My math prof does not have a proof for this: Existence of improper integrals for $lim_{kto infty}...












1












$begingroup$


In our analysis lecture our professor talked about the following assumption to which he said, that he does not have a proof:




If for all sequences $(a_k)_{kin mathbb{N}},(b_k)_{kin mathbb{N}}$
with $a_kto -infty, b_kto infty$ already $lim_{kto infty}
int_{a_k}^{b_k} f(x),mathrm{d}x=alpha$ holds (meaning that the
limit is distinct and it exists), then the following improper integrals
exist as well: $int_0^infty f(x),mathrm{d}x$ and $int_{-infty}^0
f(x)mathrm{d}x$ and it is $alpha=int_{-infty}^0
f(x)mathrm{d}x+int_0^infty f(x),mathrm{d}x$.




Now the problem:



We only presuppose the convergence of $int_{a_k}^{b_k} f(x),mathrm{d}x$
. In doing so, constant sequences $a_k$ or $b_k$ are excluded since $a_kto -infty$ and $b_kto infty$ have to hold.



How does the convergence of $ int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ follow now? Needless to say, the $0$ can of course be replaced by another constant.



If we assume the existence of two sequences $a_k$ or $b_k$, for which that does not apply, then it's probably possible to do a proof by contradiction (if one skillfully considers subsequences). Most likely a lot of cases have to be distinguished, because one does not know why both part-integrals $int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ not converge. They could be unbounded, or oscillating (or jump back and forth) or both.



Does someone here know how to prove the assumption or know where I can find such a proof on the internet? (Because my professor said that he doesn't have an obvious solution to this question)






integration sequences-and-series analysis proof-writing improper-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
    $endgroup$
    – Michael
    Dec 18 '18 at 12:37








  • 1




    $begingroup$
    I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
    $endgroup$
    – David K
    Dec 18 '18 at 13:20


















1












$begingroup$


In our analysis lecture our professor talked about the following assumption to which he said, that he does not have a proof:




If for all sequences $(a_k)_{kin mathbb{N}},(b_k)_{kin mathbb{N}}$
with $a_kto -infty, b_kto infty$ already $lim_{kto infty}
int_{a_k}^{b_k} f(x),mathrm{d}x=alpha$ holds (meaning that the
limit is distinct and it exists), then the following improper integrals
exist as well: $int_0^infty f(x),mathrm{d}x$ and $int_{-infty}^0
f(x)mathrm{d}x$ and it is $alpha=int_{-infty}^0
f(x)mathrm{d}x+int_0^infty f(x),mathrm{d}x$.




Now the problem:



We only presuppose the convergence of $int_{a_k}^{b_k} f(x),mathrm{d}x$
. In doing so, constant sequences $a_k$ or $b_k$ are excluded since $a_kto -infty$ and $b_kto infty$ have to hold.



How does the convergence of $ int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ follow now? Needless to say, the $0$ can of course be replaced by another constant.



If we assume the existence of two sequences $a_k$ or $b_k$, for which that does not apply, then it's probably possible to do a proof by contradiction (if one skillfully considers subsequences). Most likely a lot of cases have to be distinguished, because one does not know why both part-integrals $int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ not converge. They could be unbounded, or oscillating (or jump back and forth) or both.



Does someone here know how to prove the assumption or know where I can find such a proof on the internet? (Because my professor said that he doesn't have an obvious solution to this question)






integration sequences-and-series analysis proof-writing improper-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
    $endgroup$
    – Michael
    Dec 18 '18 at 12:37








  • 1




    $begingroup$
    I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
    $endgroup$
    – David K
    Dec 18 '18 at 13:20
















1












1








1


1



$begingroup$


In our analysis lecture our professor talked about the following assumption to which he said, that he does not have a proof:




If for all sequences $(a_k)_{kin mathbb{N}},(b_k)_{kin mathbb{N}}$
with $a_kto -infty, b_kto infty$ already $lim_{kto infty}
int_{a_k}^{b_k} f(x),mathrm{d}x=alpha$ holds (meaning that the
limit is distinct and it exists), then the following improper integrals
exist as well: $int_0^infty f(x),mathrm{d}x$ and $int_{-infty}^0
f(x)mathrm{d}x$ and it is $alpha=int_{-infty}^0
f(x)mathrm{d}x+int_0^infty f(x),mathrm{d}x$.




