Interesting Integration with respect to $[x]$ where $[cdot]$ is greatest integer function
$begingroup$
$$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .
Attempt
$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.
integration definite-integrals self-learning
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .
Attempt
$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.
integration definite-integrals self-learning
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
$endgroup$
add a comment |
$begingroup$
$$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .
Attempt
$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.
integration definite-integrals self-learning
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
$endgroup$
$$int_{0}^{3} {(x^2+1)}d[x]$$ is equal to ___________ .
Attempt
$[x]$ is constant for every interval. So for that intervals $d[x]=0$ so intergral given is zero. Am I right? Though the answer given is $dfrac{17}{2}$.
integration definite-integrals self-learning
integration definite-integrals self-learning
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
edited Dec 2 '18 at 7:41
Tianlalu
3,08621038
3,08621038
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
asked Nov 28 '18 at 17:36
jayant98jayant98
513116
513116
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You must be careful about how you define your integral. I am guessing that you are defining
$$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.
Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$
In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
$$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
for a continuous function $f$.
Having described this, I hope it is now not so hard to compute the entire integral.
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
$endgroup$
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
add a comment |
$begingroup$
In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
$$
int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
$$
and apply it to your integral, yielding
$$
int_0^3 (x^2+1) frac{d[x] }{dx} dx.
$$
The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.
answered Nov 28 '18 at 17:53
EddyEddy
884612
$endgroup$
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
You must be careful about how you define your integral. I am guessing that you are defining
$$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.
Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$
In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
$$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
for a continuous function $f$.
Having described this, I hope it is now not so hard to compute the entire integral.
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
$endgroup$
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
add a comment |
$begingroup$
You must be careful about how you define your integral. I am guessing that you are defining
$$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.
Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$
In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
$$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
for a continuous function $f$.
Having described this, I hope it is now not so hard to compute the entire integral.
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
$endgroup$
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
add a comment |
$begingroup$
You must be careful about how you define your integral. I am guessing that you are defining
$$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.
Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$
In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
$$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
for a continuous function $f$.
Having described this, I hope it is now not so hard to compute the entire integral.
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
$endgroup$
You must be careful about how you define your integral. I am guessing that you are defining
$$ int_a^b f(x) dg(x) = lim sum_i f(c_i) (g(x_{i+1}) - g(x_i)),$$
where $c_i in [x_i, x_{i+1}]$, and the limit is really a limit over partitions of the interval $[a,b]$. This is typically called the Riemann-Stieltjes integral.
Then in your case, $d lfloor x rfloor$ is $0$ when its defined, but at integer values one must be a bit delicate. For instance, looking just at
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim sum (x_i^2 + 1)Big( lfloor x_{i+1} rfloor - lfloor x_i rfloorBig),$$
and all summands are zero except for the one summand where $x_i < 1$ and $x_{i+1} geq 1$. For that one term, $lfloor x_{i+1} rfloor - lfloor x_i rfloor = 1$. And as the partitions of $[1/2, 3/2]$ become finer, the endpoints $x_i$ and $x_{i+1}$ (by which I mean the two partition points surrounding $1$ in the corresponding partition --- there is a minor abuse of notation here) both approach $1$. Thus
$$ int_{1/2}^{3/2} (x^2 + 1) dlfloor x rfloor = lim (x_i^2 + 1) (1) = 1^2 + 1 = 2.$$
In fact, what this really is is exactly the value of $x^2 + 1$ at $1$, and more generally
$$ int_a^b f(x) d lfloor x rfloor = sum_{a < n leq b} f(n)$$
for a continuous function $f$.
Having described this, I hope it is now not so hard to compute the entire integral.
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
edited Nov 28 '18 at 18:21
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
answered Nov 28 '18 at 17:50
davidlowryduda♦davidlowryduda
74.5k7118253
74.5k7118253
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
add a comment |
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
So for whole we can say that for x=2 the value will be there and so for the x=3. And following this I got the answer as 17. Is it now correct?
$endgroup$
– jayant98
Nov 28 '18 at 18:17
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
$begingroup$
@jayant98 Yes, it should be $17$.
$endgroup$
– davidlowryduda♦
Nov 28 '18 at 18:23
add a comment |
$begingroup$
In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
$$
int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
$$
and apply it to your integral, yielding
$$
int_0^3 (x^2+1) frac{d[x] }{dx} dx.
$$
The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.
answered Nov 28 '18 at 17:53
EddyEddy
884612
$endgroup$
add a comment |
$begingroup$
In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
$$
int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
$$
and apply it to your integral, yielding
$$
int_0^3 (x^2+1) frac{d[x] }{dx} dx.
$$
The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.
answered Nov 28 '18 at 17:53
EddyEddy
884612
$endgroup$
add a comment |
$begingroup$
In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
$$
int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
$$
and apply it to your integral, yielding
$$
int_0^3 (x^2+1) frac{d[x] }{dx} dx.
$$
The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.
answered Nov 28 '18 at 17:53
EddyEddy
884612
$endgroup$
In your reasoning you forget that at integers the value of $d[x] $ diverges. To write the integral in a tractable way you should use that
$$
int_a^b f(x) dy = int_a^b f(x) frac{dy}{dx} dx
$$
and apply it to your integral, yielding
$$
int_0^3 (x^2+1) frac{d[x] }{dx} dx.
$$
The value of $frac{d[x] }{dx}$ is a sum of Dirac deltas. However, there is a delta at each of the endpoints of the integration domain, and the value of an integral who's integrand contains a delta at an endpoint is not defined, so you cannot assign a value to this integral.
answered Nov 28 '18 at 17:53
EddyEddy
884612
answered Nov 28 '18 at 17:53
EddyEddy
884612
answered Nov 28 '18 at 17:53
EddyEddy
884612
answered Nov 28 '18 at 17:53
EddyEddy
884612
884612
add a comment |
add a comment |
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