Definition of $int_a^b f(x) ,mathrm{d}x$ for $f$ continuous on $[ a,b )$ and unbounded at the right-hand...
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How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?
Could anyone provide an example?
This is the definition I have for a closed and bounded interval:
Let $f$ be a continuous function on an interval $[a,b]$ Then
$$
intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
$$
I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
$$
sumlimits_{k=0}^{n-1}f(ldots)
$$
I am not really sure where to start though.
real-analysis definite-integrals
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
asked Dec 5 at 12:12
kaisa
999
add a comment |
up vote
1
down vote
favorite
How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?
Could anyone provide an example?
This is the definition I have for a closed and bounded interval:
Let $f$ be a continuous function on an interval $[a,b]$ Then
$$
intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
$$
I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
$$
sumlimits_{k=0}^{n-1}f(ldots)
$$
I am not really sure where to start though.
real-analysis definite-integrals
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
asked Dec 5 at 12:12
kaisa
999
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?
Could anyone provide an example?
This is the definition I have for a closed and bounded interval:
Let $f$ be a continuous function on an interval $[a,b]$ Then
$$
intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
$$
I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
$$
sumlimits_{k=0}^{n-1}f(ldots)
$$
I am not really sure where to start though.
real-analysis definite-integrals
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
asked Dec 5 at 12:12
kaisa
999
How would I formulate a definition of the integral $int_a^b f(x) ,mathrm{d}x$ for a function continuous on $[a,b)$ and unbounded at the right-hand endpoint?
Could anyone provide an example?
This is the definition I have for a closed and bounded interval:
Let $f$ be a continuous function on an interval $[a,b]$ Then
$$
intlimits_a^b f(x) ,mathrm{d}x = limlimits_{ntoinfty} dfrac{b-a}{n}sumlimits_{k=0}^{n-1} f Bigg( a+dfrac{k}{n}(b-a) Bigg)
$$
I am thinking I need to use the left-hand endpoint to do this because the right-hand endpoint is unbounded. Thus the definition will have
$$
sumlimits_{k=0}^{n-1}f(ldots)
$$
I am not really sure where to start though.
real-analysis definite-integrals
real-analysis definite-integrals
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
asked Dec 5 at 12:12
kaisa
999
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
asked Dec 5 at 12:12
kaisa
999
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
edited Dec 5 at 12:41
Björn Friedrich
2,57961731
2,57961731
asked Dec 5 at 12:12
kaisa
999
asked Dec 5 at 12:12
kaisa
999
asked Dec 5 at 12:12
kaisa
999
999
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$
To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.
Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.
Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
answered Dec 5 at 14:00
Masacroso
12.8k41746
add a comment |
up vote
3
down vote
Answer.
Definition Improper Riemann integral
$$
int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
$$
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$
To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.
Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.
Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
answered Dec 5 at 14:00
Masacroso
12.8k41746
add a comment |
up vote
4
down vote
accepted
The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$
To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.
Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.
Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
answered Dec 5 at 14:00
Masacroso
12.8k41746
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$
To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.
Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.
Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
answered Dec 5 at 14:00
Masacroso
12.8k41746
The integral of Riemann doesn't exist for unbounded functions. This is the reason why mathematicians created the improper integral of Riemann.
The improper integral of Riemann is defined as a limit of proper integrals, in your case
$$int_a^b f(x), dx:=lim_{yto b^-}int_a^y f(x), dx$$
To show why the proper integral of Riemann doesn't exists for unbounded functions WLOG suppose that we have a function $f:[a,b]toBbb R$ such that it is continuous to the left of some point $cin[a,b]$ but $lim_{xto c^-}f(x)=infty$.
Then we can choose a sequence of partitions $(P_k)$ of $[a,b]$ with decreasing mesh $Delta_{P_k}=delta_k$ for some sequence $(delta_k)downarrow 0$, such that in each $P_k$ there is an interval of the kind $(c-delta_k,c]$. Then, because $f$ is continuous and unbounded to the left of $c$, for each $k$ there is some $x_kin(c-delta_k,c]$ such that $f(x_k)ge k/delta_k$.
Then if, for the interval $(c-delta_k,c]$, we choose alternatively the tag $c$, when $k$ is odd, and the tag $x_k$ when $k$ is even, we find that the sequence of Riemann sums is not convergent, it will have two cluster points, hence the integral of Riemann is not defined for this function in $[a,b]$.
Note that the (proper) integral of Riemann is defined only in compact intervals, so an integral over a set of the kind $[a,b)$ is outside of the original definition of Riemann. To make sense we need to integrate on $[a,b]$ or $[a,c]$ for some $cin[a,b)$.
If we try to apply the original definition of Riemann in an interval of the kind $[a,b)$ where $lim_{xto b^-}f(x)=infty$ and $f$ is continuous in $[a,b)$ then we will get an integral that diverges to infinity.
answered Dec 5 at 14:00
Masacroso
12.8k41746
edited Dec 5 at 17:49
answered Dec 5 at 14:00
Masacroso
12.8k41746
answered Dec 5 at 14:00
Masacroso
12.8k41746
answered Dec 5 at 14:00
Masacroso
12.8k41746
12.8k41746
add a comment |
add a comment |
up vote
3
down vote
Answer.
Definition Improper Riemann integral
$$
int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
$$
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
add a comment |
up vote
3
down vote
Answer.
Definition Improper Riemann integral
$$
int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
$$
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
add a comment |
up vote
3
down vote
up vote
3
down vote
Answer.
Definition Improper Riemann integral
$$
int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
$$
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
Answer.
Definition Improper Riemann integral
$$
int_a^b f(x),dx=lim_{varepsilonsearrow 0}int_a^{b-varepsilon} f(x),dx
$$
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
answered Dec 5 at 12:36
Yiorgos S. Smyrlis
62.3k1383162
62.3k1383162
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
add a comment |
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
This is the thing to use. Riemann sums may or may not converge to the right value.
– GEdgar
Dec 5 at 15:19
add a comment |
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