Now the problem:



We only presuppose the convergence of $int_{a_k}^{b_k} f(x),mathrm{d}x$
. In doing so, constant sequences $a_k$ or $b_k$ are excluded since $a_kto -infty$ and $b_kto infty$ have to hold.



How does the convergence of $ int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ follow now? Needless to say, the $0$ can of course be replaced by another constant.



If we assume the existence of two sequences $a_k$ or $b_k$, for which that does not apply, then it's probably possible to do a proof by contradiction (if one skillfully considers subsequences). Most likely a lot of cases have to be distinguished, because one does not know why both part-integrals $int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ not converge. They could be unbounded, or oscillating (or jump back and forth) or both.



Does someone here know how to prove the assumption or know where I can find such a proof on the internet? (Because my professor said that he doesn't have an obvious solution to this question)






integration sequences-and-series analysis proof-writing improper-integrals







share|cite|improve this question











$endgroup$




In our analysis lecture our professor talked about the following assumption to which he said, that he does not have a proof:




If for all sequences $(a_k)_{kin mathbb{N}},(b_k)_{kin mathbb{N}}$
with $a_kto -infty, b_kto infty$ already $lim_{kto infty}
int_{a_k}^{b_k} f(x),mathrm{d}x=alpha$ holds (meaning that the
limit is distinct and it exists), then the following improper integrals
exist as well: $int_0^infty f(x),mathrm{d}x$ and $int_{-infty}^0
f(x)mathrm{d}x$ and it is $alpha=int_{-infty}^0
f(x)mathrm{d}x+int_0^infty f(x),mathrm{d}x$.




Now the problem:



We only presuppose the convergence of $int_{a_k}^{b_k} f(x),mathrm{d}x$
. In doing so, constant sequences $a_k$ or $b_k$ are excluded since $a_kto -infty$ and $b_kto infty$ have to hold.



How does the convergence of $ int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ follow now? Needless to say, the $0$ can of course be replaced by another constant.



If we assume the existence of two sequences $a_k$ or $b_k$, for which that does not apply, then it's probably possible to do a proof by contradiction (if one skillfully considers subsequences). Most likely a lot of cases have to be distinguished, because one does not know why both part-integrals $int_{0}^{b_k} f(x),mathrm{d}x$ and $int_{a_k}^{0} f(x),mathrm{d}x$ not converge. They could be unbounded, or oscillating (or jump back and forth) or both.



Does someone here know how to prove the assumption or know where I can find such a proof on the internet? (Because my professor said that he doesn't have an obvious solution to this question)






integration sequences-and-series analysis proof-writing improper-integrals




integration sequences-and-series analysis proof-writing improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 12:48







NotEinstein

















asked Dec 18 '18 at 12:09









NotEinsteinNotEinstein

2146




2146








  • 1




    $begingroup$
    Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
    $endgroup$
    – Michael
    Dec 18 '18 at 12:37








  • 1




    $begingroup$
    I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
    $endgroup$
    – David K
    Dec 18 '18 at 13:20
















  • 1




    $begingroup$
    Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
    $endgroup$
    – Michael
    Dec 18 '18 at 12:37








  • 1




    $begingroup$
    I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
    $endgroup$
    – David K
    Dec 18 '18 at 13:20










1




1




$begingroup$
Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
$endgroup$
– Michael
Dec 18 '18 at 12:37






$begingroup$
Suppose there is an $epsilon>0$ such that for all $n in mathbb{N}$ there is a value $c_n>n$ such that $|int_n^{c_n} f(x)dx|>epsilon$. So consider $int_{-n}^{c_n} f(x)dx = int_{-n}^{n} f(x)dx + int_n^{c_n}f(x)dx$ as $nrightarrowinfty$.
$endgroup$
– Michael
Dec 18 '18 at 12:37






1




1




$begingroup$
I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
$endgroup$
– David K
Dec 18 '18 at 13:20






$begingroup$
I think it would depend on how you treat certain improper integrals. If you allow Cauchy principal value integrals, then you could have (for example) $int_{-1}^1f(x),dx=0$ while $int_{-1}^0f(x),dx$ and $int_0^1f(x),dx$ both are undefined.
$endgroup$
– David K
Dec 18 '18 at 13:20












