Calculate $lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x)$
$begingroup$
the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $
so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore
$l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $
by the dominated convergence theorem :
$l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$
I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$
any hints ?
integration measure-theory improper-integrals lebesgue-integral lebesgue-measure
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $
so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore
$l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $
by the dominated convergence theorem :
$l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$
I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$
any hints ?
integration measure-theory improper-integrals lebesgue-integral lebesgue-measure
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
$endgroup$
add a comment |
$begingroup$
the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $
so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore
$l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $
by the dominated convergence theorem :
$l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$
I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$
any hints ?
integration measure-theory improper-integrals lebesgue-integral lebesgue-measure
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
$endgroup$
the exponential function being increasing we have $| exp((cos^n x) -x)| leq exp(1 -x) in L^1([0,+infty[) $
so $x to exp((cos^n x) -x)$ is Riemann absolutely convergent therefore
$l = lim_{n to infty} int_{mathbb{R_{+}}} exp((cos x^n) -x) dlambda(x) =lim_{n to infty} int_{0}^{+infty} exp((cos^n x) -x) dx $
by the dominated convergence theorem :
$l = int_{0}^{+infty} lim_{n to infty} exp((cos^n x) -x) dx$
I don't know how to deal with this limit, as $x$ is in $mathbb{R_{+}}$ I can't even use a taylor expression around $0$
any hints ?
integration measure-theory improper-integrals lebesgue-integral lebesgue-measure
integration measure-theory improper-integrals lebesgue-integral lebesgue-measure
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
asked Jan 9 at 18:26
rapidracimrapidracim
1,7191419
1,7191419
add a comment |
add a comment |
2 Answers
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$begingroup$
For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:
$$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and
$$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.
As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that
$$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
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$begingroup$
Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:
$$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and
$$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.
As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that
$$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:
$$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and
$$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.
As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that
$$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
add a comment |
$begingroup$
For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:
$$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and
$$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.
As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that
$$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
$endgroup$
For $x in mathbb R_+ setminus{kpi + pi/2 ; k in mathbb N}$ you have:
$$exp((cos^n x) -x) to e^{-x}$$ as $n to infty$ and
$$0 le exp((cos^n x) -x) le e cdot e^{-x}$$ for all $x in mathbb R_+$.
As ${kpi + pi/2 ; k in mathbb N}$ is a null set (for Lebesgue measure) and $int_{mathbb R_+} e^{-x} dx$ converges, you can apply Lebesgue dominated convergence theorem and conclude that
$$lim_{n to infty} int_{mathbb{R_{+}}} exp((cos^n x) -x) dlambda(x) = int_{mathbb R_+} e^{-x} dx$$
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
edited Jan 9 at 21:36
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
answered Jan 9 at 18:44
mathcounterexamples.netmathcounterexamples.net
26.9k22157
26.9k22157
add a comment |
add a comment |
$begingroup$
Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
$endgroup$
add a comment |
$begingroup$
Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
$endgroup$
add a comment |
$begingroup$
Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
$endgroup$
Notice that as $n to infty$ all values of $cos^n(x)$ go to zero except the exact points at which $cos(x)=1$ which have infinitesimally small width so have a value of zero when integrating over them. The integral then becomes simply the integral of $exp(-x)$.
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
answered Jan 9 at 18:36
Peter ForemanPeter Foreman
2,5441214
2,5441214
add a comment |
add a comment |
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