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How to evaluate this integral $$ \int_{-\infty}^\infty\frac{\sin x}{x(x^2 + 1)}dx $$ I am having a problem to solve this because of two poles when I solve it by integration first from $-R$ to $R$ and then along a semicircle in the upper half plane.

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    $\begingroup$ Use Residue theorem. Note that only $0$ and $i$ lie in the upper half plane. $\endgroup$ – Paras Khosla Jan 25 at 14:35
  • $\begingroup$ You need to show your work, so that you can get direction towards the approach that'll get you to the solution. $\endgroup$ – Paras Khosla Jan 25 at 14:50
  • $\begingroup$ And when i try to evalute ∫e∧iz/ z(z²⁺1) problem is that z=0 pole comes on boundry of our closed contour $\endgroup$ – sweety tarika Jan 25 at 15:35
  • $\begingroup$ $z=0$ should not be a problem because in fact $\text{Res} \bigl(\frac{\sin z}{z(z^2+1)}, 0\bigr)=0$ $\endgroup$ – Paras Khosla Jan 25 at 16:06
  • $\begingroup$ see z=0 is not inside the contour...its on boundry $\endgroup$ – sweety tarika Jan 25 at 16:35
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According to your choice of the contour: $C$ is the upper half-plane and $\Gamma$ is the semicircular arc of radius $R$ (say).

$$\int_C \frac{\sin z}{z(z^2+1)}\mathrm dz=\text{P.V.} \int_{-\infty}^{+\infty} \frac{\sin x}{x(x^2+1)}\mathrm dx+\lim_{R \to \infty}\int_{\Gamma}\frac{\sin z}{z(z^2+1)}\mathrm dz$$

Now, use the fact that if $f(z)=\frac{g(z)}{h(z)}$ where $f$ and $g$ are analytic near $z_0$ and $h$ has a simple zero at $z_0$, then $\text{Res}(f(z), z_0) = \frac{g(z_0)}{h'(z_0)}$. Note that $g$ is $\sin z$ and $h$ is $z(z^2+1)$. For the evaluation of the integral over the arc $\Gamma$, use the ML-inequality and you should be good to go.

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  • $\begingroup$ First of all thanks, but sir we cant use ML inequalty because its unbounded $\endgroup$ – sweety tarika Jan 25 at 15:30
  • $\begingroup$ No the function is not unbounded. ML Inequality can be used. In fact, $\text{P.V.} \int_{-\infty}^{\infty} \frac{\sin z}{z(z^2+1)}\mathrm dz=e^{-1}(e-1)\pi$. $\endgroup$ – Paras Khosla Jan 25 at 16:25
  • $\begingroup$ can you show how you solve? $\endgroup$ – sweety tarika Jan 25 at 16:27
  • $\begingroup$ can you show how you solve? $\endgroup$ – sweety tarika Jan 25 at 16:27
  • $\begingroup$ i am new user, i dont know how to use this app $\endgroup$ – sweety tarika Jan 25 at 16:31
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Let us avoid contour-hunting. The integrand function is even, hence it is enough to compute $$ F(a)=\int_{0}^{+\infty}\frac{\sin(ax)}{x(x^2+1)}\,dx $$ (defined over $\text{Re}(a)>0$) at $a=1$. We may notice that $$ (\mathcal{L} F)(s) = \int_{0}^{+\infty}\frac{dx}{(s^2+x^2)(1+x^2)}=\frac{\pi}{2s(s+1)} $$ hence $$F(a) = \frac{\pi}{2}\mathcal{L}^{-1}\left(\frac{1}{s}-\frac{1}{s+1}\right)(a)=\frac{\pi}{2}(1-e^{-a}) $$ and $$ \int_{-\infty}^{+\infty}\frac{\sin x}{x(x^2+1)}\,dx = \color{blue}{\pi(1-e^{-1})}.$$ There is no need to introduce a principal value: the $\text{sinc}$ function is continuous and bounded over $\mathbb{R}$.

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  • $\begingroup$ Wowww...its good aproach $\endgroup$ – sweety tarika Jan 25 at 17:50
  • $\begingroup$ Sir,you directly write value of integration...when i try to solve by putting x² =t ...it become lengthy $\endgroup$ – sweety tarika Jan 25 at 17:55
  • $\begingroup$ is there any quick method for this integration? $\endgroup$ – sweety tarika Jan 25 at 17:55
  • $\begingroup$ @sweetytarika: quicker than the shown one, you mean? You may use Fourier transforms instead of Laplace transforms, it remains a four-liner or so. $\endgroup$ – Jack D'Aurizio Jan 25 at 18:10
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Here is another approach that like @Jack D'Aurizio approach avoids contour integration altogether.

Let $$F(a) = \int_{-\infty}^\infty \frac{\sin (ax)}{x(x^2 + 1)} \, dx = 2 \int_0^\infty \frac{\sin (ax)}{x(x^2 + 1)} \, dx, \qquad a > 0.$$ We are required to find $F(1)$.

Using Feynman's trick of differentiating under the integral sign we have $$F'(a) = 2 \int_0^\infty \frac{\cos (ax)}{x^2 + 1} \, dx,$$ and $$F''(a) = - 2 \int_0^\infty \frac{x \sin (ax)}{x^2 + 1} \, dx.$$ Observe that $$F''(a) - F(a) = -2 \int_0^\infty \frac{\sin (ax)}{x} \, dx = -\pi.$$ Here the well-known result of $$\int_0^\infty \frac{\sin (ax)}{x} \, dx = \frac{\pi}{2}, \qquad a > 0,$$ has been used.

One solving the differential equaltion $F''(a) - F(a) = -\pi$ we have $$F(a) = C_1 e^a + C_2 e^{-a} + \pi.$$ The two unknown constants $C_1$ and $C_2$ can be found by noting that $F(0) = 0$ and $F'(0) = \pi$. Doing so one finds $$F(a) = \pi(1 - e^{-a}).$$ Finally, setting $a = 1$ one has $$\int_{-\infty}^\infty \frac{\sin x}{x(x^2 + 1)} \, dx = \pi(1 - e^{-1}),$$ as required.

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  • $\begingroup$ in begining how you took F(a⁾=2 ,except this step every step is meaningfull $\endgroup$ – sweety tarika Jan 26 at 11:40
  • $\begingroup$ If you mean the factor of 2 appearing in front of the integral on the first line, that comes from the integrand being an even function between symmetric limits. Is this what you mean? $\endgroup$ – omegadot Jan 26 at 11:42
  • $\begingroup$ ohkk, got it..thanks $\endgroup$ – sweety tarika Jan 26 at 11:49

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