Integration involving a Dirac delta function

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How can the following integral can be solved
$$I(t_1,t_2)=int_0^t_1dt'e^-a(t_1-t')int_0^t_2dt''e^-a(t_2-t'')delta(t''-t')$$
where there are no assumptions regarding $t_1$ or $t_2$ (both cases of $t_1>t_2$ or $t_2>t_1$ are valid).

are there any limitations for using the Dirac delta here?

dirac-delta

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asked Jul 22 at 11:51

jarhead

1148

add a comment | 

up vote
0
down vote

favorite

How can the following integral can be solved
$$I(t_1,t_2)=int_0^t_1dt'e^-a(t_1-t')int_0^t_2dt''e^-a(t_2-t'')delta(t''-t')$$
where there are no assumptions regarding $t_1$ or $t_2$ (both cases of $t_1>t_2$ or $t_2>t_1$ are valid).

are there any limitations for using the Dirac delta here?

dirac-delta

share|cite|improve this question

asked Jul 22 at 11:51

jarhead

1148

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

How can the following integral can be solved
$$I(t_1,t_2)=int_0^t_1dt'e^-a(t_1-t')int_0^t_2dt''e^-a(t_2-t'')delta(t''-t')$$
where there are no assumptions regarding $t_1$ or $t_2$ (both cases of $t_1>t_2$ or $t_2>t_1$ are valid).

are there any limitations for using the Dirac delta here?

dirac-delta

share|cite|improve this question

asked Jul 22 at 11:51

jarhead

1148

How can the following integral can be solved
$$I(t_1,t_2)=int_0^t_1dt'e^-a(t_1-t')int_0^t_2dt''e^-a(t_2-t'')delta(t''-t')$$
where there are no assumptions regarding $t_1$ or $t_2$ (both cases of $t_1>t_2$ or $t_2>t_1$ are valid).

are there any limitations for using the Dirac delta here?

dirac-delta

share|cite|improve this question

asked Jul 22 at 11:51

jarhead

1148

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 11:51

jarhead

1148

asked Jul 22 at 11:51

jarhead

1148

asked Jul 22 at 11:51

jarhead

1148

1148

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
2
down vote



accepted

First I do the inner integral:
$$
int_0^t_2 dt'' , e^-a(t_2-t'') , delta(t''-t')
= int_-infty^infty dt'' , chi_[0, t_2](t'') , e^-a(t_2-t'') , delta(t''-t')
= chi_[0, t_2](t') , e^-a(t_2-t')
$$
Here, for any set $A$, $chi_A(t) = 1$ if $tin A$, $chi_A(t)=0$ otherwise.

Then the outer integral:
$$
I(t_1,t_2)
= int_0^t_1 dt' , e^-a(t_1-t') , chi_[0, t_2](t') , e^-a(t_2-t') \
= e^-a(t_1+t_2) int_0^t_1 dt' , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, t_1](t') , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, min(t_1,t_2)](t') , e^2at' \
= e^-a(t_1+t_2) int_0^min(t_1,t_2) dt' , e^2at' \
= e^-a(t_1+t_2) left[frac12a e^2at' right]_0^min(t_1,t_2) \
= frac12a e^-a(t_1+t_2) left[e^2amin(t_1,t_2) -1 right]
$$

share|cite|improve this answer

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ is just some set. I expanded on it somewhat in my post.
  – md2perpe
  Jul 22 at 12:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
  – md2perpe
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
  – md2perpe
  Jul 22 at 14:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
  – Maxim
  Jul 22 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
  – Maxim
  Jul 22 at 16:09
  
  

  
  
  
  

 | 
show 11 more comments

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

First I do the inner integral:
$$
int_0^t_2 dt'' , e^-a(t_2-t'') , delta(t''-t')
= int_-infty^infty dt'' , chi_[0, t_2](t'') , e^-a(t_2-t'') , delta(t''-t')
= chi_[0, t_2](t') , e^-a(t_2-t')
$$
Here, for any set $A$, $chi_A(t) = 1$ if $tin A$, $chi_A(t)=0$ otherwise.

