Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$
$begingroup$
Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$
$$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$
Let $u=x_1+x_2$ and $,du=,dx_1$.
$$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$
How do I compute the inner integral?
calculus integration improper-integrals
edited Nov 21 '14 at 3:34
dustin
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
$endgroup$
add a comment |
$begingroup$
Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$
$$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$
Let $u=x_1+x_2$ and $,du=,dx_1$.
$$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$
How do I compute the inner integral?
calculus integration improper-integrals
edited Nov 21 '14 at 3:34
dustin
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
$endgroup$
add a comment |
$begingroup$
Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$
$$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$
Let $u=x_1+x_2$ and $,du=,dx_1$.
$$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$
How do I compute the inner integral?
calculus integration improper-integrals
edited Nov 21 '14 at 3:34
dustin
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
$endgroup$
Solving for C when we have $Cint_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2=1$
$$int_0^infty int_0^infty frac{e^frac{-(x_1+x_2)}{2}}{x_1+x_2} ,dx_1 ,dx_2$$
Let $u=x_1+x_2$ and $,du=,dx_1$.
$$int_0^infty int_{x_2}^infty frac{e^frac{-u}{2}}{u} ,du ,dx_2 $$
How do I compute the inner integral?
calculus integration improper-integrals
calculus integration improper-integrals
edited Nov 21 '14 at 3:34
dustin
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
edited Nov 21 '14 at 3:34
dustin
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
edited Nov 21 '14 at 3:34
dustin
6,72892969
edited Nov 21 '14 at 3:34
dustin
6,72892969
edited Nov 21 '14 at 3:34
dustin
6,72892969
6,72892969
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
asked Nov 21 '14 at 3:00
Username UnknownUsername Unknown
1,25242058
1,25242058
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
begin{align}
4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
&= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
&= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
&=cdots
end{align}
If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.
When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$
Mouse over for solution:
begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
$endgroup$
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
add a comment |
$begingroup$
Integration by parts:))
$$
int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
$$
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
begin{align}
4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
&= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
&= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
&=cdots
end{align}
If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.
When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$
Mouse over for solution:
begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
$endgroup$
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
add a comment |
$begingroup$
From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
begin{align}
4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
&= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
&= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
&=cdots
end{align}
If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.
When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$
Mouse over for solution:
begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
$endgroup$
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
add a comment |
$begingroup$
From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
begin{align}
4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
&= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
&= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
&=cdots
end{align}
If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.
When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$
Mouse over for solution:
begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
$endgroup$
From my initial assessment, the substitution $x_1 = x^2$ and $x_2 = y^2$ should work. After that let $x^2 + y^2 = r^2$ where the Jacobian is then $rdrdtheta$. Then you can make a u sub where $u = r^2/2$. At this point, you can integrate your u integral and you left with a $theta$ integral of $sin(theta)cos(theta)$
begin{align}
4iintfrac{exp(-r^2/2)}{r^2}r^3sin(theta)cos(theta)drdtheta
&= 4iintexp(-u)sin(theta)cos(theta)dudtheta\
&= -4intexp(-u)Bigl|_0^{infty}sin(theta)cos(theta)dtheta\
&=cdots
end{align}
If you obtained $frac{1}{16pi^2}$, I would suggest you try again. If you still cant figure it out, show what you have done.
When we convert to $r$ and $theta$, the bounds become $0<r<infty$ and $0<theta<pi/2$
Mouse over for solution:
begin{align}&= 4int_0^{pi/2}sin(theta)cos(theta)dtheta\&=2int_0^{pi/2}sin(2theta)dtheta\&= -cos(2theta)Bigr|_0^{pi/2}\&=1 + 1\&=2end{align}
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
edited Nov 21 '14 at 16:25
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
answered Nov 21 '14 at 3:17
dustindustin
6,72892969
6,72892969
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
add a comment |
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
Perfect so $c=frac{1}{16pi^2}$
$endgroup$
– Username Unknown
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
$begingroup$
@UsernameUnknown You should get $C=1/2$
$endgroup$
– dustin
Nov 21 '14 at 3:37
add a comment |
$begingroup$
Integration by parts:))
$$
int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
$$
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
$endgroup$
add a comment |
$begingroup$
Integration by parts:))
$$
int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
$$
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
$endgroup$
add a comment |
$begingroup$
Integration by parts:))
$$
int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
$$
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
$endgroup$
Integration by parts:))
$$
int_0^{infty}int_{x_2}^{infty}frac{e^{-frac u2}}udu, dx_2=int_0^{infty}x_2frac{e^{-frac {x_2}2}}{x_2}dx_2=2.
$$
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
edited Dec 8 '18 at 7:16
Brahadeesh
6,32942363
6,32942363
answered Dec 7 '18 at 23:20
AAKAAK
365
answered Dec 7 '18 at 23:20
AAKAAK
365
answered Dec 7 '18 at 23:20
AAKAAK
365
365
add a comment |
add a comment |
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