What will be the pdf of $X+Y$ if $X$ and $Y$ are iid from Cauchy? [duplicate]
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = int_{-infty}^{infty} f(y) f(x-y) mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=frac {1} {pi ({1+ x^2})},x in Bbb R$. Then the whole integration becomes $$frac {1} {pi^2} int_{-infty}^{infty} frac {mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.
probability probability-theory random-variables independence density-function
asked Nov 30 at 19:57
Dbchatto67
51215
marked as duplicate by StubbornAtom, Lord Shark the Unknown, Leucippus, Martin Sleziak, José Carlos Santos Dec 1 at 10:02
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.
|
show 4 more comments
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = int_{-infty}^{infty} f(y) f(x-y) mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=frac {1} {pi ({1+ x^2})},x in Bbb R$. Then the whole integration becomes $$frac {1} {pi^2} int_{-infty}^{infty} frac {mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.
probability probability-theory random-variables independence density-function
asked Nov 30 at 19:57
Dbchatto67
51215
marked as duplicate by StubbornAtom, Lord Shark the Unknown, Leucippus, Martin Sleziak, José Carlos Santos Dec 1 at 10:02
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.
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
1
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47
|
show 4 more comments
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = int_{-infty}^{infty} f(y) f(x-y) mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=frac {1} {pi ({1+ x^2})},x in Bbb R$. Then the whole integration becomes $$frac {1} {pi^2} int_{-infty}^{infty} frac {mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.
probability probability-theory random-variables independence density-function
asked Nov 30 at 19:57
Dbchatto67
51215
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = int_{-infty}^{infty} f(y) f(x-y) mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=frac {1} {pi ({1+ x^2})},x in Bbb R$. Then the whole integration becomes $$frac {1} {pi^2} int_{-infty}^{infty} frac {mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.
This question already has an answer here:
Proving the sum of two independent Cauchy Random Variables is Cauchy
2 answers
probability probability-theory random-variables independence density-function
probability probability-theory random-variables independence density-function
asked Nov 30 at 19:57
Dbchatto67
51215
asked Nov 30 at 19:57
Dbchatto67
51215
edited Dec 1 at 4:30
asked Nov 30 at 19:57
Dbchatto67
51215
asked Nov 30 at 19:57
Dbchatto67
51215
asked Nov 30 at 19:57
Dbchatto67
51215
51215
marked as duplicate by StubbornAtom, Lord Shark the Unknown, Leucippus, Martin Sleziak, José Carlos Santos Dec 1 at 10:02
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 StubbornAtom, Lord Shark the Unknown, Leucippus, Martin Sleziak, José Carlos Santos Dec 1 at 10:02
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.
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
1
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47
|
show 4 more comments
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
1
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
1
1
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47
|
show 4 more comments
2 Answers
2
active
oldest
votes
What about using characteristic functions? Let $phi$ be the characteristic function of a Cauchy distribution. We know that
$$phi(t) =frac 1piint_mathbb{R} frac{e^{itx}} {1+x^2}, dx$$
Such integrals are many times solved on this site, for instance here, the result is
$$phi(t) = e^{-|t|} $$
So the characteristic function of $X+Y$ is $phi$ squared, due to the independence of $X$ and $Y$. We conclude
$$phi_{X+Y} (t) =phi(t) phi(t) =e^{-2|t|}=e^{-|2t|}$$
Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.
