Computing an Integral by means of measure theory












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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







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  • 2




    $begingroup$
    What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 17:23






  • 2




    $begingroup$
    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
    $endgroup$
    – Masacroso
    Dec 6 '18 at 17:27


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 17:23






  • 2




    $begingroup$
    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
    $endgroup$
    – Masacroso
    Dec 6 '18 at 17:27
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 6 '18 at 17:51







user249018

















asked Dec 6 '18 at 17:20









user249018user249018

393127




393127








  • 2




    $begingroup$
    What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 17:23






  • 2




    $begingroup$
    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
    $endgroup$
    – Masacroso
    Dec 6 '18 at 17:27
















  • 2




    $begingroup$
    What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
    $endgroup$
    – Eclipse Sun
    Dec 6 '18 at 17:23






  • 2




    $begingroup$
    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
    $endgroup$
    – Masacroso
    Dec 6 '18 at 17:27










2




2




$begingroup$
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
$endgroup$
– Eclipse Sun
Dec 6 '18 at 17:23




$begingroup$
What do you mean by "measure theoretical means"? And why doesn't integration by parts work here?
$endgroup$
– Eclipse Sun
Dec 6 '18 at 17:23




2




2




$begingroup$
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
$endgroup$
– Masacroso
Dec 6 '18 at 17:27






$begingroup$
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
$endgroup$
– Masacroso
Dec 6 '18 at 17:27












1 Answer
1






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oldest

votes


















7












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but $k$ is a constant and not a variable.
    $endgroup$
    – user249018
    Dec 6 '18 at 17:49






  • 2




    $begingroup$
    Don't be afraid to turn your constants into variables.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 18:18











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but $k$ is a constant and not a variable.
    $endgroup$
    – user249018
    Dec 6 '18 at 17:49






  • 2




    $begingroup$
    Don't be afraid to turn your constants into variables.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 18:18
















7












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks, but $k$ is a constant and not a variable.
    $endgroup$
    – user249018
    Dec 6 '18 at 17:49






  • 2




    $begingroup$
    Don't be afraid to turn your constants into variables.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 18:18














7












7








7





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 6 '18 at 17:31









Robert IsraelRobert Israel

319k23209460




319k23209460












  • $begingroup$
    Thanks, but $k$ is a constant and not a variable.
    $endgroup$
    – user249018
    Dec 6 '18 at 17:49






  • 2




    $begingroup$
    Don't be afraid to turn your constants into variables.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 18:18


















  • $begingroup$
    Thanks, but $k$ is a constant and not a variable.
    $endgroup$
    – user249018
    Dec 6 '18 at 17:49






  • 2




    $begingroup$
    Don't be afraid to turn your constants into variables.
    $endgroup$
    – Robert Israel
    Dec 6 '18 at 18:18
















$begingroup$
Thanks, but $k$ is a constant and not a variable.
$endgroup$
– user249018
Dec 6 '18 at 17:49




$begingroup$
Thanks, but $k$ is a constant and not a variable.
$endgroup$
– user249018
Dec 6 '18 at 17:49




2




2




$begingroup$
Don't be afraid to turn your constants into variables.
$endgroup$
– Robert Israel
Dec 6 '18 at 18:18




$begingroup$
Don't be afraid to turn your constants into variables.
$endgroup$
– Robert Israel
Dec 6 '18 at 18:18


















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