1 Answer
1






active

oldest

votes


















4












$begingroup$

Let $b_k, c_k geq 0$ be sequences such that $lim_{ktoinfty} b_k = lim_{ktoinfty} c_k = infty$ and suppose that $b_k < c_k$ for all $k in mathbb{N}$. By assumption,



$$lim_{ktoinfty} int_{b_k}^{c_k} f(x)~mathrm dx = lim_{ktoinfty}left( int_{-k}^{c_k} f(x)~mathrm dx - int_{-k}^{b_k} f(x)~mathrm dx right) = alpha - alpha = 0.$$



By the Cauchy criterion (cf here), this shows that $$lim_{ytoinfty} int_{0}^{y} f(x)~mathrm d x $$
exists.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
    $endgroup$
    – David K
    Dec 18 '18 at 13:32










  • $begingroup$
    Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
    $endgroup$
    – user159517
    Dec 18 '18 at 19:19










  • $begingroup$
    You want $-k$ instead of $-n$ above. Nice solution, +1.
    $endgroup$
    – zhw.
    Dec 18 '18 at 19:31










  • $begingroup$
    @zhw of course, thanks.
    $endgroup$
    – user159517
    Dec 18 '18 at 19:32












  • $begingroup$
    I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
    $endgroup$
    – David K
    Dec 18 '18 at 23:23











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Let $b_k, c_k geq 0$ be sequences such that $lim_{ktoinfty} b_k = lim_{ktoinfty} c_k = infty$ and suppose that $b_k < c_k$ for all $k in mathbb{N}$. By assumption,



$$lim_{ktoinfty} int_{b_k}^{c_k} f(x)~mathrm dx = lim_{ktoinfty}left( int_{-k}^{c_k} f(x)~mathrm dx - int_{-k}^{b_k} f(x)~mathrm dx right) = alpha - alpha = 0.$$



By the Cauchy criterion (cf here), this shows that $$lim_{ytoinfty} int_{0}^{y} f(x)~mathrm d x $$
exists.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
    $endgroup$
    – David K
    Dec 18 '18 at 13:32










  • $begingroup$
    Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
    $endgroup$
    – user159517
    Dec 18 '18 at 19:19










  • $begingroup$
    You want $-k$ instead of $-n$ above. Nice solution, +1.
    $endgroup$
    – zhw.
    Dec 18 '18 at 19:31










  • $begingroup$
    @zhw of course, thanks.
    $endgroup$
    – user159517
    Dec 18 '18 at 19:32












  • $begingroup$
    I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
    $endgroup$
    – David K
    Dec 18 '18 at 23:23
















4












$begingroup$

Let $b_k, c_k geq 0$ be sequences such that $lim_{ktoinfty} b_k = lim_{ktoinfty} c_k = infty$ and suppose that $b_k < c_k$ for all $k in mathbb{N}$. By assumption,



$$lim_{ktoinfty} int_{b_k}^{c_k} f(x)~mathrm dx = lim_{ktoinfty}left( int_{-k}^{c_k} f(x)~mathrm dx - int_{-k}^{b_k} f(x)~mathrm dx right) = alpha - alpha = 0.$$



By the Cauchy criterion (cf here), this shows that $$lim_{ytoinfty} int_{0}^{y} f(x)~mathrm d x $$
exists.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
    $endgroup$
    – David K
    Dec 18 '18 at 13:32










  • $begingroup$
    Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
    $endgroup$
    – user159517
    Dec 18 '18 at 19:19










  • $begingroup$
    You want $-k$ instead of $-n$ above. Nice solution, +1.
    $endgroup$
    – zhw.
    Dec 18 '18 at 19:31










  • $begingroup$
    @zhw of course, thanks.
    $endgroup$
    – user159517
    Dec 18 '18 at 19:32












  • $begingroup$
    I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
    $endgroup$
    – David K
    Dec 18 '18 at 23:23














