Calculate a Riemann Stieltjes Integral $int_0^3 x d([x] - x) = 3/2$.
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I need to prove that
$$int_0^3 x d([x] - x) = dfrac32$$
However I can not think of a change of variable that can be used so I tried to approach it using Riemann Sums
$sum_i=0^3(alpha(i)-alpha(i-1) )i$
But I can not prove that the result is 3/2.
I took it from the Apóstol on Mathematical Analysis.
real-analysis integration change-of-variable stieltjes-integral
edited Sep 2 at 6:45
Nosrati
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
add a comment |Â
up vote
2
down vote
favorite
I need to prove that
$$int_0^3 x d([x] - x) = dfrac32$$
However I can not think of a change of variable that can be used so I tried to approach it using Riemann Sums
$sum_i=0^3(alpha(i)-alpha(i-1) )i$
But I can not prove that the result is 3/2.
I took it from the Apóstol on Mathematical Analysis.
real-analysis integration change-of-variable stieltjes-integral
edited Sep 2 at 6:45
Nosrati
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
1
What about integrating by parts?
– xbh
Sep 2 at 2:07
1
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
1
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
1
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to prove that
$$int_0^3 x d([x] - x) = dfrac32$$
However I can not think of a change of variable that can be used so I tried to approach it using Riemann Sums
$sum_i=0^3(alpha(i)-alpha(i-1) )i$
But I can not prove that the result is 3/2.
I took it from the Apóstol on Mathematical Analysis.
real-analysis integration change-of-variable stieltjes-integral
edited Sep 2 at 6:45
Nosrati
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
I need to prove that
$$int_0^3 x d([x] - x) = dfrac32$$
However I can not think of a change of variable that can be used so I tried to approach it using Riemann Sums
$sum_i=0^3(alpha(i)-alpha(i-1) )i$
But I can not prove that the result is 3/2.
I took it from the Apóstol on Mathematical Analysis.
real-analysis integration change-of-variable stieltjes-integral
real-analysis integration change-of-variable stieltjes-integral
edited Sep 2 at 6:45
Nosrati
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
edited Sep 2 at 6:45
Nosrati
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
edited Sep 2 at 6:45
Nosrati
22.1k61747
edited Sep 2 at 6:45
Nosrati
22.1k61747
edited Sep 2 at 6:45
Nosrati
22.1k61747
22.1k61747
asked Sep 2 at 1:45
Fer Stein
317
asked Sep 2 at 1:45
Fer Stein
317
asked Sep 2 at 1:45
Fer Stein
317
317
1
What about integrating by parts?
– xbh
Sep 2 at 2:07
1
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
1
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
1
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11
add a comment |Â
1
What about integrating by parts?
– xbh
Sep 2 at 2:07
1
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
1
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
1
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11
1
1
What about integrating by parts?
– xbh
Sep 2 at 2:07
What about integrating by parts?
– xbh
Sep 2 at 2:07
1
1
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
1
1
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
1
1
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
We discussed this type of integrals here, see the details there and you find it easy
$$int_0^3 x d([x] - x) =int_0^3 x d[x] - int_0^3xdx = 1+2+3-dfrac92 = dfrac32$$
answered Sep 2 at 4:06
Nosrati
22.1k61747
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
add a comment |Â
up vote
1
down vote
By Wiki: Integration by parts for R-S, we have
$$int_0^3 x d([x] - x) = 3 ([3] - 3) - 0 ([0] - 0) - int_0^3 [x] - x dx$$
$$= 0 + 0 - int_0^3 [x] - x dx$$
$$= int_0^3 x - [x] dx,$$
an ordinary Riemann integral.
