Computing an Integral by means of measure theory
I need to compute the integral $$int _0 ^ infty xe^{-kx} dx, ,kgeq1$$ by means of measure theory. Integration by parts will not work here. Also I used the Taylor series of the exponential function with the idea in mind to use the fact that for positive functions one can interchange sum symbol and integral, but it did not lead me further.
I will appreciate your help. Thanks.
real-analysis measure-theory
asked Dec 6 at 17:20
user249018
338127
add a comment |
I need to compute the integral $$int _0 ^ infty xe^{-kx} dx, ,kgeq1$$ by means of measure theory. Integration by parts will not work here. Also I used the Taylor series of the exponential function with the idea in mind to use the fact that for positive functions one can interchange sum symbol and integral, but it did not lead me further.
I will appreciate your help. Thanks.
real-analysis measure-theory
asked Dec 6 at 17:20
user249018
338127
2
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
2
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27
add a comment |
I need to compute the integral $$int _0 ^ infty xe^{-kx} dx, ,kgeq1$$ by means of measure theory. Integration by parts will not work here. Also I used the Taylor series of the exponential function with the idea in mind to use the fact that for positive functions one can interchange sum symbol and integral, but it did not lead me further.
I will appreciate your help. Thanks.
real-analysis measure-theory
asked Dec 6 at 17:20
user249018
338127
I need to compute the integral $$int _0 ^ infty xe^{-kx} dx, ,kgeq1$$ by means of measure theory. Integration by parts will not work here. Also I used the Taylor series of the exponential function with the idea in mind to use the fact that for positive functions one can interchange sum symbol and integral, but it did not lead me further.
I will appreciate your help. Thanks.
real-analysis measure-theory
real-analysis measure-theory
asked Dec 6 at 17:20
user249018
338127
asked Dec 6 at 17:20
user249018
338127
edited Dec 6 at 17:51
asked Dec 6 at 17:20
user249018
338127
asked Dec 6 at 17:20
user249018
338127
asked Dec 6 at 17:20
user249018
338127
338127
2
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
2
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27
add a comment |
2
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
2
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27
2
2
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
2
2
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27
add a comment |
1 Answer
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Hint: $$int_0^infty x e^{-kx}; dx = - frac{d}{dk} int_0^infty e^{-kx}; dx $$
Use "measure theoretical means" to justify the interchange of derivative and integral.
answered Dec 6 at 17:31
Robert Israel
317k23206457
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
Hint: $$int_0^infty x e^{-kx}; dx = - frac{d}{dk} int_0^infty e^{-kx}; dx $$
Use "measure theoretical means" to justify the interchange of derivative and integral.
answered Dec 6 at 17:31
Robert Israel
317k23206457
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
add a comment |
Hint: $$int_0^infty x e^{-kx}; dx = - frac{d}{dk} int_0^infty e^{-kx}; dx $$
Use "measure theoretical means" to justify the interchange of derivative and integral.
answered Dec 6 at 17:31
Robert Israel
317k23206457
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
add a comment |
Hint: $$int_0^infty x e^{-kx}; dx = - frac{d}{dk} int_0^infty e^{-kx}; dx $$
Use "measure theoretical means" to justify the interchange of derivative and integral.
answered Dec 6 at 17:31
Robert Israel
317k23206457
Hint: $$int_0^infty x e^{-kx}; dx = - frac{d}{dk} int_0^infty e^{-kx}; dx $$
Use "measure theoretical means" to justify the interchange of derivative and integral.
answered Dec 6 at 17:31
Robert Israel
317k23206457
answered Dec 6 at 17:31
Robert Israel
317k23206457
answered Dec 6 at 17:31
Robert Israel
317k23206457
answered Dec 6 at 17:31
Robert Israel
317k23206457
317k23206457
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
add a comment |
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
Thanks, but $k$ is a constant and not a variable.
– user249018
Dec 6 at 17:49
2
2
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
Don't be afraid to turn your constants into variables.
– Robert Israel
Dec 6 at 18:18
add a comment |
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2
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
– Eclipse Sun
Dec 6 at 17:23
2
maybe you can use the sequence of functions defined by $$f_n(x):=chi_{[0,n)}(x), xleft(1-frac{kx}{n}right)^n$$ It is just an idea
– Masacroso
Dec 6 at 17:27