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How to prove that $int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite when $lim_f(t)exp(-t^2) = 0$?
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Given a function $f$ which is continuously differentiable over the real line, and given that $displaystyle lim_f(t)exp(-t^2) = 0$ how does one prove that $displaystyle int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite?
real-analysis improper-integrals
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
asked Aug 8 at 16:40
Anant Joshi
383111
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up vote
3
down vote
favorite
Given a function $f$ which is continuously differentiable over the real line, and given that $displaystyle lim_f(t)exp(-t^2) = 0$ how does one prove that $displaystyle int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite?
real-analysis improper-integrals
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
asked Aug 8 at 16:40
Anant Joshi
383111
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a function $f$ which is continuously differentiable over the real line, and given that $displaystyle lim_f(t)exp(-t^2) = 0$ how does one prove that $displaystyle int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite?
real-analysis improper-integrals
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
asked Aug 8 at 16:40
Anant Joshi
383111
Given a function $f$ which is continuously differentiable over the real line, and given that $displaystyle lim_f(t)exp(-t^2) = 0$ how does one prove that $displaystyle int_-infty^inftyf'(t)exp(-t^2)textdt$ is finite?
real-analysis improper-integrals
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
asked Aug 8 at 16:40
Anant Joshi
383111
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
edited Aug 9 at 0:09
Asaf Karagila♦
293k31409736
293k31409736
asked Aug 8 at 16:40
Anant Joshi
383111
asked Aug 8 at 16:40
Anant Joshi
383111
asked Aug 8 at 16:40
Anant Joshi
383111
383111
add a comment |Â
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
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up vote
5
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The result is wrong as stated
By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$
$A$ converges to zero as $a to infty$ by hypothesis.
Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_f(t)exp(-t^2) = 0$.
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
add a comment |Â
up vote
3
down vote
This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.
More explicitly,
$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)
answered Aug 8 at 16:58
DinosaurEgg
3157
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
add a comment |Â
up vote
6
down vote
accepted
I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
I think you can't. Just looking at the positive side of your integral, consider
$$f:tmapsto frace^t^2sqrtt+1$$
You do have $lim_tto+infty f(t)e^-t^2=0$, but
$$f'(t)e^-t^2 = frac2tsqrtt+1-frac1sqrtt+1t+1 sim_tto+infty 2sqrt t$$
which is not integrable on $[0,+infty[$.
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
answered Aug 8 at 16:54
Nicolas FRANCOIS
3,3121415
3,3121415
add a comment |Â
add a comment |Â
up vote
5
down vote
The result is wrong as stated
By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$
$A$ converges to zero as $a to infty$ by hypothesis.
Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_f(t)exp(-t^2) = 0$.
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
add a comment |Â
up vote
5
down vote
The result is wrong as stated
By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$
$A$ converges to zero as $a to infty$ by hypothesis.
Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_f(t)exp(-t^2) = 0$.
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The result is wrong as stated
By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$
$A$ converges to zero as $a to infty$ by hypothesis.
Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_f(t)exp(-t^2) = 0$.
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
The result is wrong as stated
By integration by parts for $a >0$
$$int_- a^a f^prime(t) e^-t^2 dt = underbraceleft[f(t)e^-t^2right]_-a^a_A+2underbraceint_- a^a tf(t) e^-t^2 dt_B $$
$A$ converges to zero as $a to infty$ by hypothesis.
Now take a function $f$ that vanishes for $xle 1$ and such that $int_1^infty tf(t) e^-t^2 dt$ diverges. For example $$f : x mapsto t^alpha e^t^2$$ with $-2 <alpha <0$. This will provide a counterexample as $B$ diverges for $a to infty$ while $f$ satisfies the hypothesis $displaystyle lim_f(t)exp(-t^2) = 0$.
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
edited Aug 8 at 17:25
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
answered Aug 8 at 17:11
mathcounterexamples.net
25.5k21755
25.5k21755
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
add a comment |Â
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
for $alpha<0$ and real however this function is not continuously differentiable everywhere on the real axis. Not that it really matters, but just saying.
– DinosaurEgg
Aug 8 at 17:29
add a comment |Â
up vote
3
down vote
This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.
More explicitly,
$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)
answered Aug 8 at 16:58
DinosaurEgg
3157
add a comment |Â
up vote
3
down vote
This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.
More explicitly,
$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)
answered Aug 8 at 16:58
DinosaurEgg
3157
add a comment |Â
up vote
3
down vote
up vote
3
down vote
This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.
More explicitly,
$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)
answered Aug 8 at 16:58
DinosaurEgg
3157
This cannot be true for all functions $f(t)$. Consider $f(t)=fracsin tte^t^2$. This is continuously differentiable on the real line, and $f'(t)e^-t^2=2sin t-fracsin tt^2+fraccos tt$. The $sin t$ part makes the integral oscillate indefinitely and thus diverge.
More explicitly,
$int^infty_-infty(2sin t-fracsin tt^2+fraccos tt)dt=2int^infty_-inftydtsin t+[fracsin tt]^infty_-infty=2int^infty_-inftydtsin t $ -(DNE)
answered Aug 8 at 16:58
DinosaurEgg
3157
edited Aug 8 at 19:02
answered Aug 8 at 16:58
DinosaurEgg
3157
answered Aug 8 at 16:58
DinosaurEgg
3157
answered Aug 8 at 16:58
DinosaurEgg
3157
3157
add a comment |Â
add a comment |Â
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