[SOLVED]sin^2x/x^2 complex analysis

dwsmith

Well-known member
$$\int_{0}^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{\pi}{2}$$.

[Hint: Consider the integral of $(1 - e^{2ix})/x^2)$.]

If we look at the complex sine, we have that $\sin z = \frac{e^{iz}-e^{-iz}}{2i}$. Then
$$\sin^2z = \frac{e^{-2iz}-e^{2iz}}{4}$$
so
$$\frac{\sin^2 z}{z^2} = \frac{e^{-2iz}-e^{2iz}}{4z^2}$$

I can't obtain the hint it is saying. What am I doing wrong?

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chisigma

Well-known member
Is...

$\displaystyle \sin^{2} z= (\frac{e^{i z}-e^{- i z}}{2i})^{2}= - \frac{e^{2 i z} -2 + e^{-2 i z}}{4}$ (1)

Kind regards

$\chi$ $\sigma$

dwsmith

Well-known member
Is...

$\displaystyle \sin^{2} z= (\frac{e^{i z}-e^{- i z}}{2i})^{2}= - \frac{e^{2 i z} -2 + e^{-2 i z}}{4}$ (1)

Kind regards

$\chi$ $\sigma$
From that, I don't see how we get the hint.

chisigma

Well-known member
In my opinion the best way to compute the definite integral $\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin^{2} x}{x^{2}}\ dx$ is the application of the Parseval's identity of the Fourier Transform...

$\displaystyle \int_{- \infty}^{+ \infty} |f(t)|^{2}\ dt = \frac{1}{2 \pi}\ \int_{- \infty}^{+ \infty} |F(\omega)|^{2}\ d \omega$ (1)

... to the function...

$\displaystyle f(t)=\begin{cases}1&\text{if}\ |t|<1\\ 0 &\text{if}\ |t|>1\end{cases}$ (2)

Kind regards

$\chi$ $\sigma$

dwsmith

Well-known member
In my opinion the best way to compute the definite integral $\displaystyle \int_{- \infty}^{+ \infty} \frac{\sin^{2} x}{x^{2}}\ dx$ is the application of the Parseval's identity of the Fourier Transform...$\displaystyle \int_{- \infty}^{+ \infty} |f(t)|^{2}\ dt = \frac{1}{2 \pi}\ \int_{- \infty}^{+ \infty} |F(\omega)|^{2}\ d \omega$ (1)... to the function... $\displaystyle f(t)=\begin{cases}1&\text{if}\ |t|1\end{cases}$ (2)Kind regards $\chi$ $\sigma$
In the fall when I am taking Fourier Transforms & Series, this method may be great, but at the moment, it isn't.I need to use Residue Theory, Cauchy Principal Value Theorem, and complex integration.

dwsmith

Well-known member
Since $\frac{\sin^2 z}{z^2}$ is even,
$$\int_{0}^{\infty}\frac{\sin^2 x}{x^2}dx = \frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^2 x}{x^2}dx$$
and $\sin^2z = \frac{1 - \cos 2z}{2}$.

Then we have
$$\frac{1}{4}\int_{-\infty}^{\infty}\frac{1 - \cos 2z}{z^2}dx$$

How does $\cos 2z$ which is $\frac{e^{2iz}+e^{-2iz}}{2}$ go from that to simply $e^{-2iz}$?

