calculate some integral of sum of two variables [duplicate]Proving the sum of two independent Cauchy Random Variables is CauchyDoes this multivariate function have only one maximum?change of variables for 3 dimensional integralCalculating mean vector of a multivariate distributionDifferential derivatives of dependent variablesDistributing integral into product of integralsPDF of the difference between two independent beta random variablesLimits of Integration Over Iterated Region2D Integration With Quadratic Arg. of Delta FunctionHow to take derivative with respect to a function that is a linear function the variables in the original function?Compute an indefinite integral through change of variables
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calculate some integral of sum of two variables [duplicate]
Proving the sum of two independent Cauchy Random Variables is CauchyDoes this multivariate function have only one maximum?change of variables for 3 dimensional integralCalculating mean vector of a multivariate distributionDifferential derivatives of dependent variablesDistributing integral into product of integralsPDF of the difference between two independent beta random variablesLimits of Integration Over Iterated Region2D Integration With Quadratic Arg. of Delta FunctionHow to take derivative with respect to a function that is a linear function the variables in the original function?Compute an indefinite integral through change of variables
$begingroup$
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Let $X_1$ and $X_2$ be two i.i.d. continuous random variables which have probability density function $$f(x) = frac1pifrac11+x^2.$$ Now I want to calculate the density of $X_1 + X_2$.
$$
f_X_1+X_2(z)
= int_-infty^infty f_X_1(x_1)f_X_2(z-x_1)dx_1
= int_-infty^infty frac1pi frac11+x_1^2
frac1pi frac11+(z-x_1)^2 dx_1
$$
Is there any efficient way to calculate this integral?
multivariable-calculus
asked Mar 29 at 1:23
EricEric
386
$endgroup$
marked as duplicate by Mike Earnest, Lord Shark the Unknown, Jyrki Lahtonen, Leucippus, Eevee Trainer Mar 30 at 8:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Let $X_1$ and $X_2$ be two i.i.d. continuous random variables which have probability density function $$f(x) = frac1pifrac11+x^2.$$ Now I want to calculate the density of $X_1 + X_2$.
$$
f_X_1+X_2(z)
= int_-infty^infty f_X_1(x_1)f_X_2(z-x_1)dx_1
= int_-infty^infty frac1pi frac11+x_1^2
frac1pi frac11+(z-x_1)^2 dx_1
$$
Is there any efficient way to calculate this integral?
multivariable-calculus
asked Mar 29 at 1:23
EricEric
386
$endgroup$
marked as duplicate by Mike Earnest, Lord Shark the Unknown, Jyrki Lahtonen, Leucippus, Eevee Trainer Mar 30 at 8:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56
add a comment |
$begingroup$
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Let $X_1$ and $X_2$ be two i.i.d. continuous random variables which have probability density function $$f(x) = frac1pifrac11+x^2.$$ Now I want to calculate the density of $X_1 + X_2$.
$$
f_X_1+X_2(z)
= int_-infty^infty f_X_1(x_1)f_X_2(z-x_1)dx_1
= int_-infty^infty frac1pi frac11+x_1^2
frac1pi frac11+(z-x_1)^2 dx_1
$$
Is there any efficient way to calculate this integral?
multivariable-calculus
asked Mar 29 at 1:23
EricEric
386
$endgroup$
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Let $X_1$ and $X_2$ be two i.i.d. continuous random variables which have probability density function $$f(x) = frac1pifrac11+x^2.$$ Now I want to calculate the density of $X_1 + X_2$.
$$
f_X_1+X_2(z)
= int_-infty^infty f_X_1(x_1)f_X_2(z-x_1)dx_1
= int_-infty^infty frac1pi frac11+x_1^2
frac1pi frac11+(z-x_1)^2 dx_1
$$
Is there any efficient way to calculate this integral?
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
multivariable-calculus
multivariable-calculus
asked Mar 29 at 1:23
EricEric
386
asked Mar 29 at 1:23
EricEric
386
edited Mar 29 at 2:41
Eric
asked Mar 29 at 1:23
EricEric
386
asked Mar 29 at 1:23
EricEric
386
asked Mar 29 at 1:23
EricEric
386
386
marked as duplicate by Mike Earnest, Lord Shark the Unknown, Jyrki Lahtonen, Leucippus, Eevee Trainer Mar 30 at 8:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Mike Earnest, Lord Shark the Unknown, Jyrki Lahtonen, Leucippus, Eevee Trainer Mar 30 at 8:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56
add a comment |
1
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56
1
1
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You made a mistake, the correct integral is
$$
frac1pi^2 int_mathbbR frac11+x^2 frac11 + (z-x)^2 dx
$$
and I would look around trig substitution or partial fractions...
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
$endgroup$
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
add a comment |
$begingroup$
You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You made a mistake, the correct integral is
$$
frac1pi^2 int_mathbbR frac11+x^2 frac11 + (z-x)^2 dx
$$
and I would look around trig substitution or partial fractions...
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
$endgroup$
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
add a comment |
$begingroup$
You made a mistake, the correct integral is
$$
frac1pi^2 int_mathbbR frac11+x^2 frac11 + (z-x)^2 dx
$$
and I would look around trig substitution or partial fractions...
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
$endgroup$
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
add a comment |
$begingroup$
You made a mistake, the correct integral is
$$
frac1pi^2 int_mathbbR frac11+x^2 frac11 + (z-x)^2 dx
$$
and I would look around trig substitution or partial fractions...
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
$endgroup$
You made a mistake, the correct integral is
$$
frac1pi^2 int_mathbbR frac11+x^2 frac11 + (z-x)^2 dx
$$
and I would look around trig substitution or partial fractions...
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
answered Mar 29 at 1:51
gt6989bgt6989b
35.5k22557
35.5k22557
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
add a comment |
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
thanks for pointing it out
$endgroup$
– Eric
Mar 29 at 2:07
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
$begingroup$
I think it is not possible to get the original one.
$endgroup$
– Eric
Mar 29 at 2:40
add a comment |
$begingroup$
You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
add a comment |
$begingroup$
You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
add a comment |
$begingroup$
You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
You could try partial fractions. An integral of (1+(x1)^2)^-1 on its own is fine to work out, it is an arctan integral. Likewise, an integral of (1+(z-x1)^2)^-1 is also very doable.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
answered Mar 29 at 1:31
George DewhirstGeorge Dewhirst
5234
5234
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
George Dewhirst is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
add a comment |
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
$begingroup$
Also double check your expression for fx2(z-x1)
$endgroup$
– George Dewhirst
Mar 29 at 1:34
add a comment |
1
$begingroup$
Does this integral exist? What happens when $x_1 = z$?
$endgroup$
– Ryan Goulden
Mar 29 at 1:35
$begingroup$
@RyanGoulden this one does not but the correct one does
$endgroup$
– gt6989b
Mar 29 at 1:56