Then the outer integral:
$$
I(t_1,t_2)
= int_0^t_1 dt' , e^-a(t_1-t') , chi_[0, t_2](t') , e^-a(t_2-t') \
= e^-a(t_1+t_2) int_0^t_1 dt' , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, t_1](t') , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, min(t_1,t_2)](t') , e^2at' \
= e^-a(t_1+t_2) int_0^min(t_1,t_2) dt' , e^2at' \
= e^-a(t_1+t_2) left[frac12a e^2at' right]_0^min(t_1,t_2) \
= frac12a e^-a(t_1+t_2) left[e^2amin(t_1,t_2) -1 right]
$$

share|cite|improve this answer

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ is just some set. I expanded on it somewhat in my post.
  – md2perpe
  Jul 22 at 12:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
  – md2perpe
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
  – md2perpe
  Jul 22 at 14:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
  – Maxim
  Jul 22 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
  – Maxim
  Jul 22 at 16:09
  
  

  
  
  
  

 | 
show 11 more comments

up vote
2
down vote



accepted

First I do the inner integral:
$$
int_0^t_2 dt'' , e^-a(t_2-t'') , delta(t''-t')
= int_-infty^infty dt'' , chi_[0, t_2](t'') , e^-a(t_2-t'') , delta(t''-t')
= chi_[0, t_2](t') , e^-a(t_2-t')
$$
Here, for any set $A$, $chi_A(t) = 1$ if $tin A$, $chi_A(t)=0$ otherwise.

Then the outer integral:
$$
I(t_1,t_2)
= int_0^t_1 dt' , e^-a(t_1-t') , chi_[0, t_2](t') , e^-a(t_2-t') \
= e^-a(t_1+t_2) int_0^t_1 dt' , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, t_1](t') , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, min(t_1,t_2)](t') , e^2at' \
= e^-a(t_1+t_2) int_0^min(t_1,t_2) dt' , e^2at' \
= e^-a(t_1+t_2) left[frac12a e^2at' right]_0^min(t_1,t_2) \
= frac12a e^-a(t_1+t_2) left[e^2amin(t_1,t_2) -1 right]
$$

share|cite|improve this answer

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ is just some set. I expanded on it somewhat in my post.
  – md2perpe
  Jul 22 at 12:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
  – md2perpe
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
  – md2perpe
  Jul 22 at 14:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
  – Maxim
  Jul 22 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
  – Maxim
  Jul 22 at 16:09
  
  

  
  
  
  

 | 
show 11 more comments

up vote
2
down vote



accepted

up vote
2
down vote



accepted

First I do the inner integral:
$$
int_0^t_2 dt'' , e^-a(t_2-t'') , delta(t''-t')
= int_-infty^infty dt'' , chi_[0, t_2](t'') , e^-a(t_2-t'') , delta(t''-t')
= chi_[0, t_2](t') , e^-a(t_2-t')
$$
Here, for any set $A$, $chi_A(t) = 1$ if $tin A$, $chi_A(t)=0$ otherwise.

Then the outer integral:
$$
I(t_1,t_2)
= int_0^t_1 dt' , e^-a(t_1-t') , chi_[0, t_2](t') , e^-a(t_2-t') \
= e^-a(t_1+t_2) int_0^t_1 dt' , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, t_1](t') , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, min(t_1,t_2)](t') , e^2at' \
= e^-a(t_1+t_2) int_0^min(t_1,t_2) dt' , e^2at' \
= e^-a(t_1+t_2) left[frac12a e^2at' right]_0^min(t_1,t_2) \
= frac12a e^-a(t_1+t_2) left[e^2amin(t_1,t_2) -1 right]
$$

share|cite|improve this answer

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

First I do the inner integral:
$$
int_0^t_2 dt'' , e^-a(t_2-t'') , delta(t''-t')
= int_-infty^infty dt'' , chi_[0, t_2](t'') , e^-a(t_2-t'') , delta(t''-t')
= chi_[0, t_2](t') , e^-a(t_2-t')
$$
Here, for any set $A$, $chi_A(t) = 1$ if $tin A$, $chi_A(t)=0$ otherwise.