Random Variable

Well-known member
MHB Math Helper
$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{2} \ \text{PV} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{4} \ \text{Re} \ \text{PV}\int_{0}^{\infty} \frac{1-e^{2ix}}{x^{2}} \ dx$

let $f\displaystyle (z) = \frac{1-e^{2iz}}{z^{2}}$

$\displaystyle \int_{-R}^{-r} f(x) \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} f(x) \ dx + \int_{C_{R}} f(z) \ dz = 2 \pi i (0) = 0$

where $C_{r}$ is a small upper half circle of radius $r$ about the origin and $C_{R}$ is a large upper half circle of radius $R$

$f(z)$ has a simple pole at the origin

$\displaystyle \frac{1- e^{2iz}}{z^{2}} = \frac{1 -(1 +2iz-2z^{2} + \ldots) }{z^{2}} = -\frac{2i}{z} + 2 + \ldots$

now let $r$ go to zero and $R$ go to infinity

$\displaystyle \text{PV}\ \int_{-\infty}^{\infty} f(x) \ dx + \lim_{r \to 0} \int_{C_{r}} f(z) \ dz + \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = 0$

since $z=0$ is a simple pole $\displaystyle \lim_{r \to 0} \int_{C_{r}} f(z) \ dz = - i \pi \text{Res}[f,0] = -i \pi (-2i) = -2 \pi$

and $\displaystyle \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = 0$ by Jordan's lemma

so $\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{4} \text{Re} \ \text{PV} \int_{-\infty}^{\infty} f(x) \ dx = \frac{1}{4} (2 \pi) = \frac{\pi}{2}$

dwsmith

Well-known member
$\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{2} \ \text{PV} \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{4} \ \text{Re} \ \text{PV}\int_{0}^{\infty} \frac{1-e^{2ix}}{x^{2}} \ dx$

let $f\displaystyle (z) = \frac{1-e^{2iz}}{z^{2}}$

$\displaystyle \int_{-R}^{-r} f(x) \ dx + \int_{C_{r}} f(z) \ dz + \int_{r}^{R} f(x) \ dx + \int_{C_{R}} f(z) \ dz = 2 \pi i (0) = 0$

where $C_{r}$ is a small upper half circle of radius $r$ about the origin and $C_{R}$ is a large upper half circle of radius $R$

$f(z)$ has a simple pole at the origin

$\displaystyle \frac{1- e^{2iz}}{z^{2}} = \frac{1 -(1 +2iz-2z^{2} + \ldots) }{z^{2}} = -\frac{2i}{z} + 2 + \ldots$

now let $r$ go to zero and $R$ go to infinity

$\displaystyle \text{PV}\ \int_{-\infty}^{\infty} f(x) \ dx + \lim_{r \to 0} \int_{C_{r}} f(z) \ dz + \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = 0$

since $z=0$ is a simple pole $\displaystyle \lim_{r \to 0} \int_{C_{r}} f(z) \ dz = - i \pi \text{Res}[f,0] = -i \pi (-2i) = -2 \pi$

and $\displaystyle \lim_{R \to \infty} \int_{C_{R}} f(z) \ dz = 0$ by Jordan's lemma

so $\displaystyle \int_{0}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = \frac{1}{4} \text{Re} \ \text{PV} \int_{-\infty}^{\infty} f(x) \ dx = \frac{1}{4} (2 \pi) = \frac{\pi}{2}$
This is great but it didn't answer the question I had in post 6. If it does, it isn't explicitly apparent.

Random Variable

Well-known member
MHB Math Helper
$\displaystyle \text{PV} \int_{-\infty}^{\infty} \frac{1-e^{2ix}}{x^{2}} \ dx = \text{PV} \int_{-\infty}^{\infty} \frac{1-\cos 2x - i \sin 2x}{x^{2}} \ dx$

so $\displaystyle \text{Re} \ \text{PV} \int_{-\infty}^{\infty} \frac{1-e^{2ix}}{x^{2}} \ dx = \text{PV} \int_{-\infty}^{\infty} \frac{1-\cos 2x}{x^{2}} \ = \ \text{PV} \ 2 \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx = 2 \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx$ (since $\displaystyle \int_{-\infty}^{\infty} \frac{\sin^{2} x}{x^{2}} \ dx$ is a convergent integral)

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dwsmith

Well-known member
One last quick question. Why is your expansion of e neglecting division by division by n! for the respected term of the series?
Never mind I forgot to square the 2