answered Nov 30 at 20:50
Shashi
7,0361528
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
add a comment |
You can try partial fraction decomposition:
begin{align}frac1{(1+y^2)(1+(x-y)^2)} &= frac{x+2y}{x(x^2+4)(1+y^2)} + frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \
&= frac{1}{(x^2+4)(1+y^2)}+
frac{2y}{x(x^2+4)(1+y^2)} + frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + frac{1}{(x^2+4)(1+(x-y)^2)}\
&= frac1{x^2+4}left(frac1{1+y^2} + frac1{1+(x-y)^2}right) + frac1{x(x^2+4)}left(frac{y}{1+y^2} + frac{x-y}{1+(x-y)^2}right)
end{align}
so we have
begin{align}
frac1{pi^2}int_{-infty}^infty frac{dy}{(1+y^2)(1+(x-y)^2)} &=
frac1{pi^2(x^2+4)}left[int_{-infty}^inftyleft(frac1{1+y^2} + frac1{1+(x-y)^2}right)dy + frac1xint_{-infty}^inftyleft(frac{y}{1+y^2} + frac{2(x-y)}{1+(x-y)^2}right)dy
right]\
&= frac1{pi^2(x^2+4)}left[Big(arctan(1+y^2) + arctan(1+(x-y)^2)Big)Big|_{-infty}^infty + frac1xBig(ln(1+y^2) - ln(1+(x-y)^2)Big)Big|_{-infty}^infty
right]\
&= frac1{pi^2(x^2+4)}left[2pi + frac2x lim_{y to infty} ln left(frac{1+y^2}{1+(x-y)^2}right)right]\
&= frac{2}{pi(x^2+4)}
end{align}
answered Nov 30 at 20:43
mechanodroid
26.1k62245
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
What about using characteristic functions? Let $phi$ be the characteristic function of a Cauchy distribution. We know that
$$phi(t) =frac 1piint_mathbb{R} frac{e^{itx}} {1+x^2}, dx$$
Such integrals are many times solved on this site, for instance here, the result is
$$phi(t) = e^{-|t|} $$
So the characteristic function of $X+Y$ is $phi$ squared, due to the independence of $X$ and $Y$. We conclude
$$phi_{X+Y} (t) =phi(t) phi(t) =e^{-2|t|}=e^{-|2t|}$$
Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.
answered Nov 30 at 20:50
Shashi
7,0361528
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
add a comment |
What about using characteristic functions? Let $phi$ be the characteristic function of a Cauchy distribution. We know that
$$phi(t) =frac 1piint_mathbb{R} frac{e^{itx}} {1+x^2}, dx$$
Such integrals are many times solved on this site, for instance here, the result is
$$phi(t) = e^{-|t|} $$
So the characteristic function of $X+Y$ is $phi$ squared, due to the independence of $X$ and $Y$. We conclude
$$phi_{X+Y} (t) =phi(t) phi(t) =e^{-2|t|}=e^{-|2t|}$$
Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.
answered Nov 30 at 20:50
Shashi
7,0361528
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
add a comment |
What about using characteristic functions? Let $phi$ be the characteristic function of a Cauchy distribution. We know that
$$phi(t) =frac 1piint_mathbb{R} frac{e^{itx}} {1+x^2}, dx$$
Such integrals are many times solved on this site, for instance here, the result is
$$phi(t) = e^{-|t|} $$
So the characteristic function of $X+Y$ is $phi$ squared, due to the independence of $X$ and $Y$. We conclude
$$phi_{X+Y} (t) =phi(t) phi(t) =e^{-2|t|}=e^{-|2t|}$$
Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.
answered Nov 30 at 20:50
Shashi
7,0361528
What about using characteristic functions? Let $phi$ be the characteristic function of a Cauchy distribution. We know that
$$phi(t) =frac 1piint_mathbb{R} frac{e^{itx}} {1+x^2}, dx$$
Such integrals are many times solved on this site, for instance here, the result is
$$phi(t) = e^{-|t|} $$
So the characteristic function of $X+Y$ is $phi$ squared, due to the independence of $X$ and $Y$. We conclude
$$phi_{X+Y} (t) =phi(t) phi(t) =e^{-2|t|}=e^{-|2t|}$$
Indeed, we may also conclude from this that $(X+Y) /2$ is Cauchy distributed.
answered Nov 30 at 20:50
Shashi
7,0361528
answered Nov 30 at 20:50
Shashi
7,0361528
answered Nov 30 at 20:50
Shashi
7,0361528
answered Nov 30 at 20:50
Shashi
7,0361528
7,0361528
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
add a comment |
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
The process you have mentioned is somewhat related to Fourier transform. Isn't it so?
– Dbchatto67
Nov 30 at 21:11
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
@Dbchatto67 yep. Fourier transform of the probability density. However you should be aware that the characteristic function uniquely defines the distribution of the random variable. Have you covered moment generating functions or characteristic functions? It's worth learning them since they are useful.
– Shashi
Nov 30 at 21:15
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
Yeah I covered all of them. Our professor in Indian Statistical Institute has included all these things in our first semester syllabus.
