$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$
-
- Posts: 14
- Joined: Fri Dec 04, 2015 4:54 pm
$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$
Evaluate the integral:
$$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$$
$$\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$$
Re: $\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$
\begin{align*}&\int_0^{\infty} \frac{\sin^2 \tan x}{x^2}\,dx\\&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\sin^2 \tan x}{x^2}\,dx\\&= \frac{1}{2}\sum\limits_{n=-\infty}^{\infty} \int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan (x+n\pi)}{(x+n\pi)^2}\,dx\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \sin^2 \tan x \left(\sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}\right)\,dx\\&= \frac{1}{2}\int_{-\pi/2}^{\pi/2} \frac{\sin^2 \tan x}{\sin^2 x}\,dx\\&= \int_{0}^{\pi/2} \frac{\sin^2 \cot x}{\cos^2 x}\,dx\\&= \int_0^{\infty} \sin^2 \frac{1}{y}\,dy = \frac{\pi}{2}\end{align*}
where, we used the partial fraction expansion: $\displaystyle \frac{1}{\sin^2 x} = \sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}$ for $x \notin \mathbb{Z}$.
where, we used the partial fraction expansion: $\displaystyle \frac{1}{\sin^2 x} = \sum\limits_{n=-\infty}^{\infty} \frac{1}{(x+n\pi)^2}$ for $x \notin \mathbb{Z}$.
Last edited by r9m on Sat Dec 19, 2015 6:26 pm, edited 1 time in total.
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: $\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$
I give some proofs for some results used above and I am presenting some other that can be extracted via them pretty easily.
- It holds that $\displaystyle \pi \cot \pi x = \frac{1}{x} + \sum_{n=1}^{\infty}\left ( \frac{1}{x+n} - \frac{1}{x-n} \right )= \lim_{N \rightarrow +\infty}\sum_{n=-N}^{N}\frac{1}{x+n}, \;\; x \in \mathbb{R} \setminus \mathbb{Z}$.
Proof: A standard proof can be found through Fourier Series. One can expand in Fourier series the function $f(x)= \cos a x , \;a \notin \mathbb{Z}, \;\; |x| \leq \pi$. Here is another proof. One begines from the Weierstrass product , that is $ \sin \pi x = \pi x \prod \limits_{n=1}^{\infty}\left ( 1- \frac{x^2}{n^2} \right )$. Taking $\log$s at both sides we have that:
$$\begin{align*}
\ln \sin \pi x &= \ln \pi x + \sum_{n=1}^{\infty}\ln \left ( 1- \frac{x^2}{n^2} \right )\\
&= \ln \pi x + \sum_{n=1}^{\infty}\left [ \ln \left ( 1-\frac{x}{n} \right ) - \ln \left ( 1+ \frac{x}{n} \right ) \right ]
\end{align*}$$
Hence $\displaystyle \frac{\pi \cos \pi x}{\sin \pi x} = \frac{1}{x}+ \sum_{n=1}^{\infty}\left ( \frac{1}{x+n} - \frac{1}{x-n} \right )= \lim_{N \rightarrow +\infty}\sum_{n=-N}^{N}\frac{1}{x+n}$, that is what we wanted. - It holds that $\displaystyle \sum_{n=-\infty}^{\infty}\frac{1}{\left ( z+n \right )^2} = \frac{\pi^2}{\sin^2 \pi z}$.
Proof: Just differentiate the above identity. - It holds that $\displaystyle \sum_{n=-\infty}^{\infty}\frac{(-1)^{n+1}}{n+z}= - \frac{\pi}{\sin \pi z}$. (I leave the derivation of this identity as an exercise)
Imagination is much more important than knowledge.
Re: $\int_0^\infty \frac{\sin^2 (\tan x)}{x^2}\, dx$
I think I have another way using contours.
$$\sin^{2}(u)=\frac{1-\cos(2u)}{2}$$
Now, consider $$f(z)=\frac{1-e^{2i\tan(z)}}{2z^{2}}$$
and use a rectangular contour with height y. There is a pole at z=0.
[/centre]
There are also what is called 'essential singularities' at $z=\pi k +\frac{\pi}{2}$ due to
$\tan(\pi k+\frac{\pi}{2})=\text{undefined}, \;\ k=0,1,2,3,....$
An essential singularity is neither a pole nor a removable singularity. But, they contribute nothing anyway. As do the vertical sides.
Continuing.....
along the x-axis:
$$\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx$$
along the top:
$$-\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x+iy)}}{2(x+iy)^{2}}dx$$
The residue at $z=0$ is
$$\lim_{z\to 0}\frac{d}{dz}\left[z^{2}\cdot \frac{1-e^{2i\tan(z)}}{2z^{2}}\right]=-i$$
$$\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx-\underbrace{\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x+iy)}}{2(x+iy)^{2}}dx}_{\text{0}}-\pi i (-i)=0$$
$$2\int_{0}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx-\pi =0$$
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}(\tan(x))}{x^{2}}dx=\frac{\pi}{2}$$
$$\sin^{2}(u)=\frac{1-\cos(2u)}{2}$$
Now, consider $$f(z)=\frac{1-e^{2i\tan(z)}}{2z^{2}}$$
and use a rectangular contour with height y. There is a pole at z=0.
[/centre]
There are also what is called 'essential singularities' at $z=\pi k +\frac{\pi}{2}$ due to
$\tan(\pi k+\frac{\pi}{2})=\text{undefined}, \;\ k=0,1,2,3,....$
An essential singularity is neither a pole nor a removable singularity. But, they contribute nothing anyway. As do the vertical sides.
Continuing.....
along the x-axis:
$$\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx$$
along the top:
$$-\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x+iy)}}{2(x+iy)^{2}}dx$$
The residue at $z=0$ is
$$\lim_{z\to 0}\frac{d}{dz}\left[z^{2}\cdot \frac{1-e^{2i\tan(z)}}{2z^{2}}\right]=-i$$
$$\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx-\underbrace{\int_{-\infty}^{\infty}\frac{1-e^{2i\tan(x+iy)}}{2(x+iy)^{2}}dx}_{\text{0}}-\pi i (-i)=0$$
$$2\int_{0}^{\infty}\frac{1-e^{2i\tan(x)}}{2x^{2}}dx-\pi =0$$
$$\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}(\tan(x))}{x^{2}}dx=\frac{\pi}{2}$$
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 10 guests