Hello, i have a problem that i do not know where to begin with. It says to prove, using the Laplace Transform, that
$\displaystyle \int_{0}^{\infty} \frac {sin^{2}(u)}{u^{2}}\,du = \frac {\pi}{2}$
Is it done using Euler's formula?
Hello, i have a problem that i do not know where to begin with. It says to prove, using the Laplace Transform, that
$\displaystyle \int_{0}^{\infty} \frac {sin^{2}(u)}{u^{2}}\,du = \frac {\pi}{2}$
Is it done using Euler's formula?
The main idea with using Laplace transforms is to transform the given equation into an algebraic one and simplify the algebraic equation (maybe by breaking it down into several fractions) and then transform it back using an inverse Laplace transform.
Do you have tables of Laplace and inverse Laplace transforms to help you carry out those operations?
I suspect that the problem requires the use of the Fourier Transform and not of the Laplace Transform... that's why the 'Parseval Identity' extablishes that if $\displaystyle \varphi (\lambda)= \mathcal{F} \{f(t)\}$ is...
$\displaystyle \displaystyle \int_{-\infty}^{+ \infty} |f(t)|^{2}\ dt = \frac{1}{2 \pi}\ \int_{-\infty}^{+ \infty} |\varphi(\lambda)|^{2}\ d \lambda$ (1)
... and from (1) derive directly that is...
$\displaystyle \displaystyle \int_{0}^{\infty} (\frac{ \sin \lambda}{\lambda})^{2}\ d \lambda = \frac{\pi}{2}$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
mr_fantastic told you how to solve it.
to find $\displaystyle L\left(\dfrac{sin^2(t)}{t^2}\right)$ you want to use the following formula two times:
If f(t) of class A & $\displaystyle L(f(t))=F(s)$, then :
$\displaystyle \displaystyle L\left(\dfrac{f(t)}{t}\right)=\int_s^{\infty} F(\beta) \, d\beta$
Re-writting the function as $\displaystyle \displaystyle \dfrac{ \left(\dfrac{sin^2(t)}{t} \right) }{t}$ and using the formula two times will solve your problem.
Let see what you will get.
A praticable way to realize the idea may be the following. Let's start writing...
$\displaystyle \displaystyle \int_{0}^{t} \frac{\sin^{2} \tau}{\tau^{2}}\ d \tau = Si(2t) + \frac{\cos 2t -1}{2t}$ (1)
.... where Si(*) is the 'Sine Integral Function'. Now is...
$\displaystyle \displaystyle \mathcal{L} \{\frac{\cos 2t -1}{2} \} = \frac{1}{2} \{\frac{s}{s^{2}+4} - \frac{1}{s} \}$ (2)
... so that...
$\displaystyle \displaystyle \mathcal{L} \{\frac{\cos 2t -1}{2t} \} = \frac{1}{2} \int_{s}^{\infty} (\frac{u}{u^{2}+4} - \frac{1}{u} )\ du= -\frac{1}{4} \ln (1+\frac{4}{s^{2}})$ (3)
... and is...
$\displaystyle \displaystyle \mathcal{L} \{Si(2t)\}= \frac{\tan^{-1} \frac{2}{s}}{s}$ (4)
Now we can combine (1), (2), (3) and (4) and obtain...
$\displaystyle \displaystyle \int_{0}^{\infty} \frac{\sin^{2} t}{t^{2}}\ dt = \lim_{s \rightarrow 0+} \{\tan^{-1} \frac{2}{s} - \frac {s}{4}\ \ln (1+\frac{4}{s^{2}})\} = \frac{\pi}{2}$ (5)
In my opinion however the use of the Fourier Tranform and the 'Parseval's Identity' allows to obtain the result in a more confortable and also elegant way...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$