answered Sep 2 at 3:12
BCLC
1
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
We discussed this type of integrals here, see the details there and you find it easy
$$int_0^3 x d([x] - x) =int_0^3 x d[x] - int_0^3xdx = 1+2+3-dfrac92 = dfrac32$$
answered Sep 2 at 4:06
Nosrati
22.1k61747
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
add a comment |Â
up vote
1
down vote
accepted
We discussed this type of integrals here, see the details there and you find it easy
$$int_0^3 x d([x] - x) =int_0^3 x d[x] - int_0^3xdx = 1+2+3-dfrac92 = dfrac32$$
answered Sep 2 at 4:06
Nosrati
22.1k61747
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
We discussed this type of integrals here, see the details there and you find it easy
$$int_0^3 x d([x] - x) =int_0^3 x d[x] - int_0^3xdx = 1+2+3-dfrac92 = dfrac32$$
answered Sep 2 at 4:06
Nosrati
22.1k61747
We discussed this type of integrals here, see the details there and you find it easy
$$int_0^3 x d([x] - x) =int_0^3 x d[x] - int_0^3xdx = 1+2+3-dfrac92 = dfrac32$$
answered Sep 2 at 4:06
Nosrati
22.1k61747
answered Sep 2 at 4:06
Nosrati
22.1k61747
answered Sep 2 at 4:06
Nosrati
22.1k61747
answered Sep 2 at 4:06
Nosrati
22.1k61747
22.1k61747
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
add a comment |Â
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
Why $x d[x]$ and not $[x] dx$?
– BCLC
Sep 2 at 8:52
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
oh yes. I just separate the integral and didn't write the rest @BCLC
– Nosrati
Sep 2 at 9:14
add a comment |Â
up vote
1
down vote
By Wiki: Integration by parts for R-S, we have
$$int_0^3 x d([x] - x) = 3 ([3] - 3) - 0 ([0] - 0) - int_0^3 [x] - x dx$$
$$= 0 + 0 - int_0^3 [x] - x dx$$
$$= int_0^3 x - [x] dx,$$
an ordinary Riemann integral.
answered Sep 2 at 3:12
BCLC
1
add a comment |Â
up vote
1
down vote
By Wiki: Integration by parts for R-S, we have
$$int_0^3 x d([x] - x) = 3 ([3] - 3) - 0 ([0] - 0) - int_0^3 [x] - x dx$$
$$= 0 + 0 - int_0^3 [x] - x dx$$
$$= int_0^3 x - [x] dx,$$
an ordinary Riemann integral.
answered Sep 2 at 3:12
BCLC
1
add a comment |Â
up vote
1
down vote
up vote
1
down vote
By Wiki: Integration by parts for R-S, we have
$$int_0^3 x d([x] - x) = 3 ([3] - 3) - 0 ([0] - 0) - int_0^3 [x] - x dx$$
$$= 0 + 0 - int_0^3 [x] - x dx$$
$$= int_0^3 x - [x] dx,$$
an ordinary Riemann integral.
answered Sep 2 at 3:12
BCLC
1
By Wiki: Integration by parts for R-S, we have
$$int_0^3 x d([x] - x) = 3 ([3] - 3) - 0 ([0] - 0) - int_0^3 [x] - x dx$$
$$= 0 + 0 - int_0^3 [x] - x dx$$
$$= int_0^3 x - [x] dx,$$
an ordinary Riemann integral.
answered Sep 2 at 3:12
BCLC
1
answered Sep 2 at 3:12
BCLC
1
answered Sep 2 at 3:12
BCLC
1
answered Sep 2 at 3:12
BCLC
1
1
add a comment |Â
add a comment |Â
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1
What about integrating by parts?
– xbh
Sep 2 at 2:07
1
I can't think of how to do so.
– Fer Stein
Sep 2 at 2:11
1
Apostol contains such method. Refer to Section 7.5.
– xbh
Sep 2 at 2:13
@xbh Wiki: Integration by parts for R-S ?
– BCLC
Sep 2 at 3:10
1
@BCLC Yes. The OP is dealing with R-S integral, so…
– xbh
Sep 2 at 3:11