An exemple of integral of distributions












0












$begingroup$


Need to solve this integral:
$$I=int_{-1}^{1}dx(lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2}f(x)+pivartheta(x)frac{df(x)}{dx}(x)) $$
I think I should recognize the limit as a distribution, but I can't find which one with a precise argument, I think it is $ delta$ Dirac function. So if it is correct this is my solution:
$$I=int_{-1}^{1}delta(x)f(x)dx+piint_{-1}^{1}vartheta(x)frac{df(x)}{dx}(x)dx=$$
$$f(0)+pi(f(1)-f(0))$$



Is this correct? If the identification of the limit with delta function is correct, can you please tell me why?



Thanks a lot






integration limits dirac-delta step-function







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  • $begingroup$
    $$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
    $endgroup$
    – md2perpe
    Dec 31 '18 at 13:54
















0












$begingroup$


Need to solve this integral:
$$I=int_{-1}^{1}dx(lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2}f(x)+pivartheta(x)frac{df(x)}{dx}(x)) $$
I think I should recognize the limit as a distribution, but I can't find which one with a precise argument, I think it is $ delta$ Dirac function. So if it is correct this is my solution:
$$I=int_{-1}^{1}delta(x)f(x)dx+piint_{-1}^{1}vartheta(x)frac{df(x)}{dx}(x)dx=$$
$$f(0)+pi(f(1)-f(0))$$



Is this correct? If the identification of the limit with delta function is correct, can you please tell me why?



Thanks a lot






integration limits dirac-delta step-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
    $endgroup$
    – md2perpe
    Dec 31 '18 at 13:54














0












0








0





$begingroup$


Need to solve this integral:
$$I=int_{-1}^{1}dx(lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2}f(x)+pivartheta(x)frac{df(x)}{dx}(x)) $$
I think I should recognize the limit as a distribution, but I can't find which one with a precise argument, I think it is $ delta$ Dirac function. So if it is correct this is my solution:
$$I=int_{-1}^{1}delta(x)f(x)dx+piint_{-1}^{1}vartheta(x)frac{df(x)}{dx}(x)dx=$$
$$f(0)+pi(f(1)-f(0))$$



Is this correct? If the identification of the limit with delta function is correct, can you please tell me why?



Thanks a lot






integration limits dirac-delta step-function







share|cite|improve this question











$endgroup$




Need to solve this integral:
$$I=int_{-1}^{1}dx(lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2}f(x)+pivartheta(x)frac{df(x)}{dx}(x)) $$
I think I should recognize the limit as a distribution, but I can't find which one with a precise argument, I think it is $ delta$ Dirac function. So if it is correct this is my solution:
$$I=int_{-1}^{1}delta(x)f(x)dx+piint_{-1}^{1}vartheta(x)frac{df(x)}{dx}(x)dx=$$
$$f(0)+pi(f(1)-f(0))$$



Is this correct? If the identification of the limit with delta function is correct, can you please tell me why?



Thanks a lot






integration limits dirac-delta step-function




integration limits dirac-delta step-function






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share|cite|improve this question













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edited Dec 31 '18 at 13:56







pter26

















asked Dec 31 '18 at 13:01









pter26pter26

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317111












  • $begingroup$
    $$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
    $endgroup$
    – md2perpe
    Dec 31 '18 at 13:54


















  • $begingroup$
    $$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
    $endgroup$
    – md2perpe
    Dec 31 '18 at 13:54
















$begingroup$
$$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
$endgroup$
– md2perpe
Dec 31 '18 at 13:54




$begingroup$
$$lim_{varepsilonto 0^+}frac{varepsilon}{varepsilon^2+x^2} = pi delta(x)$$
$endgroup$
– md2perpe
Dec 31 '18 at 13:54










1 Answer
1






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oldest

votes


















1












$begingroup$

Almost. The limit is $pi delta(x)$ since
$$
int_{-infty}^{infty} frac{epsilon}{epsilon^2+x^2} phi(x) , dx
= { x = epsilon y }
= int_{-infty}^{infty} frac{epsilon}{epsilon^2+epsilon^2 y^2} phi(epsilon y) , epsilon , dy
= int_{-infty}^{infty} frac{1}{1+y^2} phi(epsilon y) , dy \
to int_{-infty}^{infty} frac{1}{1+y^2} phi(0) , dy
= left( int_{-infty}^{infty} frac{1}{1+y^2} , dy right) phi(0)
= pi phi(0)
= int_{-infty}^{infty} pi delta(x) , phi(x) , dx
$$
for all $phi in C_c^infty.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:02













