Use Fubini's Theorem to calculate the integral $int_{0}^{pi/2}int_{y}^{pi/2}yfrac{sin x}{x}dx dy$
Here is my attempt:
As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and
By change of limit we get:
$int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$
$implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$
and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.
Thanks for any help!!
real-analysis measure-theory lebesgue-integral
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
add a comment |
Here is my attempt:
As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and
By change of limit we get:
$int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$
$implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$
and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.
Thanks for any help!!
real-analysis measure-theory lebesgue-integral
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
add a comment |
Here is my attempt:
As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and
By change of limit we get:
$int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$
$implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$
and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.
Thanks for any help!!
real-analysis measure-theory lebesgue-integral
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
Here is my attempt:
As $[0,pi/2] times [0,pi/2]$ is a finite measure space(hence $sigma$ finite) and also its $x times y$ integrable therefore hypothesis of the fubini's theorem are satisfied and
By change of limit we get:
$int_{0}^{pi/2}int_{x}^{pi/2}yfrac{sin x}{x}dx dy$
$implies$ $frac{pi^2}{8}int_{0}^{pi/2}frac{sin x}{x}-int_{0}^{pi/2}frac{xsin x}{2}dx$
and then I don't know how to integrate the first integral $int_0^{pi/2}frac{sin x}{x}dx$.
Thanks for any help!!
real-analysis measure-theory lebesgue-integral
real-analysis measure-theory lebesgue-integral
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
edited Dec 31 '18 at 10:57
Davide Giraudo
125k16150261
125k16150261
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
asked Dec 29 '18 at 1:04
InfinityInfinity
312112
312112
add a comment |
add a comment |
1 Answer
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I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:
$$0 leq y leq x leq frac pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$
Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$
$int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:
$$int_0^{frac pi 2}frac{xsin x}{2}dx$$
I will leave the final integration by parts for you to do.
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
add a comment |
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1 Answer
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votes
I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:
$$0 leq y leq x leq frac pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$
Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$
$int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:
$$int_0^{frac pi 2}frac{xsin x}{2}dx$$
I will leave the final integration by parts for you to do.
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
add a comment |
I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:
$$0 leq y leq x leq frac pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$
Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$
$int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:
$$int_0^{frac pi 2}frac{xsin x}{2}dx$$
I will leave the final integration by parts for you to do.
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
add a comment |
I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:
$$0 leq y leq x leq frac pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$
Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$
$int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:
$$int_0^{frac pi 2}frac{xsin x}{2}dx$$
I will leave the final integration by parts for you to do.
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
I think you messed up the change of order. In the first integral, we have $0leq y leq frac pi 2$. Then, in the second integral, we have $y leq x leq fracpi 2$. In total, this gives us:
$$0 leq y leq x leq frac pi 2$$
Now, if we want to switch the order, we need to remember abide by this. First, $0leq xleq fracpi 2$, which determines the limits of the new outer integral. Then, we have $0leq y leq x$, which determines the limits of the new inner integral. Thus, the new integral becomes:
$$int_0^{frac pi 2}int_0^x yfrac{sin x}{x}dydx$$
Now, we can take $frac{sin x}{x}$ out of the inner integral since it has nothing to do with $y$, so we get:
$$int_0^{frac pi 2}frac{sin x}{x}int_0^x ydydx$$
$int_0^x ydy=frac{x^2}{2}$, so when we substitute back in and multiply it with $frac{sin x}{x}$, we get:
$$int_0^{frac pi 2}frac{xsin x}{2}dx$$
I will leave the final integration by parts for you to do.
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
edited Dec 29 '18 at 14:51
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 29 '18 at 1:12
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
add a comment |
add a comment |
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