– Dbchatto67
Nov 30 at 21:24
add a comment |
You can try partial fraction decomposition:
begin{align}frac1{(1+y^2)(1+(x-y)^2)} &= frac{x+2y}{x(x^2+4)(1+y^2)} + frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \
&= frac{1}{(x^2+4)(1+y^2)}+
frac{2y}{x(x^2+4)(1+y^2)} + frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + frac{1}{(x^2+4)(1+(x-y)^2)}\
&= frac1{x^2+4}left(frac1{1+y^2} + frac1{1+(x-y)^2}right) + frac1{x(x^2+4)}left(frac{y}{1+y^2} + frac{x-y}{1+(x-y)^2}right)
end{align}
so we have
begin{align}
frac1{pi^2}int_{-infty}^infty frac{dy}{(1+y^2)(1+(x-y)^2)} &=
frac1{pi^2(x^2+4)}left[int_{-infty}^inftyleft(frac1{1+y^2} + frac1{1+(x-y)^2}right)dy + frac1xint_{-infty}^inftyleft(frac{y}{1+y^2} + frac{2(x-y)}{1+(x-y)^2}right)dy
right]\
&= frac1{pi^2(x^2+4)}left[Big(arctan(1+y^2) + arctan(1+(x-y)^2)Big)Big|_{-infty}^infty + frac1xBig(ln(1+y^2) - ln(1+(x-y)^2)Big)Big|_{-infty}^infty
right]\
&= frac1{pi^2(x^2+4)}left[2pi + frac2x lim_{y to infty} ln left(frac{1+y^2}{1+(x-y)^2}right)right]\
&= frac{2}{pi(x^2+4)}
end{align}
answered Nov 30 at 20:43
mechanodroid
26.1k62245
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
add a comment |
You can try partial fraction decomposition:
begin{align}frac1{(1+y^2)(1+(x-y)^2)} &= frac{x+2y}{x(x^2+4)(1+y^2)} + frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \
&= frac{1}{(x^2+4)(1+y^2)}+
frac{2y}{x(x^2+4)(1+y^2)} + frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + frac{1}{(x^2+4)(1+(x-y)^2)}\
&= frac1{x^2+4}left(frac1{1+y^2} + frac1{1+(x-y)^2}right) + frac1{x(x^2+4)}left(frac{y}{1+y^2} + frac{x-y}{1+(x-y)^2}right)
end{align}
so we have
begin{align}
frac1{pi^2}int_{-infty}^infty frac{dy}{(1+y^2)(1+(x-y)^2)} &=
frac1{pi^2(x^2+4)}left[int_{-infty}^inftyleft(frac1{1+y^2} + frac1{1+(x-y)^2}right)dy + frac1xint_{-infty}^inftyleft(frac{y}{1+y^2} + frac{2(x-y)}{1+(x-y)^2}right)dy
right]\
&= frac1{pi^2(x^2+4)}left[Big(arctan(1+y^2) + arctan(1+(x-y)^2)Big)Big|_{-infty}^infty + frac1xBig(ln(1+y^2) - ln(1+(x-y)^2)Big)Big|_{-infty}^infty
right]\
&= frac1{pi^2(x^2+4)}left[2pi + frac2x lim_{y to infty} ln left(frac{1+y^2}{1+(x-y)^2}right)right]\
&= frac{2}{pi(x^2+4)}
end{align}
answered Nov 30 at 20:43
mechanodroid
26.1k62245
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
add a comment |
You can try partial fraction decomposition:
begin{align}frac1{(1+y^2)(1+(x-y)^2)} &= frac{x+2y}{x(x^2+4)(1+y^2)} + frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \
&= frac{1}{(x^2+4)(1+y^2)}+
frac{2y}{x(x^2+4)(1+y^2)} + frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + frac{1}{(x^2+4)(1+(x-y)^2)}\
&= frac1{x^2+4}left(frac1{1+y^2} + frac1{1+(x-y)^2}right) + frac1{x(x^2+4)}left(frac{y}{1+y^2} + frac{x-y}{1+(x-y)^2}right)
end{align}
so we have
begin{align}
frac1{pi^2}int_{-infty}^infty frac{dy}{(1+y^2)(1+(x-y)^2)} &=
