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Integral of odd exponents of Cos(x) is bounded
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In a 2010 exam question I found, no solution is provided. The question is:
Show that for all $Kin mathbbN $ the integral converges:
$int_1^inftyfraccos^2k+1xxdx$
I want to say that for every odd power of $cos x$, since the positivity stays the same, the integral $int cos^2k+1xdx$ is bounded, and then use Dirichlet's Criterion.
I know how to actively divide the integral into the periodic regions, show that over every period the integral is 0 and bound between the maximum and minimum in the regions.
But is there a neater way to show this?
real-analysis improper-integrals
asked Jul 27 at 8:30
ikoikoia
1399
add a comment |Â
up vote
1
down vote
favorite
In a 2010 exam question I found, no solution is provided. The question is:
Show that for all $Kin mathbbN $ the integral converges:
$int_1^inftyfraccos^2k+1xxdx$
I want to say that for every odd power of $cos x$, since the positivity stays the same, the integral $int cos^2k+1xdx$ is bounded, and then use Dirichlet's Criterion.
I know how to actively divide the integral into the periodic regions, show that over every period the integral is 0 and bound between the maximum and minimum in the regions.
But is there a neater way to show this?
real-analysis improper-integrals
asked Jul 27 at 8:30
ikoikoia
1399
oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In a 2010 exam question I found, no solution is provided. The question is:
Show that for all $Kin mathbbN $ the integral converges:
$int_1^inftyfraccos^2k+1xxdx$
I want to say that for every odd power of $cos x$, since the positivity stays the same, the integral $int cos^2k+1xdx$ is bounded, and then use Dirichlet's Criterion.
I know how to actively divide the integral into the periodic regions, show that over every period the integral is 0 and bound between the maximum and minimum in the regions.
But is there a neater way to show this?
real-analysis improper-integrals
asked Jul 27 at 8:30
ikoikoia
1399
In a 2010 exam question I found, no solution is provided. The question is:
Show that for all $Kin mathbbN $ the integral converges:
$int_1^inftyfraccos^2k+1xxdx$
I want to say that for every odd power of $cos x$, since the positivity stays the same, the integral $int cos^2k+1xdx$ is bounded, and then use Dirichlet's Criterion.
I know how to actively divide the integral into the periodic regions, show that over every period the integral is 0 and bound between the maximum and minimum in the regions.
But is there a neater way to show this?
real-analysis improper-integrals
asked Jul 27 at 8:30
ikoikoia
1399
asked Jul 27 at 8:30
ikoikoia
1399
asked Jul 27 at 8:30
ikoikoia
1399
asked Jul 27 at 8:30
ikoikoia
1399
1399
oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32
add a comment |Â
oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32
oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32
oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32
add a comment |Â
1 Answer
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1
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You can explicitly compute the integral $int_a^bcos^nxdx$. Define $J(n) = intcos^nxdx$ (indefinite integral). Then you get
$$J(n) = sin x fraccos^n-1xn + fracn-1nJ(n-2) .$$
This gives a recursion ending with $J(1) = sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = sin x cdot p_n-1(cos x) + a_0x$$
where $p_n-1$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ int_1^bcos^2k+1xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $int_1^inftyfraccos^2k+1xxdx$ converges. In fact, the sequence
$$a_m = int_mpi - pi/2^mpi + pi/2fraccos^2k+1xxdx $$
converges to $0$ and the series
$$Sigma_m=1^infty a_m$$
is alternating and therefore converges.
answered Jul 27 at 11:09
Paul Frost
3,611420
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You can explicitly compute the integral $int_a^bcos^nxdx$. Define $J(n) = intcos^nxdx$ (indefinite integral). Then you get
$$J(n) = sin x fraccos^n-1xn + fracn-1nJ(n-2) .$$
This gives a recursion ending with $J(1) = sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = sin x cdot p_n-1(cos x) + a_0x$$
where $p_n-1$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ int_1^bcos^2k+1xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $int_1^inftyfraccos^2k+1xxdx$ converges. In fact, the sequence
$$a_m = int_mpi - pi/2^mpi + pi/2fraccos^2k+1xxdx $$
converges to $0$ and the series
$$Sigma_m=1^infty a_m$$
is alternating and therefore converges.
