Prove that integral of $ 1-cosleft(frac{1}{x^2}right) $ is finite
I need to prove that
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$
My attempt:
$$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.
So I think that, there exist $N in mathbb{N} $ such that:
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$
but I'm not sure. I would appreciate any tips or hints.
calculus integration proof-verification
edited Dec 2 '18 at 14:34
Larry
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
add a comment |
I need to prove that
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$
My attempt:
$$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.
So I think that, there exist $N in mathbb{N} $ such that:
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$
but I'm not sure. I would appreciate any tips or hints.
calculus integration proof-verification
edited Dec 2 '18 at 14:34
Larry
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
add a comment |
I need to prove that
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$
My attempt:
$$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.
So I think that, there exist $N in mathbb{N} $ such that:
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$
but I'm not sure. I would appreciate any tips or hints.
calculus integration proof-verification
edited Dec 2 '18 at 14:34
Larry
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
I need to prove that
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right) dx < infty $$
My attempt:
$$ forall xin[0,infty] hspace{1cm} 0 < 1 - cosleft(frac{1}{x^2}right) < 2 tag{1}. $$
Using L'Hôpital's rule I can show that:
$$ 1 - cosleft(frac{1}{x^2}right) underset{x to infty}{sim} frac{1}{2x^4} tag{2} $$
Which means that $ 1 - cosleft(frac{1}{x^2}right)$ behaves like $ frac{1}{2x^4} $ when $ x to infty $.
So I think that, there exist $N in mathbb{N} $ such that:
$$ int_0^{infty} left(1 - cosleft(frac{1}{x^2}right)right)dx < int_0^{N} 2dx + int_N^{infty} frac{1}{x^4}dx < infty $$
but I'm not sure. I would appreciate any tips or hints.
calculus integration proof-verification
calculus integration proof-verification
edited Dec 2 '18 at 14:34
Larry
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
edited Dec 2 '18 at 14:34
Larry
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
edited Dec 2 '18 at 14:34
Larry
1,7232823
edited Dec 2 '18 at 14:34
Larry
1,7232823
edited Dec 2 '18 at 14:34
Larry
1,7232823
1,7232823
asked Dec 2 '18 at 14:01
Wywana
535
asked Dec 2 '18 at 14:01
Wywana
535
asked Dec 2 '18 at 14:01
Wywana
535
535
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
Use the substitution $1/x=t$, so the integral becomes
$$
int_0^{infty} frac{1-cos(t^2)}{t^2},dt
$$
Since
$$
lim_{tto0}frac{1-cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0le 1-cos(t^2)le 2
$$
we have that, for $tge1$,
$$
0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
$$
Now show convergence of
$$
int_1^infty frac{2}{t^2},dt
$$
answered Dec 2 '18 at 15:22
egreg
178k1484201
add a comment |
$$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
leads to $I=color{red}{sqrt{frac{pi}{2}}}$.
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
add a comment |
$int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$
answered Dec 2 '18 at 14:22
John_Wick
1,381111
add a comment |
One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get
begin{multline}
int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
\ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
Use the substitution $1/x=t$, so the integral becomes
$$
int_0^{infty} frac{1-cos(t^2)}{t^2},dt
$$
Since
$$
lim_{tto0}frac{1-cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0le 1-cos(t^2)le 2
$$
we have that, for $tge1$,
$$
0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
$$
Now show convergence of
$$
int_1^infty frac{2}{t^2},dt
$$
answered Dec 2 '18 at 15:22
egreg
178k1484201
add a comment |
Use the substitution $1/x=t$, so the integral becomes
$$
int_0^{infty} frac{1-cos(t^2)}{t^2},dt
$$
Since
$$
lim_{tto0}frac{1-cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0le 1-cos(t^2)le 2
$$
we have that, for $tge1$,
$$
0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
$$
Now show convergence of
$$
int_1^infty frac{2}{t^2},dt
$$
answered Dec 2 '18 at 15:22
egreg
178k1484201
add a comment |
Use the substitution $1/x=t$, so the integral becomes
$$
int_0^{infty} frac{1-cos(t^2)}{t^2},dt
$$
Since
$$
lim_{tto0}frac{1-cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0le 1-cos(t^2)le 2
$$
we have that, for $tge1$,
$$
0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
$$
Now show convergence of
$$
int_1^infty frac{2}{t^2},dt
$$
answered Dec 2 '18 at 15:22
egreg
178k1484201
Use the substitution $1/x=t$, so the integral becomes
$$
int_0^{infty} frac{1-cos(t^2)}{t^2},dt
$$
Since
$$
lim_{tto0}frac{1-cos(t^2)}{t^2}=0
$$
convergence is not an issue at $0$. Since
$$
0le 1-cos(t^2)le 2
$$
we have that, for $tge1$,
$$
0lefrac{1-cos(t^2)}{t^2}lefrac{2}{t^2}
$$
Now show convergence of
$$
int_1^infty frac{2}{t^2},dt
$$
answered Dec 2 '18 at 15:22
egreg
178k1484201
answered Dec 2 '18 at 15:22
egreg
178k1484201
answered Dec 2 '18 at 15:22
egreg
178k1484201
answered Dec 2 '18 at 15:22
egreg
178k1484201
178k1484201
add a comment |
add a comment |
$$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
leads to $I=color{red}{sqrt{frac{pi}{2}}}$.
