Mixing
Fourier transform to solve $int_{- infty}^{+infty}frac{sin^3(x)}{x^3},dx$
$begingroup$
I think that I am almost there:
giving $P(x)=1$ for $x in ]-1/2;1/2[$ and $P(x)=0$ elsewhere.
I managed to find that the convolution product $P*P*P=frac{3}{4}-x^2$ for $x in [-1/2;1/2]$
I found that Reverse Fourier transform of $P$ is $p=frac{1}{2pi}times frac{sin(x/2)}{x/2}$
knowing that Fourier Transform of
$ptimes p times p$ is $frac{1}{4pi^2}P*P*P$
We then get :
$int_{- infty}^{+infty} p(x)times p(x) times p(x)times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
$int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
Then for $k=0$, $int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}dx =frac{1}{4pi^2}frac{3}{4}$
then I find $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{2}$
but this is wrong because $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{4}$
Where is the problem?
Thanks
fourier-transform
edited Jan 12 at 22:45
J.G.
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
$endgroup$
add a comment |
$begingroup$
I think that I am almost there:
giving $P(x)=1$ for $x in ]-1/2;1/2[$ and $P(x)=0$ elsewhere.
I managed to find that the convolution product $P*P*P=frac{3}{4}-x^2$ for $x in [-1/2;1/2]$
I found that Reverse Fourier transform of $P$ is $p=frac{1}{2pi}times frac{sin(x/2)}{x/2}$
knowing that Fourier Transform of
$ptimes p times p$ is $frac{1}{4pi^2}P*P*P$
We then get :
$int_{- infty}^{+infty} p(x)times p(x) times p(x)times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
$int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
Then for $k=0$, $int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}dx =frac{1}{4pi^2}frac{3}{4}$
then I find $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{2}$
but this is wrong because $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{4}$
Where is the problem?
Thanks
fourier-transform
edited Jan 12 at 22:45
J.G.
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
$endgroup$
$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50
add a comment |
$begingroup$
I think that I am almost there:
giving $P(x)=1$ for $x in ]-1/2;1/2[$ and $P(x)=0$ elsewhere.
I managed to find that the convolution product $P*P*P=frac{3}{4}-x^2$ for $x in [-1/2;1/2]$
I found that Reverse Fourier transform of $P$ is $p=frac{1}{2pi}times frac{sin(x/2)}{x/2}$
knowing that Fourier Transform of
$ptimes p times p$ is $frac{1}{4pi^2}P*P*P$
We then get :
$int_{- infty}^{+infty} p(x)times p(x) times p(x)times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
$int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
Then for $k=0$, $int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}dx =frac{1}{4pi^2}frac{3}{4}$
then I find $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{2}$
but this is wrong because $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{4}$
Where is the problem?
Thanks
fourier-transform
edited Jan 12 at 22:45
J.G.
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
$endgroup$
I think that I am almost there:
giving $P(x)=1$ for $x in ]-1/2;1/2[$ and $P(x)=0$ elsewhere.
I managed to find that the convolution product $P*P*P=frac{3}{4}-x^2$ for $x in [-1/2;1/2]$
I found that Reverse Fourier transform of $P$ is $p=frac{1}{2pi}times frac{sin(x/2)}{x/2}$
knowing that Fourier Transform of
$ptimes p times p$ is $frac{1}{4pi^2}P*P*P$
We then get :
$int_{- infty}^{+infty} p(x)times p(x) times p(x)times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
$int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}times e^{ikx}dx =frac{1}{4pi^2}P*P*P(k)$
Then for $k=0$, $int_{- infty}^{+infty} frac{1}{8pi^3} frac{sin^3(x)}{x^3}dx =frac{1}{4pi^2}frac{3}{4}$
then I find $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{2}$
but this is wrong because $int_{- infty}^{+infty} frac{sin^3(x)}{x^3},dx =frac{3pi}{4}$
Where is the problem?
Thanks
fourier-transform
fourier-transform
edited Jan 12 at 22:45
J.G.
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
edited Jan 12 at 22:45
J.G.
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
edited Jan 12 at 22:45
J.G.
26.3k22541
edited Jan 12 at 22:45
J.G.
26.3k22541
edited Jan 12 at 22:45
J.G.
26.3k22541
26.3k22541
asked Jan 12 at 22:33
favulipofavulipo
1
asked Jan 12 at 22:33
favulipofavulipo
1
asked Jan 12 at 22:33
favulipofavulipo
1
1
$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50
add a comment |
$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50
$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50
$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Your mistake is you forgot $p=operatorname{sinc}frac{x}{2}notequivoperatorname{sinc}x$. (In case the notation is unfamiliar, $operatorname{sinc}x=frac{sin x}{x}$, with $operatorname{sinc}0=1$ for continuity.) We have $int_{Bbb R}operatorname{sinc}^3frac{x}{2} dx=2int_{Bbb R}operatorname{sinc}^3y dy$ by a trivial substitution.
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
Your mistake is you forgot $p=operatorname{sinc}frac{x}{2}notequivoperatorname{sinc}x$. (In case the notation is unfamiliar, $operatorname{sinc}x=frac{sin x}{x}$, with $operatorname{sinc}0=1$ for continuity.) We have $int_{Bbb R}operatorname{sinc}^3frac{x}{2} dx=2int_{Bbb R}operatorname{sinc}^3y dy$ by a trivial substitution.
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
add a comment |
$begingroup$
Your mistake is you forgot $p=operatorname{sinc}frac{x}{2}notequivoperatorname{sinc}x$. (In case the notation is unfamiliar, $operatorname{sinc}x=frac{sin x}{x}$, with $operatorname{sinc}0=1$ for continuity.) We have $int_{Bbb R}operatorname{sinc}^3frac{x}{2} dx=2int_{Bbb R}operatorname{sinc}^3y dy$ by a trivial substitution.
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
$endgroup$
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
add a comment |
$begingroup$
Your mistake is you forgot $p=operatorname{sinc}frac{x}{2}notequivoperatorname{sinc}x$. (In case the notation is unfamiliar, $operatorname{sinc}x=frac{sin x}{x}$, with $operatorname{sinc}0=1$ for continuity.) We have $int_{Bbb R}operatorname{sinc}^3frac{x}{2} dx=2int_{Bbb R}operatorname{sinc}^3y dy$ by a trivial substitution.
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
$endgroup$
Your mistake is you forgot $p=operatorname{sinc}frac{x}{2}notequivoperatorname{sinc}x$. (In case the notation is unfamiliar, $operatorname{sinc}x=frac{sin x}{x}$, with $operatorname{sinc}0=1$ for continuity.) We have $int_{Bbb R}operatorname{sinc}^3frac{x}{2} dx=2int_{Bbb R}operatorname{sinc}^3y dy$ by a trivial substitution.
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
answered Jan 12 at 22:48
J.G.J.G.
26.3k22541
26.3k22541
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
add a comment |
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
$begingroup$
Thanks JG I now understand my mistake. All is clear.
$endgroup$
– favulipo
Jan 14 at 21:04
add a comment |
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$begingroup$
By the way, if you'd like an alternative to Fourier analysis, you can make a double integral using $x^{-3}=frac12 int_0^infty y^2exp (-xy)dy$. The result $sin x=frac{exp ix-exp -ix}{2i}$ lets you integrate out $x$, leaving a $y$-integral of a rational function of $y$.
$endgroup$
– J.G.
Jan 12 at 22:50