Then the outer integral:
$$
I(t_1,t_2)
= int_0^t_1 dt' , e^-a(t_1-t') , chi_[0, t_2](t') , e^-a(t_2-t') \
= e^-a(t_1+t_2) int_0^t_1 dt' , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, t_1](t') , e^2at' , chi_[0, t_2](t') \
= e^-a(t_1+t_2) int_-infty^infty dt' , chi_[0, min(t_1,t_2)](t') , e^2at' \
= e^-a(t_1+t_2) int_0^min(t_1,t_2) dt' , e^2at' \
= e^-a(t_1+t_2) left[frac12a e^2at' right]_0^min(t_1,t_2) \
= frac12a e^-a(t_1+t_2) left[e^2amin(t_1,t_2) -1 right]
$$

share|cite|improve this answer

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 15:39

edited Jul 22 at 15:39

edited Jul 22 at 15:39

answered Jul 22 at 12:18

md2perpe

5,85511022

answered Jul 22 at 12:18

md2perpe

5,85511022

answered Jul 22 at 12:18

md2perpe

5,85511022

5,85511022

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ is just some set. I expanded on it somewhat in my post.
  – md2perpe
  Jul 22 at 12:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
  – md2perpe
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
  – md2perpe
  Jul 22 at 14:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
  – Maxim
  Jul 22 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
  – Maxim
  Jul 22 at 16:09
  
  

  
  
  
  

 | 
show 11 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $A$ is just some set. I expanded on it somewhat in my post.
  – md2perpe
  Jul 22 at 12:23
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
  – md2perpe
  Jul 22 at 12:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
  – md2perpe
  Jul 22 at 14:06
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
  – Maxim
  Jul 22 at 15:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
  – Maxim
  Jul 22 at 16:09
  
  

  
  
  
  

1

1

$A$ is just some set. I expanded on it somewhat in my post.
– md2perpe
Jul 22 at 12:23

$A$ is just some set. I expanded on it somewhat in my post.
– md2perpe
Jul 22 at 12:23

1

1

It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
– md2perpe
Jul 22 at 12:27

It's similar: $chi_(0, infty)(t) = H(t).$ But $chi_(a,b)(t) = H(t-a) - H(t-b).$
– md2perpe
Jul 22 at 12:27

1

1

Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
– md2perpe
Jul 22 at 14:06

Look at the integral before. There we have both $chi_[0, t_1](t')$ and $chi_[0, t_2](t')$. For the product of these not vanish, we must have both $t' in [0, t_1]$ and $t' in [0, t_2],$ i.e. $t' in [0, t_1] cap [0, t_2] = [0, min(t_1,t_2)].$
– md2perpe
Jul 22 at 14:06

1

1

In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
– Maxim
Jul 22 at 15:20

In $I(t_1, t_2)$, $e^-2 a t'$ in the second line should be $e^2 a t'$. Ultimately $I(t_1, t_2)$ should come out as $$cases frac 1 2a e^-a (t_1 + t_2) (e^2 a min(t_1, t_2) - 1) & $t_1 > 0 land t_2 > 0$ \ -frac 1 2a e^-a (t_1 + t_2) ( e^2 a max(t_1, t_2) - 1)& $t_1 < 0 land t_2 < 0$ \ 0 & otherwise.$$
– Maxim
Jul 22 at 15:20

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1

Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
– Maxim
Jul 22 at 16:09

Let's make up a signed indicator function, $chi_(0,a)(t) = [0 < t < a] - [a < t < 0]$. Then your derivation goes through unchanged till the point where you take the minimum. For $t_1 t_2 < 0$, $chi_(0, t_1)(t) chi_(0, t_2)(t) = 0$. For negative $t_1$ and $t_2$, the value which is closer to zero will be $max(t_1, t_2)$ instead of $min(t_1, t_2)$.
– Maxim
Jul 22 at 16:09

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