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

Almost. The limit is $pi delta(x)$ since
$$
int_{-infty}^{infty} frac{epsilon}{epsilon^2+x^2} phi(x) , dx
= { x = epsilon y }
= int_{-infty}^{infty} frac{epsilon}{epsilon^2+epsilon^2 y^2} phi(epsilon y) , epsilon , dy
= int_{-infty}^{infty} frac{1}{1+y^2} phi(epsilon y) , dy \
to int_{-infty}^{infty} frac{1}{1+y^2} phi(0) , dy
= left( int_{-infty}^{infty} frac{1}{1+y^2} , dy right) phi(0)
= pi phi(0)
= int_{-infty}^{infty} pi delta(x) , phi(x) , dx
$$
for all $phi in C_c^infty.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:02


















1












$begingroup$

Almost. The limit is $pi delta(x)$ since
$$
int_{-infty}^{infty} frac{epsilon}{epsilon^2+x^2} phi(x) , dx
= { x = epsilon y }
= int_{-infty}^{infty} frac{epsilon}{epsilon^2+epsilon^2 y^2} phi(epsilon y) , epsilon , dy
= int_{-infty}^{infty} frac{1}{1+y^2} phi(epsilon y) , dy \
to int_{-infty}^{infty} frac{1}{1+y^2} phi(0) , dy
= left( int_{-infty}^{infty} frac{1}{1+y^2} , dy right) phi(0)
= pi phi(0)
= int_{-infty}^{infty} pi delta(x) , phi(x) , dx
$$
for all $phi in C_c^infty.$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:02
















1












1








1





$begingroup$

Almost. The limit is $pi delta(x)$ since
$$
int_{-infty}^{infty} frac{epsilon}{epsilon^2+x^2} phi(x) , dx
= { x = epsilon y }
= int_{-infty}^{infty} frac{epsilon}{epsilon^2+epsilon^2 y^2} phi(epsilon y) , epsilon , dy
= int_{-infty}^{infty} frac{1}{1+y^2} phi(epsilon y) , dy \
to int_{-infty}^{infty} frac{1}{1+y^2} phi(0) , dy
= left( int_{-infty}^{infty} frac{1}{1+y^2} , dy right) phi(0)
= pi phi(0)
= int_{-infty}^{infty} pi delta(x) , phi(x) , dx
$$
for all $phi in C_c^infty.$






share|cite|improve this answer









$endgroup$



Almost. The limit is $pi delta(x)$ since
$$
int_{-infty}^{infty} frac{epsilon}{epsilon^2+x^2} phi(x) , dx
= { x = epsilon y }
= int_{-infty}^{infty} frac{epsilon}{epsilon^2+epsilon^2 y^2} phi(epsilon y) , epsilon , dy
= int_{-infty}^{infty} frac{1}{1+y^2} phi(epsilon y) , dy \
to int_{-infty}^{infty} frac{1}{1+y^2} phi(0) , dy
= left( int_{-infty}^{infty} frac{1}{1+y^2} , dy right) phi(0)
= pi phi(0)
= int_{-infty}^{infty} pi delta(x) , phi(x) , dx
$$
for all $phi in C_c^infty.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 31 '18 at 13:58









md2perpemd2perpe

7,78111028




7,78111028








  • 1




    $begingroup$
    Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:02
















  • 1




    $begingroup$
    Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
    $endgroup$
    – Jean Marie
    Dec 31 '18 at 19:02










1




1




$begingroup$
Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
$endgroup$
– Jean Marie
Dec 31 '18 at 19:02






$begingroup$
Has the OP observed that the graphical representation of function $f_{varepsilon}$ defined by $f_{varepsilon}(x):=varepsilon/(varepsilon^2+x^2)$ is obtained from the curve of $f(x):=1/(1+x^2)$ (Cauchy function) by squeezing it in a ratio $1:varepsilon$ along the $x$-axis while elongating it in an inverse ratio along the $y$ axis, thus preserving the area under its curve.
$endgroup$
– Jean Marie
Dec 31 '18 at 19:02




















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