Integral
Let's solve this integral:
$int_{-infty}^{+infty}sinc^5(t)dt$ where $sinc(t)=(sin(pi*t))/(pi*t)$
$int_{-infty}^{+infty}sinc^5(t)dt$ where $sinc(t)=(sin(pi*t))/(pi*t)$
Risposte
Now i give my solution:
$int_{-infty}^{+infty}sinc^5(t)dt=int_{-infty}^{+infty}sinc^3(t)*sinc^2(t)dt=int_{-infty}^{+infty}T(f)*Delta(f)df$ in which i have exploited Parseval's theorem, therefore $Delta(f)=F[sinc^2(t)]=(1-|f|)Pi[f/2],T(f)=F[sinc^3(t)]$ where $F(*)$ is Fourier's transformate.
Now $T(f)=F[sinc^3(t)]=F[sinc^2(t)*sinc(t)]=Delta(f)**Pi(f)$ in which i have exploited the property that to the product in time domain corresponds convolution in frequency domain. Now
$T(f)=int_{-infty}^{+infty}Delta(f-theta)*Pi(theta)d theta=int_{-1/2}^{+1/2}Delta(f-theta)d theta=int_{f-1/2}^{f+1/2}Delta(theta)d theta$. Through simple steps we obtain
$T(f)={(0,,f<-3/2),(int_{-1}^{f+1/2)(1+theta)d theta=1/2(f+3/2)^2,,-3/2<=f<-1/2),(int_{f-1/2}^{0}(1+theta)d theta+int_{0}^{f+1/2}(1-theta)d theta=3/4-f^2,,-1/2<=f<1/2),(int_{f-1/2}^{1}(1-theta)d theta=1/2(3/2-f)^2,,1/23/2):}$
So
$int_{-infty}^{+infty}sinc^5(t)dt=int_{-1}^{-1/2}(1+f)*1/2(f+3/2)^2df+int_{-1/2}^{0}(1+f)(3/4-f^2)df+int_{0}^{1/2}(1-f)(3/4-f^2)df+int_{1/2}^{1}(1-f)1/2(3/2-f)^2df$=
$2int_{-1}^{-1/2}(1+f)*1/2(f+3/2)^2df+2*int_{-1/2}^{0}(1+f)(3/4-f^2)df=int_{-1}^{-1/2}(1+f)(f+3/2)^2df+2*int_{-1/2}^{0}(1+f)(3/4-f^2)df$=
$[1/3(1+f)(3/2+f)^3-1/12*(3/2+f)^4]_{-1}^{-1/2}+2[(3/4f-1/3f^3)(1+f)-(3/8f^2-1/12f^4]_{-1/2}^{0}=(1/12+1/192)+2(3/16-1/48+3/32-1/192)$=
$17/192+98/192=115/192$
An other method to solve the same integral is acceptable.
$int_{-infty}^{+infty}sinc^5(t)dt=int_{-infty}^{+infty}sinc^3(t)*sinc^2(t)dt=int_{-infty}^{+infty}T(f)*Delta(f)df$ in which i have exploited Parseval's theorem, therefore $Delta(f)=F[sinc^2(t)]=(1-|f|)Pi[f/2],T(f)=F[sinc^3(t)]$ where $F(*)$ is Fourier's transformate.
Now $T(f)=F[sinc^3(t)]=F[sinc^2(t)*sinc(t)]=Delta(f)**Pi(f)$ in which i have exploited the property that to the product in time domain corresponds convolution in frequency domain. Now
$T(f)=int_{-infty}^{+infty}Delta(f-theta)*Pi(theta)d theta=int_{-1/2}^{+1/2}Delta(f-theta)d theta=int_{f-1/2}^{f+1/2}Delta(theta)d theta$. Through simple steps we obtain
$T(f)={(0,,f<-3/2),(int_{-1}^{f+1/2)(1+theta)d theta=1/2(f+3/2)^2,,-3/2<=f<-1/2),(int_{f-1/2}^{0}(1+theta)d theta+int_{0}^{f+1/2}(1-theta)d theta=3/4-f^2,,-1/2<=f<1/2),(int_{f-1/2}^{1}(1-theta)d theta=1/2(3/2-f)^2,,1/2
So
$int_{-infty}^{+infty}sinc^5(t)dt=int_{-1}^{-1/2}(1+f)*1/2(f+3/2)^2df+int_{-1/2}^{0}(1+f)(3/4-f^2)df+int_{0}^{1/2}(1-f)(3/4-f^2)df+int_{1/2}^{1}(1-f)1/2(3/2-f)^2df$=
$2int_{-1}^{-1/2}(1+f)*1/2(f+3/2)^2df+2*int_{-1/2}^{0}(1+f)(3/4-f^2)df=int_{-1}^{-1/2}(1+f)(f+3/2)^2df+2*int_{-1/2}^{0}(1+f)(3/4-f^2)df$=
$[1/3(1+f)(3/2+f)^3-1/12*(3/2+f)^4]_{-1}^{-1/2}+2[(3/4f-1/3f^3)(1+f)-(3/8f^2-1/12f^4]_{-1/2}^{0}=(1/12+1/192)+2(3/16-1/48+3/32-1/192)$=
$17/192+98/192=115/192$
An other method to solve the same integral is acceptable.
excuse me, but...Excuse-moi is French, isn't it?
Excuse-moi.
this post has been written in english, therefore if you want to partecipate to it you should write in english.
then, i desire to have the result of this integral and all the steps.
then, i desire to have the result of this integral and all the steps.
of course, but the spirit of this topic is that only english is allowed
"nicola de rosa":
Let's solve this integral:
$int_{-infty}^{+infty}sinc^5(t)dt$ where $sinc(t)=(sin(pi*t))/(pi*t)$
Un classico, molti conti. Basta applicare un paio di volte il teorema di Borel per la Fourier-trasformata di un prodotto: $F(uv) = F(u)*F(v)$, dove $F(\cdot)$ indica l'operatore di Fourier in $L^1_{loc}(RR)$ e * è un prodotto di convoluzione - indovinereste quale?
when exams come... lack of time! 
nothing?
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