4












4








4





$begingroup$

Let $b_k, c_k geq 0$ be sequences such that $lim_{ktoinfty} b_k = lim_{ktoinfty} c_k = infty$ and suppose that $b_k < c_k$ for all $k in mathbb{N}$. By assumption,



$$lim_{ktoinfty} int_{b_k}^{c_k} f(x)~mathrm dx = lim_{ktoinfty}left( int_{-k}^{c_k} f(x)~mathrm dx - int_{-k}^{b_k} f(x)~mathrm dx right) = alpha - alpha = 0.$$



By the Cauchy criterion (cf here), this shows that $$lim_{ytoinfty} int_{0}^{y} f(x)~mathrm d x $$
exists.






share|cite|improve this answer











$endgroup$



Let $b_k, c_k geq 0$ be sequences such that $lim_{ktoinfty} b_k = lim_{ktoinfty} c_k = infty$ and suppose that $b_k < c_k$ for all $k in mathbb{N}$. By assumption,



$$lim_{ktoinfty} int_{b_k}^{c_k} f(x)~mathrm dx = lim_{ktoinfty}left( int_{-k}^{c_k} f(x)~mathrm dx - int_{-k}^{b_k} f(x)~mathrm dx right) = alpha - alpha = 0.$$



By the Cauchy criterion (cf here), this shows that $$lim_{ytoinfty} int_{0}^{y} f(x)~mathrm d x $$
exists.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 18 '18 at 19:32

























answered Dec 18 '18 at 13:27









user159517user159517

4,575931




4,575931








  • 1




    $begingroup$
    How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
    $endgroup$
    – David K
    Dec 18 '18 at 13:32










  • $begingroup$
    Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
    $endgroup$
    – user159517
    Dec 18 '18 at 19:19










  • $begingroup$
    You want $-k$ instead of $-n$ above. Nice solution, +1.
    $endgroup$
    – zhw.
    Dec 18 '18 at 19:31










  • $begingroup$
    @zhw of course, thanks.
    $endgroup$
    – user159517
    Dec 18 '18 at 19:32












  • $begingroup$
    I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
    $endgroup$
    – David K
    Dec 18 '18 at 23:23














  • 1




    $begingroup$
    How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
    $endgroup$
    – David K
    Dec 18 '18 at 13:32










  • $begingroup$
    Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
    $endgroup$
    – user159517
    Dec 18 '18 at 19:19










  • $begingroup$
    You want $-k$ instead of $-n$ above. Nice solution, +1.
    $endgroup$
    – zhw.
    Dec 18 '18 at 19:31










  • $begingroup$
    @zhw of course, thanks.
    $endgroup$
    – user159517
    Dec 18 '18 at 19:32












  • $begingroup$
    I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
    $endgroup$
    – David K
    Dec 18 '18 at 23:23








1




1




$begingroup$
How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
$endgroup$
– David K
Dec 18 '18 at 13:32




$begingroup$
How would this proof treat a function such as $f(x)=1/x^3,$ $f(0)=0$?
$endgroup$
– David K
Dec 18 '18 at 13:32












$begingroup$
Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
$endgroup$
– user159517
Dec 18 '18 at 19:19




$begingroup$
Good point. I've edited my question to include a link to a definition of improper integral, one of its requirements is that the function should be integrable on every subinterval of the region that you define the improper integral over, so in our case for all $[a,b]subseteq (-infty,infty)$. It seems to me that an assumption of this kind is usually incorporated into the definition of the improper Riemann integral. Would you disagree?
$endgroup$
– user159517
Dec 18 '18 at 19:19












$begingroup$
You want $-k$ instead of $-n$ above. Nice solution, +1.
$endgroup$
– zhw.
Dec 18 '18 at 19:31




$begingroup$
You want $-k$ instead of $-n$ above. Nice solution, +1.
$endgroup$
– zhw.
Dec 18 '18 at 19:31












$begingroup$
@zhw of course, thanks.
$endgroup$
– user159517
Dec 18 '18 at 19:32






$begingroup$
@zhw of course, thanks.
$endgroup$
– user159517
Dec 18 '18 at 19:32














$begingroup$
I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
$endgroup$
– David K
Dec 18 '18 at 23:23




$begingroup$
I think you have the right approach here. The point is that it comes down to how the integrals are defined in the first place, which the question does not completely specify. Your way seems fairly standard, and if they do it the same, they'll be OK.
$endgroup$
– David K
Dec 18 '18 at 23:23


















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