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Fresnel integrals are not Lebesgue integrable
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How does one prove that the Fresnel integrals
$$
S(x) = int_0^x ! sin(t^2) , mathrmdt, qquad C(x) = int_0^x ! cos(t^2) , mathrmdt
$$
do not belong to $L^1(mathbbR)$?
real-analysis integration lebesgue-integral
asked Jul 25 at 13:16
sleepingrabbit
1197
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up vote
0
down vote
favorite
How does one prove that the Fresnel integrals
$$
S(x) = int_0^x ! sin(t^2) , mathrmdt, qquad C(x) = int_0^x ! cos(t^2) , mathrmdt
$$
do not belong to $L^1(mathbbR)$?
real-analysis integration lebesgue-integral
asked Jul 25 at 13:16
sleepingrabbit
1197
1
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How does one prove that the Fresnel integrals
$$
S(x) = int_0^x ! sin(t^2) , mathrmdt, qquad C(x) = int_0^x ! cos(t^2) , mathrmdt
$$
do not belong to $L^1(mathbbR)$?
real-analysis integration lebesgue-integral
asked Jul 25 at 13:16
sleepingrabbit
1197
How does one prove that the Fresnel integrals
$$
S(x) = int_0^x ! sin(t^2) , mathrmdt, qquad C(x) = int_0^x ! cos(t^2) , mathrmdt
$$
do not belong to $L^1(mathbbR)$?
real-analysis integration lebesgue-integral
asked Jul 25 at 13:16
sleepingrabbit
1197
asked Jul 25 at 13:16
sleepingrabbit
1197
asked Jul 25 at 13:16
sleepingrabbit
1197
asked Jul 25 at 13:16
sleepingrabbit
1197
1197
1
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40
add a comment |Â
1
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40
1
1
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Since $lim_lvert x rvert to infty lvert S(x) rvert = sqrtpi/8$, we can find $M in mathbb R$ such that for $lvert x rvert > M$, we have $$lvert S(x) rvert ge sqrtpi/16.$$ But then $$int^infty_-infty lvert S(x) rvert dx ge int^M_-M lvert S(x) rvert dx + int^infty_M sqrtpi/16 ,,dx + int^-M_-infty sqrtpi/16,, dx = infty$$ and similarly for $C(x)$.
answered Jul 25 at 13:44
User8128
10.2k1522
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Since $lim_lvert x rvert to infty lvert S(x) rvert = sqrtpi/8$, we can find $M in mathbb R$ such that for $lvert x rvert > M$, we have $$lvert S(x) rvert ge sqrtpi/16.$$ But then $$int^infty_-infty lvert S(x) rvert dx ge int^M_-M lvert S(x) rvert dx + int^infty_M sqrtpi/16 ,,dx + int^-M_-infty sqrtpi/16,, dx = infty$$ and similarly for $C(x)$.
answered Jul 25 at 13:44
User8128
10.2k1522
add a comment |Â
up vote
4
down vote
accepted
Since $lim_lvert x rvert to infty lvert S(x) rvert = sqrtpi/8$, we can find $M in mathbb R$ such that for $lvert x rvert > M$, we have $$lvert S(x) rvert ge sqrtpi/16.$$ But then $$int^infty_-infty lvert S(x) rvert dx ge int^M_-M lvert S(x) rvert dx + int^infty_M sqrtpi/16 ,,dx + int^-M_-infty sqrtpi/16,, dx = infty$$ and similarly for $C(x)$.
answered Jul 25 at 13:44
User8128
10.2k1522
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Since $lim_lvert x rvert to infty lvert S(x) rvert = sqrtpi/8$, we can find $M in mathbb R$ such that for $lvert x rvert > M$, we have $$lvert S(x) rvert ge sqrtpi/16.$$ But then $$int^infty_-infty lvert S(x) rvert dx ge int^M_-M lvert S(x) rvert dx + int^infty_M sqrtpi/16 ,,dx + int^-M_-infty sqrtpi/16,, dx = infty$$ and similarly for $C(x)$.
answered Jul 25 at 13:44
User8128
10.2k1522
Since $lim_lvert x rvert to infty lvert S(x) rvert = sqrtpi/8$, we can find $M in mathbb R$ such that for $lvert x rvert > M$, we have $$lvert S(x) rvert ge sqrtpi/16.$$ But then $$int^infty_-infty lvert S(x) rvert dx ge int^M_-M lvert S(x) rvert dx + int^infty_M sqrtpi/16 ,,dx + int^-M_-infty sqrtpi/16,, dx = infty$$ and similarly for $C(x)$.
answered Jul 25 at 13:44
User8128
10.2k1522
answered Jul 25 at 13:44
User8128
10.2k1522
answered Jul 25 at 13:44
User8128
10.2k1522
answered Jul 25 at 13:44
User8128
10.2k1522
10.2k1522
add a comment |Â
add a comment |Â
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1
Do you mean to show that $$int^infty_0lvert sin(t^2) rvert dt ,,,,,, text and ,,, int^infty_0 lvert cos(t^2) rvert dt$$ diverge or that $$int_-infty^infty lvert S(x) rvert dx ,,,,, text and ,,,,, int^infty_-infty lvert C(x) rvert dx$$ diverge?
– User8128
Jul 25 at 13:36
The latter. I would like to show that $S, C notin L^1(mathbbR)$.
– sleepingrabbit
Jul 25 at 13:40