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What is wrong with the derivation?
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I want to evaluate
$$int_0^inftyfrace^jtx(1+x)^2,dx$$
where $j=sqrt-1$ and $t$ is a real number. I did this change of variables $y=1/(1+x)$, which resulted in the integration
$$-int_0^1e^jt/y,dy=frac1jt$$
As you can see, the original integral is actually the characteristic function of a random variable with a PDF $f_X(x)=(1+x)^-2$, and one of the properties of the characteristic function is that its value at $0$ is $1$, which isn't the case in my result, and the characteristic function always exists.
What's wrong with the derivation?
EDIT 1: I evaluated $dy=fracdx(1+x)^2$, while it should be $dy=frac-dx(1+x)^2$. So, the integral becomes
$$int_0^1e^jt/y,dy=frac-1jt$$
EDIT 2: I forgot to the scaling factor $e^jt$, but the final result is still true. So, the integral is
$$e^jtint_0^1e^jt/y,dy=frac-1jt$$
EDIT 3: The indefinte integral is evaluated as
$$e^jtint e^jt/y,dy = frac-e^jtjt,y^2e^jt/y$$
integration statistics characteristic-functions
asked Aug 6 at 14:27
BlackMath
948
add a comment |Â
up vote
0
down vote
favorite
I want to evaluate
$$int_0^inftyfrace^jtx(1+x)^2,dx$$
where $j=sqrt-1$ and $t$ is a real number. I did this change of variables $y=1/(1+x)$, which resulted in the integration
$$-int_0^1e^jt/y,dy=frac1jt$$
As you can see, the original integral is actually the characteristic function of a random variable with a PDF $f_X(x)=(1+x)^-2$, and one of the properties of the characteristic function is that its value at $0$ is $1$, which isn't the case in my result, and the characteristic function always exists.
What's wrong with the derivation?
EDIT 1: I evaluated $dy=fracdx(1+x)^2$, while it should be $dy=frac-dx(1+x)^2$. So, the integral becomes
$$int_0^1e^jt/y,dy=frac-1jt$$
EDIT 2: I forgot to the scaling factor $e^jt$, but the final result is still true. So, the integral is
$$e^jtint_0^1e^jt/y,dy=frac-1jt$$
EDIT 3: The indefinte integral is evaluated as
$$e^jtint e^jt/y,dy = frac-e^jtjt,y^2e^jt/y$$
integration statistics characteristic-functions
asked Aug 6 at 14:27
BlackMath
948
1
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to evaluate
$$int_0^inftyfrace^jtx(1+x)^2,dx$$
where $j=sqrt-1$ and $t$ is a real number. I did this change of variables $y=1/(1+x)$, which resulted in the integration
$$-int_0^1e^jt/y,dy=frac1jt$$
As you can see, the original integral is actually the characteristic function of a random variable with a PDF $f_X(x)=(1+x)^-2$, and one of the properties of the characteristic function is that its value at $0$ is $1$, which isn't the case in my result, and the characteristic function always exists.
What's wrong with the derivation?
EDIT 1: I evaluated $dy=fracdx(1+x)^2$, while it should be $dy=frac-dx(1+x)^2$. So, the integral becomes
$$int_0^1e^jt/y,dy=frac-1jt$$
EDIT 2: I forgot to the scaling factor $e^jt$, but the final result is still true. So, the integral is
$$e^jtint_0^1e^jt/y,dy=frac-1jt$$
EDIT 3: The indefinte integral is evaluated as
$$e^jtint e^jt/y,dy = frac-e^jtjt,y^2e^jt/y$$
integration statistics characteristic-functions
asked Aug 6 at 14:27
BlackMath
948
I want to evaluate
$$int_0^inftyfrace^jtx(1+x)^2,dx$$
where $j=sqrt-1$ and $t$ is a real number. I did this change of variables $y=1/(1+x)$, which resulted in the integration
$$-int_0^1e^jt/y,dy=frac1jt$$
As you can see, the original integral is actually the characteristic function of a random variable with a PDF $f_X(x)=(1+x)^-2$, and one of the properties of the characteristic function is that its value at $0$ is $1$, which isn't the case in my result, and the characteristic function always exists.
What's wrong with the derivation?
EDIT 1: I evaluated $dy=fracdx(1+x)^2$, while it should be $dy=frac-dx(1+x)^2$. So, the integral becomes
$$int_0^1e^jt/y,dy=frac-1jt$$
EDIT 2: I forgot to the scaling factor $e^jt$, but the final result is still true. So, the integral is
$$e^jtint_0^1e^jt/y,dy=frac-1jt$$
EDIT 3: The indefinte integral is evaluated as
$$e^jtint e^jt/y,dy = frac-e^jtjt,y^2e^jt/y$$
integration statistics characteristic-functions
asked Aug 6 at 14:27
BlackMath
948
edited Aug 6 at 17:30
asked Aug 6 at 14:27
BlackMath
948
asked Aug 6 at 14:27
BlackMath
948
asked Aug 6 at 14:27
BlackMath
948
948
1
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04
add a comment |Â
1
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04
1
1
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
You evaluate the integral
as if the function was
$e^jty$.
edited Aug 10 at 17:01
amWhy
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
You evaluate the integral
as if the function was
$e^jty$.
edited Aug 10 at 17:01
amWhy
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
add a comment |Â
up vote
3
down vote
You evaluate the integral
as if the function was
$e^jty$.
edited Aug 10 at 17:01
amWhy
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
add a comment |Â
up vote
3
down vote
up vote
3
down vote
You evaluate the integral
as if the function was
$e^jty$.
edited Aug 10 at 17:01
amWhy
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
You evaluate the integral
as if the function was
$e^jty$.
edited Aug 10 at 17:01
amWhy
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
edited Aug 10 at 17:01
amWhy
189k25219431
edited Aug 10 at 17:01
amWhy
189k25219431
edited Aug 10 at 17:01
amWhy
189k25219431
189k25219431
answered Aug 6 at 14:32
marty cohen
69.3k446122
answered Aug 6 at 14:32
marty cohen
69.3k446122
answered Aug 6 at 14:32
marty cohen
69.3k446122
69.3k446122
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
add a comment |Â
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
Not exactly. the OP thinks they are integrating $e^jt/y$
– amWhy
Aug 6 at 14:35
1
1
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
How I evaluated the integral as it was $e^jty$? Could you explain?
– BlackMath
Aug 6 at 16:23
add a comment |Â
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1
If $y= frac 1(1+x), $ then $x = frac 1y -1$, and $dx = -frac 1y^2 dy $
– amWhy
Aug 6 at 14:31
$dy = -fracdx(1+x)^2$, so, the $fracdx(1+x)^2$ part in the integral can be substituted by $-dy$. I think I just forgot the minus sign, but otherwise, we agree. The problem still presents though.
– BlackMath
Aug 6 at 14:40
$x neq frac 1y$. Rather, $x = frac 1y - 1$, given your substitution $y= frac 11+x$.
– amWhy
Aug 6 at 14:47
Right, sorry, I forgot a term outside the integral. The final result is correct though. I will edit the original post.
– BlackMath
Aug 6 at 16:04