Show $ int_ {-infty} ^{infty} (arctan(x+a)) {{1}over {sqrt{2 pi T}}} e^{-x^2 /2T} dx$ can assume any value in...
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I would like to show
$$ int_ {-infty} ^{infty} (arctan(x+a)) {{1}over {sqrt{2 pi T}}} e^{-x^2 /2T} dx$$ can assume any value in $(-pi/2, pi/2)$ where $T>0$ is fixed and $a$ may be any real number.
In the limit $T rightarrow 0$, I see the integral has value $arctan a$ which can assume any value in the above interval. I have tried to use properties of convolution but it gets me nowhere.
Any hint is appreciated.
calculus integration improper-integrals convolution
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$begingroup$
I would like to show
$$ int_ {-infty} ^{infty} (arctan(x+a)) {{1}over {sqrt{2 pi T}}} e^{-x^2 /2T} dx$$ can assume any value in $(-pi/2, pi/2)$ where $T>0$ is fixed and $a$ may be any real number.
In the limit $T rightarrow 0$, I see the integral has value $arctan a$ which can assume any value in the above interval. I have tried to use properties of convolution but it gets me nowhere.
Any hint is appreciated.
calculus integration improper-integrals convolution
$endgroup$
add a comment |
$begingroup$
I would like to show
$$ int_ {-infty} ^{infty} (arctan(x+a)) {{1}over {sqrt{2 pi T}}} e^{-x^2 /2T} dx$$ can assume any value in $(-pi/2, pi/2)$ where $T>0$ is fixed and $a$ may be any real number.
In the limit $T rightarrow 0$, I see the integral has value $arctan a$ which can assume any value in the above interval. I have tried to use properties of convolution but it gets me nowhere.
Any hint is appreciated.
calculus integration improper-integrals convolution
$endgroup$
I would like to show
$$ int_ {-infty} ^{infty} (arctan(x+a)) {{1}over {sqrt{2 pi T}}} e^{-x^2 /2T} dx$$ can assume any value in $(-pi/2, pi/2)$ where $T>0$ is fixed and $a$ may be any real number.
In the limit $T rightarrow 0$, I see the integral has value $arctan a$ which can assume any value in the above interval. I have tried to use properties of convolution but it gets me nowhere.
Any hint is appreciated.
calculus integration improper-integrals convolution
calculus integration improper-integrals convolution
asked Dec 1 '18 at 5:59
izimathizimath
366110
366110
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1 Answer
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Let your integral be $F(a)$. First note that by dominated convergence, $lim_{ato a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $lim_{ato pm infty} F(a)= pm pi/2$.
Use IVT to conclude.
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let your integral be $F(a)$. First note that by dominated convergence, $lim_{ato a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $lim_{ato pm infty} F(a)= pm pi/2$.
Use IVT to conclude.
$endgroup$
add a comment |
$begingroup$
Let your integral be $F(a)$. First note that by dominated convergence, $lim_{ato a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $lim_{ato pm infty} F(a)= pm pi/2$.
Use IVT to conclude.
$endgroup$
add a comment |
$begingroup$
Let your integral be $F(a)$. First note that by dominated convergence, $lim_{ato a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $lim_{ato pm infty} F(a)= pm pi/2$.
Use IVT to conclude.
$endgroup$
Let your integral be $F(a)$. First note that by dominated convergence, $lim_{ato a_0} F(a)=F(a_0)$. So $F$ is continuous. Again by dominated convergence, we know $lim_{ato pm infty} F(a)= pm pi/2$.
Use IVT to conclude.
answered Dec 1 '18 at 6:13
Zachary SelkZachary Selk
593311
593311
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