Integral $int_{[1,2]times[0,pi]}log(sqrt{x})sin(2y)d(x,y)$
$begingroup$
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.
integration analysis functions iterated-integrals
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
$endgroup$
add a comment |
$begingroup$
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.
integration analysis functions iterated-integrals
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
integration analysis functions iterated-integrals
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
$endgroup$
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.
integration analysis functions iterated-integrals
integration analysis functions iterated-integrals
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
edited Dec 4 '18 at 10:13
Asaf Karagila♦
305k33435766
305k33435766
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
asked Dec 3 '18 at 22:15
Ramanujan TaylorRamanujan Taylor
184
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
$endgroup$
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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$begingroup$
Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
$endgroup$
add a comment |
$begingroup$
Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
$endgroup$
add a comment |
$begingroup$
Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
$endgroup$
Hint: the integral means $int_1^2 int_0^{pi}$. Further,
$$int_1^2int_0^pi f(x)g(y),dy,dx = left(int_1^2 f(x),dxright)left(int_0^pi g(y),dyright).$$
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
answered Dec 3 '18 at 23:27
rogerlrogerl
18k22747
18k22747
add a comment |
add a comment |
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$begingroup$
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