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
add a comment |
$$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
leads to $I=color{red}{sqrt{frac{pi}{2}}}$.
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
add a comment |
$$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
leads to $I=color{red}{sqrt{frac{pi}{2}}}$.
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
$$I=int_{0}^{+infty}left[1-cosleft(tfrac{1}{x^2}right)right],dx stackrel{xmapsto 1/x}{=} int_{0}^{+infty}frac{1-cos(x^2)}{x^2},dx $$
where the function $frac{1-cos(x^2)}{x^2}$ is continuous and bounded over $(0,1]$, non-negative and bounded by $frac{2}{x^2}$ over $[1,+infty)$. It follows that the above integral is finite. Its value can be found through the Laplace transform:
$$int_{0}^{+infty}frac{1-cos x}{2xsqrt{x}},dx!stackrel{text{IBP}}{=}!int_{0}^{+infty}frac{sin x}{sqrt{x}},dx!stackrel{mathcal{L}}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{ds}{(s^2+1)sqrt{s}}!stackrel{smapsto t^2}{=}!frac{1}{sqrt{pi}}!int_{0}^{+infty}frac{2,dt}{t^4+1} $$
leads to $I=color{red}{sqrt{frac{pi}{2}}}$.
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
answered Dec 2 '18 at 17:24
Jack D'Aurizio
287k33280657
287k33280657
add a comment |
add a comment |
$int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$
answered Dec 2 '18 at 14:22
John_Wick
1,381111
add a comment |
$int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$
answered Dec 2 '18 at 14:22
John_Wick
1,381111
add a comment |
$int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$
answered Dec 2 '18 at 14:22
John_Wick
1,381111
$int_0^infty(1-cos (1/x^2))dx= int_0^1(1-cos (1/x^2))dx+int_1^infty(1-cos (1/x^2))dxleq 2+int_1^infty2sin^2 left(frac{1}{2x^2}right)dxleq 2+int_1^infty2sin left(frac{1}{2x^2}right)dxleq2+int_1^infty2 frac{1}{2x^2}dx text{ (as } sin xleq x text{ forall } x>0) =2+int_1^infty frac{1}{x^2}dx .$
answered Dec 2 '18 at 14:22
John_Wick
1,381111
answered Dec 2 '18 at 14:22
John_Wick
1,381111
answered Dec 2 '18 at 14:22
John_Wick
1,381111
answered Dec 2 '18 at 14:22
John_Wick
1,381111
1,381111
add a comment |
add a comment |
One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get
begin{multline}
int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
\ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
add a comment |
One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get
begin{multline}
int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
\ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
add a comment |
One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get
begin{multline}
int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
\ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
One way is to use $ 1 - costheta = 2sin^2theta$ and $u = x^{-2}$ to get
begin{multline}
int_0^inftyleft[1-cos(x^{-2})right]dx = int_0^1 2sin^2(x^{-2})dx + int_1^infty 2sin^2(u)frac{du}{2u^{3/2}}
\ = 2int_0^1sin^2(x^{-2})dx +int_0^1 u^{1/2}left[frac{sin(u)}{u}right]^2 du
end{multline}
Since both integrals are of bounded functions over a finite range, they must be finite.
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
answered Dec 2 '18 at 14:49
eyeballfrog
6,048629
6,048629
add a comment |
add a comment |
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