Limit of integral sequence
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I am asked to prove that
$$lim_{ntoinfty}int_0^{frac{pi}{2}}sin(n^4cos t)dt=0$$
I think you have to calculate
$$int_0^{frac{1}{n}}sin(n^4cos t)dt$$
and
$$int_{frac{1}{n}}^{frac{pi}{2}}sin(n^4cos t)dt$$
So that you can somehow integrate the second one by parts, but I can't quite figure out how.
real-analysis limits definite-integrals
asked Nov 12 at 7:50
Andrei Cataron
1449
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up vote
0
down vote
favorite
I am asked to prove that
$$lim_{ntoinfty}int_0^{frac{pi}{2}}sin(n^4cos t)dt=0$$
I think you have to calculate
$$int_0^{frac{1}{n}}sin(n^4cos t)dt$$
and
$$int_{frac{1}{n}}^{frac{pi}{2}}sin(n^4cos t)dt$$
So that you can somehow integrate the second one by parts, but I can't quite figure out how.
real-analysis limits definite-integrals
asked Nov 12 at 7:50
Andrei Cataron
1449
Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am asked to prove that
$$lim_{ntoinfty}int_0^{frac{pi}{2}}sin(n^4cos t)dt=0$$
I think you have to calculate
$$int_0^{frac{1}{n}}sin(n^4cos t)dt$$
and
$$int_{frac{1}{n}}^{frac{pi}{2}}sin(n^4cos t)dt$$
So that you can somehow integrate the second one by parts, but I can't quite figure out how.
real-analysis limits definite-integrals
asked Nov 12 at 7:50
Andrei Cataron
1449
I am asked to prove that
$$lim_{ntoinfty}int_0^{frac{pi}{2}}sin(n^4cos t)dt=0$$
I think you have to calculate
$$int_0^{frac{1}{n}}sin(n^4cos t)dt$$
and
$$int_{frac{1}{n}}^{frac{pi}{2}}sin(n^4cos t)dt$$
So that you can somehow integrate the second one by parts, but I can't quite figure out how.
real-analysis limits definite-integrals
real-analysis limits definite-integrals
asked Nov 12 at 7:50
Andrei Cataron
1449
asked Nov 12 at 7:50
Andrei Cataron
1449
asked Nov 12 at 7:50
Andrei Cataron
1449
asked Nov 12 at 7:50
Andrei Cataron
1449
asked Nov 12 at 7:50
Andrei Cataron
1449
1449
Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21
add a comment |
Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21
Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21
Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21
add a comment |
1 Answer
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2
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$$int_{0}^{pi/2}sinleft(mcos tright),dt stackrel{tmapstofrac{pi}{2}-t}{=}int_{0}^{pi/2}sin(msin t),dtstackrel{tmapstoarcsin u}{=}int_{0}^{1}frac{sin(mu)}{sqrt{1-u^2}},du$$
and now we may invoke the Riemann-Lebesgue lemma.
Since $int_{0}^{1}frac{du}{sqrt{1-u^2}},du$ is finite, the limit of the LHS as $mto +infty$ is zero.
The LHS actually is $frac{pi}{2} H_0(m)$, with $H_0$ being a Struve function.
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$int_{0}^{pi/2}sinleft(mcos tright),dt stackrel{tmapstofrac{pi}{2}-t}{=}int_{0}^{pi/2}sin(msin t),dtstackrel{tmapstoarcsin u}{=}int_{0}^{1}frac{sin(mu)}{sqrt{1-u^2}},du$$
and now we may invoke the Riemann-Lebesgue lemma.
Since $int_{0}^{1}frac{du}{sqrt{1-u^2}},du$ is finite, the limit of the LHS as $mto +infty$ is zero.
The LHS actually is $frac{pi}{2} H_0(m)$, with $H_0$ being a Struve function.
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
add a comment |
up vote
2
down vote
$$int_{0}^{pi/2}sinleft(mcos tright),dt stackrel{tmapstofrac{pi}{2}-t}{=}int_{0}^{pi/2}sin(msin t),dtstackrel{tmapstoarcsin u}{=}int_{0}^{1}frac{sin(mu)}{sqrt{1-u^2}},du$$
and now we may invoke the Riemann-Lebesgue lemma.
Since $int_{0}^{1}frac{du}{sqrt{1-u^2}},du$ is finite, the limit of the LHS as $mto +infty$ is zero.
The LHS actually is $frac{pi}{2} H_0(m)$, with $H_0$ being a Struve function.
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
add a comment |
up vote
2
down vote
up vote
2
down vote
$$int_{0}^{pi/2}sinleft(mcos tright),dt stackrel{tmapstofrac{pi}{2}-t}{=}int_{0}^{pi/2}sin(msin t),dtstackrel{tmapstoarcsin u}{=}int_{0}^{1}frac{sin(mu)}{sqrt{1-u^2}},du$$
and now we may invoke the Riemann-Lebesgue lemma.
Since $int_{0}^{1}frac{du}{sqrt{1-u^2}},du$ is finite, the limit of the LHS as $mto +infty$ is zero.
The LHS actually is $frac{pi}{2} H_0(m)$, with $H_0$ being a Struve function.
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
$$int_{0}^{pi/2}sinleft(mcos tright),dt stackrel{tmapstofrac{pi}{2}-t}{=}int_{0}^{pi/2}sin(msin t),dtstackrel{tmapstoarcsin u}{=}int_{0}^{1}frac{sin(mu)}{sqrt{1-u^2}},du$$
and now we may invoke the Riemann-Lebesgue lemma.
Since $int_{0}^{1}frac{du}{sqrt{1-u^2}},du$ is finite, the limit of the LHS as $mto +infty$ is zero.
The LHS actually is $frac{pi}{2} H_0(m)$, with $H_0$ being a Struve function.
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
answered Nov 12 at 13:20
Jack D'Aurizio
282k33274653
282k33274653
add a comment |
add a comment |
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Riemann Lebesgue lemma is the key here.
– Paramanand Singh
Nov 12 at 8:21