# Cauchy Principal Value of $\int_{-\infty}^{\infty} \dfrac{\sin x}{x(x^2+1)}\mathrm dx$

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.

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

## 3 Answers

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.

• First of all thanks, but sir we cant use ML inequalty because its unbounded – sweety tarika Jan 25 at 15:30
• No the function is not unbounded. ML Inequality can be used. In fact, $\text{P.V.} \int_{-\infty}^{\infty} \frac{\sin z}{z(z^2+1)}\mathrm dz=e^{-1}(e-1)\pi$. – Paras Khosla Jan 25 at 16:25
• can you show how you solve? – sweety tarika Jan 25 at 16:27
• can you show how you solve? – sweety tarika Jan 25 at 16:27
• i am new user, i dont know how to use this app – sweety tarika Jan 25 at 16:31

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 $$\mathbb{R}$$.

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

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.

• in begining how you took F(a⁾=2 ,except this step every step is meaningfull – sweety tarika Jan 26 at 11:40
• 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? – omegadot Jan 26 at 11:42
• ohkk, got it..thanks – sweety tarika Jan 26 at 11:49