Integral - very simple
Compute
$int_(-infty)^(+infty) (sinx/x)^(10)cos(1000x)dx$
$int_(-infty)^(+infty) (sinx/x)^(10)cos(1000x)dx$
Risposte
Uhmm... I found this number:
$138826588/(9! 2^10) pi$
or equivalently
$34706647/(9! 2^8) pi$
That's too strange... I think that's a well-known kind of number, but I'm not sure of that
$138826588/(9! 2^10) pi$
or equivalently
$34706647/(9! 2^8) pi$
That's too strange... I think that's a well-known kind of number, but I'm not sure of that
i don't think so...
-1/x<=(sen x)/x<=1/x
1<=cos1000x<=1
so integral is 0
1<=cos1000x<=1
so integral is 0
Forget it, try to answer to this: given $f(x)=(sinx/x)^10cosx$ calculate $F(0)$
"luca.barletta":
No, I don't.
Why don't you?
Sorry but I don't understand:
$int_(-infty)^(+infty) (sinx/x)^(10)cos(x)dx = 0$
Bye
$int_(-infty)^(+infty) (sinx/x)^(10)cos(x)dx = 0$
Bye
I've chosen a function a little bit pathological. At this point the question becomes: find $F(0)$.
No, I don't.
$int_(-infty)^(+infty) (sinx/x)^(10)cos(1000x)dx = int_(-infty)^(0) (sinx/x)^(10)cos(1000x)dx+int_(0)^(infty) (sinx/x)^(10)cos(1000x)dx= -int_(0)^(infty) (sinx/x)^(10)cos(1000x)dx+int_(0)^(infty) (sinx/x)^(10)cos(1000x)dx=0$
do you agree?
do you agree?
A month later... I post the faster solution:
recall the following property of the Fourier transformation:
$int_(-infty)^(+infty) f(x)dx=F(0)$
where F(y) denote the Fourier transform of f(x).
We have a passband function $f(x)=(sinx/x)^(10)cos(1000x)$, that is $F(0)~=0$, so
$int_(-infty)^(+infty) f(x)dx~=0$
recall the following property of the Fourier transformation:
$int_(-infty)^(+infty) f(x)dx=F(0)$
where F(y) denote the Fourier transform of f(x).
We have a passband function $f(x)=(sinx/x)^(10)cos(1000x)$, that is $F(0)~=0$, so
$int_(-infty)^(+infty) f(x)dx~=0$
...an other way faster than "per parti"?
Ok. An other way to solve the same problem?
"luca.barletta":
Compute
-
$int_(-infty)^(+infty) (sinx/x)^(10)cos(1000x)dx$
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