answered Jul 27 at 11:09
Paul Frost
3,611420
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
add a comment |Â
up vote
1
down vote
accepted
You can explicitly compute the integral $int_a^bcos^nxdx$. Define $J(n) = intcos^nxdx$ (indefinite integral). Then you get
$$J(n) = sin x fraccos^n-1xn + fracn-1nJ(n-2) .$$
This gives a recursion ending with $J(1) = sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = sin x cdot p_n-1(cos x) + a_0x$$
where $p_n-1$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ int_1^bcos^2k+1xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $int_1^inftyfraccos^2k+1xxdx$ converges. In fact, the sequence
$$a_m = int_mpi - pi/2^mpi + pi/2fraccos^2k+1xxdx $$
converges to $0$ and the series
$$Sigma_m=1^infty a_m$$
is alternating and therefore converges.
answered Jul 27 at 11:09
Paul Frost
3,611420
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can explicitly compute the integral $int_a^bcos^nxdx$. Define $J(n) = intcos^nxdx$ (indefinite integral). Then you get
$$J(n) = sin x fraccos^n-1xn + fracn-1nJ(n-2) .$$
This gives a recursion ending with $J(1) = sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = sin x cdot p_n-1(cos x) + a_0x$$
where $p_n-1$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ int_1^bcos^2k+1xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $int_1^inftyfraccos^2k+1xxdx$ converges. In fact, the sequence
$$a_m = int_mpi - pi/2^mpi + pi/2fraccos^2k+1xxdx $$
converges to $0$ and the series
$$Sigma_m=1^infty a_m$$
is alternating and therefore converges.
answered Jul 27 at 11:09
Paul Frost
3,611420
You can explicitly compute the integral $int_a^bcos^nxdx$. Define $J(n) = intcos^nxdx$ (indefinite integral). Then you get
$$J(n) = sin x fraccos^n-1xn + fracn-1nJ(n-2) .$$
This gives a recursion ending with $J(1) = sin x$ or $J(0) = x$. You can conclude that $J(n)$ has the form
$$J(n) = sin x cdot p_n-1(cos x) + a_0x$$
where $p_n-1$ is a polynomial of degree $n-1$ and $a_0 = 0$ if and only if $n$ is odd. Therefore $J(2k+1)$ is bounded for all $k$ and so is
$$ int_1^bcos^2k+1xdx = J(2k+1)(b) - J(2k+1)(1) .$$
But, honestly, I am not sure whether this is neater than dividing the integral into periodic regions.
Using the dividing approach you can also directly show (without Dirichlet's criterion) that $int_1^inftyfraccos^2k+1xxdx$ converges. In fact, the sequence
$$a_m = int_mpi - pi/2^mpi + pi/2fraccos^2k+1xxdx $$
converges to $0$ and the series
$$Sigma_m=1^infty a_m$$
is alternating and therefore converges.
answered Jul 27 at 11:09
Paul Frost
3,611420
answered Jul 27 at 11:09
Paul Frost
3,611420
answered Jul 27 at 11:09
Paul Frost
3,611420
answered Jul 27 at 11:09
Paul Frost
3,611420
3,611420
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
add a comment |Â
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
How did you come up with the recursive integral function?
– ikoikoia
Jul 27 at 20:08
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
This follows from partial integration: Set $u(x) = sin x, v(x) = cos^n-1(x)$ and compute $int u'(x) cdot v(x) dx$.
– Paul Frost
Jul 27 at 20:57
add a comment |Â
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oh that 2010 exam question...
– Bernard W
Jul 27 at 8:32