Analytic integration of discontinuous function
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I'm trying to calculate the convolution between a Gaussian and a discontinuous function:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau$$
Where
$$f(tgeq0)=mathrmexp(-kt),,f(t<0)=0$$
$$g(t)=frac1sigmasqrt(2pi)mathrmexp(-frac12(fract- musigma)^2)$$
I've managed to calculate the positive part by limiting the integral domain:
$$int_0^infty f(tau),g(t-tau),mathrmdtau =frac12mathrmexp(-kt),mathrmexp(frack2(ksigma^2+2mu)),left(mathrmerfleft(fract-,k,sigma ^2-,musqrt2,sigma right)+1right)$$
Is there any way to calculate the negative part ?
Update:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau=int_-infty^0 f(tau),g(t-tau),mathrmdtau+int_0^infty f(tau),g(t-tau),mathrmdtau=0+int_0^infty f(tau),g(t-tau),mathrmdtau$$
integration definite-integrals convolution discontinuous-functions
asked Aug 27 at 15:55
7E10FC9A
33
add a comment |Â
up vote
0
down vote
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I'm trying to calculate the convolution between a Gaussian and a discontinuous function:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau$$
Where
$$f(tgeq0)=mathrmexp(-kt),,f(t<0)=0$$
$$g(t)=frac1sigmasqrt(2pi)mathrmexp(-frac12(fract- musigma)^2)$$
I've managed to calculate the positive part by limiting the integral domain:
$$int_0^infty f(tau),g(t-tau),mathrmdtau =frac12mathrmexp(-kt),mathrmexp(frack2(ksigma^2+2mu)),left(mathrmerfleft(fract-,k,sigma ^2-,musqrt2,sigma right)+1right)$$
Is there any way to calculate the negative part ?
Update:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau=int_-infty^0 f(tau),g(t-tau),mathrmdtau+int_0^infty f(tau),g(t-tau),mathrmdtau=0+int_0^infty f(tau),g(t-tau),mathrmdtau$$
integration definite-integrals convolution discontinuous-functions
asked Aug 27 at 15:55
7E10FC9A
33
What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
1
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to calculate the convolution between a Gaussian and a discontinuous function:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau$$
Where
$$f(tgeq0)=mathrmexp(-kt),,f(t<0)=0$$
$$g(t)=frac1sigmasqrt(2pi)mathrmexp(-frac12(fract- musigma)^2)$$
I've managed to calculate the positive part by limiting the integral domain:
$$int_0^infty f(tau),g(t-tau),mathrmdtau =frac12mathrmexp(-kt),mathrmexp(frack2(ksigma^2+2mu)),left(mathrmerfleft(fract-,k,sigma ^2-,musqrt2,sigma right)+1right)$$
Is there any way to calculate the negative part ?
Update:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau=int_-infty^0 f(tau),g(t-tau),mathrmdtau+int_0^infty f(tau),g(t-tau),mathrmdtau=0+int_0^infty f(tau),g(t-tau),mathrmdtau$$
integration definite-integrals convolution discontinuous-functions
asked Aug 27 at 15:55
7E10FC9A
33
I'm trying to calculate the convolution between a Gaussian and a discontinuous function:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau$$
Where
$$f(tgeq0)=mathrmexp(-kt),,f(t<0)=0$$
$$g(t)=frac1sigmasqrt(2pi)mathrmexp(-frac12(fract- musigma)^2)$$
I've managed to calculate the positive part by limiting the integral domain:
$$int_0^infty f(tau),g(t-tau),mathrmdtau =frac12mathrmexp(-kt),mathrmexp(frack2(ksigma^2+2mu)),left(mathrmerfleft(fract-,k,sigma ^2-,musqrt2,sigma right)+1right)$$
Is there any way to calculate the negative part ?
Update:
$$(f*g)(t)=int_-infty^infty f(tau),g(t-tau),mathrmdtau=int_-infty^0 f(tau),g(t-tau),mathrmdtau+int_0^infty f(tau),g(t-tau),mathrmdtau=0+int_0^infty f(tau),g(t-tau),mathrmdtau$$
integration definite-integrals convolution discontinuous-functions
asked Aug 27 at 15:55
7E10FC9A
33
edited Aug 27 at 16:34
asked Aug 27 at 15:55
7E10FC9A
33
asked Aug 27 at 15:55
7E10FC9A
33
asked Aug 27 at 15:55
7E10FC9A
33
33
What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
1
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12
add a comment |Â
What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
1
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12
What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
1
1
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12
add a comment |Â
1 Answer
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up vote
2
down vote
accepted
You have
$$ (f*g)(t) = int_-infty^infty f(tau)g(t-tau) , dtau = int_0^infty f(tau)g(t-tau) , dtau + int_-infty^0 f(tau)g(t-tau) , dtau. $$
You already computed the first integral, and the second integral is $0$, since $f(tau) = 0$ for all $tau < 0$.
answered Aug 27 at 16:21
Sobi
2,418315
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have
$$ (f*g)(t) = int_-infty^infty f(tau)g(t-tau) , dtau = int_0^infty f(tau)g(t-tau) , dtau + int_-infty^0 f(tau)g(t-tau) , dtau. $$
You already computed the first integral, and the second integral is $0$, since $f(tau) = 0$ for all $tau < 0$.
answered Aug 27 at 16:21
Sobi
2,418315
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
add a comment |Â
up vote
2
down vote
accepted
You have
$$ (f*g)(t) = int_-infty^infty f(tau)g(t-tau) , dtau = int_0^infty f(tau)g(t-tau) , dtau + int_-infty^0 f(tau)g(t-tau) , dtau. $$
You already computed the first integral, and the second integral is $0$, since $f(tau) = 0$ for all $tau < 0$.
answered Aug 27 at 16:21
Sobi
2,418315
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have
$$ (f*g)(t) = int_-infty^infty f(tau)g(t-tau) , dtau = int_0^infty f(tau)g(t-tau) , dtau + int_-infty^0 f(tau)g(t-tau) , dtau. $$
You already computed the first integral, and the second integral is $0$, since $f(tau) = 0$ for all $tau < 0$.
answered Aug 27 at 16:21
Sobi
2,418315
You have
$$ (f*g)(t) = int_-infty^infty f(tau)g(t-tau) , dtau = int_0^infty f(tau)g(t-tau) , dtau + int_-infty^0 f(tau)g(t-tau) , dtau. $$
You already computed the first integral, and the second integral is $0$, since $f(tau) = 0$ for all $tau < 0$.
answered Aug 27 at 16:21
Sobi
2,418315
answered Aug 27 at 16:21
Sobi
2,418315
answered Aug 27 at 16:21
Sobi
2,418315
answered Aug 27 at 16:21
Sobi
2,418315
2,418315
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
add a comment |Â
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
Yes, you are right. Thx.
– 7E10FC9A
Aug 27 at 16:35
add a comment |Â
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What do you mean by "even $f(t)$ is zero, the integral is not"?
– Sobi
Aug 27 at 15:58
1
Maybe this will help. If $y<0$, then $f(y)=0$, so also $f(y)g(t-y)=0$, so $int_-infty^0 f(y)g(t-y)textdy = 0$
– Rumpelstiltskin
Aug 27 at 15:58
@Sobi at t<0, g(t) still overlap with the positive part of f(t), note that the integration is not with respect to t.
– 7E10FC9A
Aug 27 at 16:11
@7E10FC9A But you're integrating with respect to $tau$, and you have $f(tau)$ in your integral. So when $tau < 0,$ then $f(tau) = 0$ and the integral is $0$.
– Sobi
Aug 27 at 16:12