Integral $int_{[1,2]times[0,pi]}log(sqrt{x})sin(2y)d(x,y)$












0












$begingroup$


I want to find out if this integral can be calculated (if it exists)



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y)$$



To be honest, I don't know how, but I think that one might has to use Fubini's theorem since this is an iterated integral. Does someone know how it's done? And can someone explain to me what is meant with the interval $[1,2]times[0,pi]$?



I did this, but I don't know how to continue.



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y) = dfrac{sinleft(2yright)lnleft(xright)}{2} = frac{sin(2y)}{2} int{ln(x)}~dx$$



Here's what the function looks like, it looks nice imo.



enter image description here






integration analysis functions iterated-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $d(x,y)$ mean? Do you mean $dxdy$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 22:18






  • 1




    $begingroup$
    The integral is obviously zero.
    $endgroup$
    – Did
    Dec 3 '18 at 23:30
















0












$begingroup$


I want to find out if this integral can be calculated (if it exists)



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y)$$



To be honest, I don't know how, but I think that one might has to use Fubini's theorem since this is an iterated integral. Does someone know how it's done? And can someone explain to me what is meant with the interval $[1,2]times[0,pi]$?



I did this, but I don't know how to continue.



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y) = dfrac{sinleft(2yright)lnleft(xright)}{2} = frac{sin(2y)}{2} int{ln(x)}~dx$$



Here's what the function looks like, it looks nice imo.



enter image description here






integration analysis functions iterated-integrals







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does $d(x,y)$ mean? Do you mean $dxdy$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 22:18






  • 1




    $begingroup$
    The integral is obviously zero.
    $endgroup$
    – Did
    Dec 3 '18 at 23:30














0












0








0





$begingroup$


I want to find out if this integral can be calculated (if it exists)



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y)$$



To be honest, I don't know how, but I think that one might has to use Fubini's theorem since this is an iterated integral. Does someone know how it's done? And can someone explain to me what is meant with the interval $[1,2]times[0,pi]$?



I did this, but I don't know how to continue.



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y) = dfrac{sinleft(2yright)lnleft(xright)}{2} = frac{sin(2y)}{2} int{ln(x)}~dx$$



Here's what the function looks like, it looks nice imo.



enter image description here






integration analysis functions iterated-integrals







share|cite|improve this question











$endgroup$




I want to find out if this integral can be calculated (if it exists)



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y)$$



To be honest, I don't know how, but I think that one might has to use Fubini's theorem since this is an iterated integral. Does someone know how it's done? And can someone explain to me what is meant with the interval $[1,2]times[0,pi]$?



I did this, but I don't know how to continue.



$$int_{[1,2]times[0,pi]} log(sqrt{x})sin(2y)~d(x,y) = dfrac{sinleft(2yright)lnleft(xright)}{2} = frac{sin(2y)}{2} int{ln(x)}~dx$$



Here's what the function looks like, it looks nice imo.



enter image description here






integration analysis functions iterated-integrals




integration analysis functions iterated-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 10:13









Asaf Karagila♦

305k33435766




305k33435766










asked Dec 3 '18 at 22:15









Ramanujan TaylorRamanujan Taylor

184




184












  • $begingroup$
    What does $d(x,y)$ mean? Do you mean $dxdy$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 22:18






  • 1




    $begingroup$
    The integral is obviously zero.
    $endgroup$
    – Did
    Dec 3 '18 at 23:30


















  • $begingroup$
    What does $d(x,y)$ mean? Do you mean $dxdy$?
    $endgroup$
    – Mostafa Ayaz
    Dec 3 '18 at 22:18






  • 1




    $begingroup$
    The integral is obviously zero.
    $endgroup$
    – Did
    Dec 3 '18 at 23:30
















$begingroup$
What does $d(x,y)$ mean? Do you mean $dxdy$?
$endgroup$
– Mostafa Ayaz
Dec 3 '18 at 22:18




$begingroup$
What does $d(x,y)$ mean? Do you mean $dxdy$?
$endgroup$
– Mostafa Ayaz
Dec 3 '18 at 22:18




1




1




$begingroup$
The integral is obviously zero.
$endgroup$
– Did
Dec 3 '18 at 23:30




$begingroup$
The integral is obviously zero.
$endgroup$
– Did
Dec 3 '18 at 23:30










1 Answer
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$begingroup$

Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$






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    $begingroup$

    Hint: the integral means $int_1^2 int_0^{pi}$. Further,
    $$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Hint: the integral means $int_1^2 int_0^{pi}$. Further,
      $$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Hint: the integral means $int_1^2 int_0^{pi}$. Further,
        $$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$






        share|cite|improve this answer









        $endgroup$



        Hint: the integral means $int_1^2 int_0^{pi}$. Further,
        $$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 23:27









        rogerlrogerl

        18k22747




        18k22747






























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