Evaluating a Lebesgue integral over $mathbb{R}^2$
0
I am trying to evaluate the Lebesgue integral $int_Omega expleft(-sqrt{x^2 + y^2}right) ,dx,dy$, where $Omega = mathbb{R}^2$.
Initially I tried re-writing the integral as $int_Omega exp{|x|} ,dx$ and using the definition of the Lebesgue measure of a ball, I obtained $pi t^2$.
However, this isn't really helpful. How can I write this integral so that I get a "real" answer (i.e. not in terms of $t$)?
real-analysis
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
add a comment |
0
I am trying to evaluate the Lebesgue integral $int_Omega expleft(-sqrt{x^2 + y^2}right) ,dx,dy$, where $Omega = mathbb{R}^2$.
Initially I tried re-writing the integral as $int_Omega exp{|x|} ,dx$ and using the definition of the Lebesgue measure of a ball, I obtained $pi t^2$.
However, this isn't really helpful. How can I write this integral so that I get a "real" answer (i.e. not in terms of $t$)?
real-analysis
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
1
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14
add a comment |
0
0
0
I am trying to evaluate the Lebesgue integral $int_Omega expleft(-sqrt{x^2 + y^2}right) ,dx,dy$, where $Omega = mathbb{R}^2$.
Initially I tried re-writing the integral as $int_Omega exp{|x|} ,dx$ and using the definition of the Lebesgue measure of a ball, I obtained $pi t^2$.
However, this isn't really helpful. How can I write this integral so that I get a "real" answer (i.e. not in terms of $t$)?
real-analysis
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
I am trying to evaluate the Lebesgue integral $int_Omega expleft(-sqrt{x^2 + y^2}right) ,dx,dy$, where $Omega = mathbb{R}^2$.
Initially I tried re-writing the integral as $int_Omega exp{|x|} ,dx$ and using the definition of the Lebesgue measure of a ball, I obtained $pi t^2$.
However, this isn't really helpful. How can I write this integral so that I get a "real" answer (i.e. not in terms of $t$)?
real-analysis
real-analysis
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
edited Nov 19 '18 at 2:30
Xander Henderson
14.1k103554
14.1k103554
asked Nov 19 '18 at 2:27
Taliant
839
asked Nov 19 '18 at 2:27
Taliant
839
asked Nov 19 '18 at 2:27
Taliant
839
839
1
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14
add a comment |
1
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14
1
1
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14
add a comment |
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1
In polar coordinates, this is $int_0^{2pi} int_0^{infty} r e^{-r} , dr , dtheta$, which is easily computable. I'm sure this is a duplicate.
– T. Bongers
Nov 19 '18 at 2:29
I am well aware that the first fold in the integral is $Gamma(2) = 1$. The difficulty for me here was using Lebesgue integration to determine the first fold in the integral.
– Taliant
Nov 19 '18 at 2:45
The function is nice enough to use Riemann integration, using the comment of T Bongers.
– herb steinberg
Nov 19 '18 at 2:48
In my case I have to use Lebesgue integration.
– Taliant
Nov 19 '18 at 2:48
You know that if the Riemann integral over an interval $[a,b]$ exists then the Lebesgue integral also exists and is the same value as the Riemann integral. So when looking at $int_0^{infty} re^{-r} dr$ we have to be a bit careful. In the Riemann sense we can integrate $int_0^s re^{-r} dr$ for each $s$ so through a monotone convergence theorem argument on $int re^{-r} chi_{[0,s]}$ we see that the Lebesgue integral of $int_0^{infty} re^{-r} dr$ will be the same as the Riemann one.
– carsandpulsars
Nov 19 '18 at 15:14