Proving that $int_0^1 x^{2n} sin (pi x) dx$ are polynomials of degree $2n$ in $pi$, divided by $pi^{2n+1}$.
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I think that the integrals $int_0^1 x^{2n} sin(pi x)dx$ are polynomials of degree at most $2n$, with all expoents being even numbers, divided by $pi^{2n+1}$
A few examples given by wolfram:
begin{align}
&int_0^1 x^2sin(pi x) dx = frac{pi^2-4}{pi^3}\
&int_0^1 x^4 sin(pi x) dx = frac{48-12pi^2+pi^4}{pi^5}\
&int_0^1 x^6sin(pi x) dx = frac{-1440+360pi^2-30pi^4+pi^6}{pi^7}
end{align}
How we prove this?
real-analysis integration
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
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add a comment |
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I think that the integrals $int_0^1 x^{2n} sin(pi x)dx$ are polynomials of degree at most $2n$, with all expoents being even numbers, divided by $pi^{2n+1}$
A few examples given by wolfram:
begin{align}
&int_0^1 x^2sin(pi x) dx = frac{pi^2-4}{pi^3}\
&int_0^1 x^4 sin(pi x) dx = frac{48-12pi^2+pi^4}{pi^5}\
&int_0^1 x^6sin(pi x) dx = frac{-1440+360pi^2-30pi^4+pi^6}{pi^7}
end{align}
How we prove this?
real-analysis integration
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
$endgroup$
add a comment |
$begingroup$
I think that the integrals $int_0^1 x^{2n} sin(pi x)dx$ are polynomials of degree at most $2n$, with all expoents being even numbers, divided by $pi^{2n+1}$
A few examples given by wolfram:
begin{align}
&int_0^1 x^2sin(pi x) dx = frac{pi^2-4}{pi^3}\
&int_0^1 x^4 sin(pi x) dx = frac{48-12pi^2+pi^4}{pi^5}\
&int_0^1 x^6sin(pi x) dx = frac{-1440+360pi^2-30pi^4+pi^6}{pi^7}
end{align}
How we prove this?
real-analysis integration
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
$endgroup$
I think that the integrals $int_0^1 x^{2n} sin(pi x)dx$ are polynomials of degree at most $2n$, with all expoents being even numbers, divided by $pi^{2n+1}$
A few examples given by wolfram:
begin{align}
&int_0^1 x^2sin(pi x) dx = frac{pi^2-4}{pi^3}\
&int_0^1 x^4 sin(pi x) dx = frac{48-12pi^2+pi^4}{pi^5}\
&int_0^1 x^6sin(pi x) dx = frac{-1440+360pi^2-30pi^4+pi^6}{pi^7}
end{align}
How we prove this?
real-analysis integration
real-analysis integration
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
asked Dec 19 '18 at 22:15
PintecoPinteco
731313
731313
add a comment |
add a comment |
1 Answer
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HINT: By integration by parts,
$$int_0^1 x^{2n}sin(pi x)dx=frac{1}{pi}+frac{2n}{pi}int_0^1 x^{2n-1}cos(pi x)dx$$
and
$$int_0^1 x^{2n-1}cos(pi x)dx=-frac{2n-1}{pi}int_0^1 x^{2n-2}sin(pi x)dx$$
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
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$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
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@ShubhamJohri Right, thanks for catching that!
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– Frpzzd
Dec 19 '18 at 22:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
HINT: By integration by parts,
$$int_0^1 x^{2n}sin(pi x)dx=frac{1}{pi}+frac{2n}{pi}int_0^1 x^{2n-1}cos(pi x)dx$$
and
$$int_0^1 x^{2n-1}cos(pi x)dx=-frac{2n-1}{pi}int_0^1 x^{2n-2}sin(pi x)dx$$
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
add a comment |
$begingroup$
HINT: By integration by parts,
$$int_0^1 x^{2n}sin(pi x)dx=frac{1}{pi}+frac{2n}{pi}int_0^1 x^{2n-1}cos(pi x)dx$$
and
$$int_0^1 x^{2n-1}cos(pi x)dx=-frac{2n-1}{pi}int_0^1 x^{2n-2}sin(pi x)dx$$
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
$endgroup$
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
add a comment |
$begingroup$
HINT: By integration by parts,
$$int_0^1 x^{2n}sin(pi x)dx=frac{1}{pi}+frac{2n}{pi}int_0^1 x^{2n-1}cos(pi x)dx$$
and
$$int_0^1 x^{2n-1}cos(pi x)dx=-frac{2n-1}{pi}int_0^1 x^{2n-2}sin(pi x)dx$$
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
$endgroup$
HINT: By integration by parts,
$$int_0^1 x^{2n}sin(pi x)dx=frac{1}{pi}+frac{2n}{pi}int_0^1 x^{2n-1}cos(pi x)dx$$
and
$$int_0^1 x^{2n-1}cos(pi x)dx=-frac{2n-1}{pi}int_0^1 x^{2n-2}sin(pi x)dx$$
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
edited Dec 19 '18 at 22:36
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
answered Dec 19 '18 at 22:20
FrpzzdFrpzzd
23k841109
23k841109
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
add a comment |
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
The last integral should have $x^{2n-2}$
$endgroup$
– Shubham Johri
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
$begingroup$
@ShubhamJohri Right, thanks for catching that!
$endgroup$
– Frpzzd
Dec 19 '18 at 22:36
add a comment |
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