frac1{pi^2(x^2+4)}left[int_{-infty}^inftyleft(frac1{1+y^2} + frac1{1+(x-y)^2}right)dy + frac1xint_{-infty}^inftyleft(frac{y}{1+y^2} + frac{2(x-y)}{1+(x-y)^2}right)dy
right]\
&= frac1{pi^2(x^2+4)}left[Big(arctan(1+y^2) + arctan(1+(x-y)^2)Big)Big|_{-infty}^infty + frac1xBig(ln(1+y^2) - ln(1+(x-y)^2)Big)Big|_{-infty}^infty
right]\
&= frac1{pi^2(x^2+4)}left[2pi + frac2x lim_{y to infty} ln left(frac{1+y^2}{1+(x-y)^2}right)right]\
&= frac{2}{pi(x^2+4)}
end{align}
answered Nov 30 at 20:43
mechanodroid
26.1k62245
You can try partial fraction decomposition:
begin{align}frac1{(1+y^2)(1+(x-y)^2)} &= frac{x+2y}{x(x^2+4)(1+y^2)} + frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \
&= frac{1}{(x^2+4)(1+y^2)}+
frac{2y}{x(x^2+4)(1+y^2)} + frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + frac{1}{(x^2+4)(1+(x-y)^2)}\
&= frac1{x^2+4}left(frac1{1+y^2} + frac1{1+(x-y)^2}right) + frac1{x(x^2+4)}left(frac{y}{1+y^2} + frac{x-y}{1+(x-y)^2}right)
end{align}
so we have
begin{align}
frac1{pi^2}int_{-infty}^infty frac{dy}{(1+y^2)(1+(x-y)^2)} &=
frac1{pi^2(x^2+4)}left[int_{-infty}^inftyleft(frac1{1+y^2} + frac1{1+(x-y)^2}right)dy + frac1xint_{-infty}^inftyleft(frac{y}{1+y^2} + frac{2(x-y)}{1+(x-y)^2}right)dy
right]\
&= frac1{pi^2(x^2+4)}left[Big(arctan(1+y^2) + arctan(1+(x-y)^2)Big)Big|_{-infty}^infty + frac1xBig(ln(1+y^2) - ln(1+(x-y)^2)Big)Big|_{-infty}^infty
right]\
&= frac1{pi^2(x^2+4)}left[2pi + frac2x lim_{y to infty} ln left(frac{1+y^2}{1+(x-y)^2}right)right]\
&= frac{2}{pi(x^2+4)}
end{align}
answered Nov 30 at 20:43
mechanodroid
26.1k62245
answered Nov 30 at 20:43
mechanodroid
26.1k62245
answered Nov 30 at 20:43
mechanodroid
26.1k62245
answered Nov 30 at 20:43
mechanodroid
26.1k62245
26.1k62245
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
add a comment |
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
I have already done it. There is no need to post this unnecessary answer. Look at my comment above @mechanodroid.
– Dbchatto67
Nov 30 at 21:14
add a comment |
Do you know contour integration?
– Shashi
Nov 30 at 20:12
Yeah. I know. Is it related to contour integration?
– Dbchatto67
Nov 30 at 20:19
Yes! Maybe you can try to do that, or partial fraction decomposition. The latter needs longer calculations I guess.
– Shashi
Nov 30 at 20:25
Yeah. I have finally found $g$ which is given by $g(x)= frac {2} {pi (x^2+4)},xin Bbb R$.
– Dbchatto67
Nov 30 at 20:31
1
The usual approach is to use characteristic functions. If $X$ and $Y$ are Cauchy with parameters $a$ and $b$, $$E(e^{itX})=e^{-a|t|}qquad E(e^{itY})=e^{-b|t|}$$ If furthermore $X$ and $Y$ are independent, $$E(e^{it(X+Y)})=E(e^{itX})E(e^{itY})=e^{-a|t|}e^{-b|t|}=e^{-(a+b)|t|}$$ hence $X+Y$ is Cauchy with parameter $a+b$. Thus, it suffices to know how to prove that $$int_mathbb Re^{itx}frac{dx}{1+x^2}=e^{-|t|}$$ Do you?
– Did
Nov 30 at 20:47