi} \frac{e^{ire^{i \theta}} ri e^{i \theta}}{(re^{i \theta})^2 + 2 re^{i \theta} + 2} \ d\theta $ and since you are in the upper half plane the numerator is bounded by $r$ since $|e^{ire^{i \theta}}| \le 1, \theta \in [0, \pi]$ (you are on the upper circle so $\theta \in [0, \pi]$ btw); since the denominator is about $r^2$, the ratio is $1/r$ integrated on a finite $[0, \pi]$ domain so goes to zero etc – Conrad Jan 11 '23 at 00:26
  • @Conrad That makes sense, thanks for the help! – Stephen Brown Jan 11 '23 at 00:36
  • Happy to be of help – Conrad Jan 11 '23 at 00:39
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    Are you obliged to use contour integration ? – Claude Leibovici Jan 11 '23 at 03:11
  • @ClaudeLeibovici No, but I don't know how else to approach it. – Stephen Brown Jan 11 '23 at 10:23

3 Answers3

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\begin{align*}I & =\int_{-\infty}^\infty\frac{\sin x}{(x+1)^2+1}dx=\int_{-\infty}^{\infty}\frac{\sin(x-1)}{x^2+1}dx \\ & =\cos(1)\int_{-\infty}^\infty\frac{\sin x}{x^2+1}dx-\sin(1)\int_{-\infty}^\infty\frac{\cos x}{x^2+1}dx \\ &=-2\sin(1)\int_{0}^\infty\frac{\cos x}{x^2+1}dx=-2\sin(1)\frac{\pi}{2e}=\frac{-\pi\sin(1)}{e}. \end{align*} (See here.)

Gary
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phi-rate
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Using algebra

Consider the more general case of $$I=\int_{-\infty}^{+\infty} \frac{\sin(t)}{(t-a)(t-b)} \, dt$$ where $(a,b)$ are complex numbers.

Using partial fraction decomposition $$I=\frac 1{a-b}\left(\int_{-\infty}^{+\infty}\frac{\sin (t)}{t-a}\,dt-\int_{-\infty}^{+\infty}\frac{\sin (t)}{t-b}\,dt\right)$$ So, we have two integrals of the form $$J_c=\int_{-\infty}^{+\infty}\frac{\sin (t)}{t-c}\,dt$$ A natural change of variable $(t=x+c)$ and the expansion of the sine gives

$$K_c=\int\frac{\sin (t)}{t-c}\,dt=\sin(c)\int \frac{\sin (x)}{x}\,dx+\cos(c)\int \frac{\cos (x)}{x}\,dx$$

$$K=\sin (c)\, \text{Ci}(x)+\cos (c)\, \text{Si}(x)$$ Back to $t$ and $I$ and using the bounds, then $\cdots\cdots\cdots$

Claude Leibovici
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Complex Analysis approach: Let $$f(z)=\frac{e^{iz}}{z^2+2z+2}$$Note that $Imm(f(z))=\frac{sin(z)}{z^2+2z+2}$. $f$ has 2 simple pole with order 1. The pole are $z_1=-1+i$ and $z_2=-1-i$ wich are the roots of $p(z)=z^2+2z+2$. So $f(z)=\frac{e^{iz}}{(z-z_1)(z-z_2)}$. Writing $f(z)=\frac{1}{z-z_1}h(z)$ note that $h(z_1)\not=0$ and $h(z)$ olomorphic in $z_1$ we have $Res(f,z_1)=h(z_1)=\frac{e^{iz_1}}{(z_1-z_2)}=\frac{e^{i(-1+i)}}{(2i)}$ Analogue for $z_2$.

Now consider $\phi(t):[-R,R]\to \Bbb C,\phi(t)=t$ and $\tau_R(t):[0,\pi]\to \Bbb C,\tau_R(t)=Re^{it}$. Be $\Gamma=$ $$ \int_{\Gamma} f(z) \ dt = \int_{\phi} f(z) \ dz + \int_{\tau_R} f(z) \ dz = 2 \pi i Res(f,z_1) $$Noticing that z_1 is the only pole in $\Gamma$.

$$ \lim_{R\to \infty} \int_{\Gamma} f(z) \ dt = \lim_{R\to \infty}\int_{\phi} f(z) \ dz +\lim_{R\to \infty} \int_{\tau_R} f(z) \ dz =2 \pi i Res(f,z_1)$$. Noticing that the second integral$\to 0$ as $R \to\infty$, you consider the first integral with now is in $\Bbb R$ and splitting it in the real part and the imaginary part then, you conclude.

giorgiokyn23
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