inverse Laplace of $slndfracs-1s+1$. [closed]

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I want to find the inverse Laplace of
$$slndfracs-1s+1$$
may I use integration?

Thanks.

laplace-transform

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edited Jul 21 at 8:12

Bernard

110k635103

asked Jul 21 at 7:53

Lolita

52318

closed as off-topic by José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh Jul 21 at 9:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  HINT. Try wolframalpha.com
  – Mariusz Iwaniuk
  Jul 21 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How do I write in Wolfram?
  – Lolita
  Jul 21 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Put: inverse laplace transform s*ln((s-1)/(s+1))
  – Mariusz Iwaniuk
  Jul 21 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it doesn't work
  – Lolita
  Jul 21 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For Me works fine :)
  – Mariusz Iwaniuk
  Jul 21 at 12:39
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

1

I want to find the inverse Laplace of
$$slndfracs-1s+1$$
may I use integration?

Thanks.

laplace-transform

share|cite|improve this question

edited Jul 21 at 8:12

Bernard

110k635103

asked Jul 21 at 7:53

Lolita

52318

closed as off-topic by José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh Jul 21 at 9:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  HINT. Try wolframalpha.com
  – Mariusz Iwaniuk
  Jul 21 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How do I write in Wolfram?
  – Lolita
  Jul 21 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Put: inverse laplace transform s*ln((s-1)/(s+1))
  – Mariusz Iwaniuk
  Jul 21 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it doesn't work
  – Lolita
  Jul 21 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For Me works fine :)
  – Mariusz Iwaniuk
  Jul 21 at 12:39
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

1

up vote
-2
down vote

favorite

1

1

I want to find the inverse Laplace of
$$slndfracs-1s+1$$
may I use integration?

Thanks.

laplace-transform

share|cite|improve this question

edited Jul 21 at 8:12

Bernard

110k635103

asked Jul 21 at 7:53

Lolita

52318

I want to find the inverse Laplace of
$$slndfracs-1s+1$$
may I use integration?

Thanks.

laplace-transform

share|cite|improve this question

edited Jul 21 at 8:12

Bernard

110k635103

asked Jul 21 at 7:53

Lolita

52318

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 8:12

Bernard

110k635103

edited Jul 21 at 8:12

Bernard

110k635103

edited Jul 21 at 8:12

Bernard

110k635103

110k635103

asked Jul 21 at 7:53

Lolita

52318

asked Jul 21 at 7:53

Lolita

52318

asked Jul 21 at 7:53

Lolita

52318

52318

closed as off-topic by José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh Jul 21 at 9:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh Jul 21 at 9:27

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – José Carlos Santos, Siong Thye Goh, Did, Henrik, Shailesh

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  HINT. Try wolframalpha.com
  – Mariusz Iwaniuk
  Jul 21 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How do I write in Wolfram?
  – Lolita
  Jul 21 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Put: inverse laplace transform s*ln((s-1)/(s+1))
  – Mariusz Iwaniuk
  Jul 21 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it doesn't work
  – Lolita
  Jul 21 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For Me works fine :)
  – Mariusz Iwaniuk
  Jul 21 at 12:39
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  HINT. Try wolframalpha.com
  – Mariusz Iwaniuk
  Jul 21 at 9:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How do I write in Wolfram?
  – Lolita
  Jul 21 at 11:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Put: inverse laplace transform s*ln((s-1)/(s+1))
  – Mariusz Iwaniuk
  Jul 21 at 12:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  it doesn't work
  – Lolita
  Jul 21 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For Me works fine :)
  – Mariusz Iwaniuk
  Jul 21 at 12:39
  

  
  
  
  

HINT. Try wolframalpha.com
– Mariusz Iwaniuk
Jul 21 at 9:03

HINT. Try wolframalpha.com
– Mariusz Iwaniuk
Jul 21 at 9:03

@MariuszIwaniuk How do I write in Wolfram?
– Lolita
Jul 21 at 11:09

@MariuszIwaniuk How do I write in Wolfram?
– Lolita
Jul 21 at 11:09

Put: inverse laplace transform s*ln((s-1)/(s+1))
– Mariusz Iwaniuk
Jul 21 at 12:29

Put: inverse laplace transform s*ln((s-1)/(s+1))
– Mariusz Iwaniuk
Jul 21 at 12:29

it doesn't work
– Lolita
Jul 21 at 12:34

it doesn't work
– Lolita
Jul 21 at 12:34

For Me works fine :)
– Mariusz Iwaniuk
Jul 21 at 12:39

For Me works fine :)
– Mariusz Iwaniuk
Jul 21 at 12:39

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

You can use the fact that:
$$ mathscrLt^2 f(t)=F''(s)$$

So $$mathscrLt^2 f(t)= F''(s) = -frac4(s^2-1)^2$$

Now you can antitrasform both sides and then divide by $t^2$

So
$$ mathscrL^-1mathscrLt^2 f(t)= mathscrL^-1 -frac4(s^2-1)^2 $$
That is equal to:
$$ t^2 f(t) = e^-t(-1-e^2t(t-1)-t)$$
Finally:
$$ f(t) = frac(e^-t(-1-e^2t(t-1)-t))t^2$$

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
  – Oscar Lanzi
  Jul 21 at 9:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks. see my problem math.stackexchange.com/questions/2858434/…
  – Lolita
  Jul 21 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
  – Maxim
  Jul 21 at 13:08
  
  

  
  
  
  

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You can use the fact that:
$$ mathscrLt^2 f(t)=F''(s)$$

So $$mathscrLt^2 f(t)= F''(s) = -frac4(s^2-1)^2$$

Now you can antitrasform both sides and then divide by $t^2$

So
$$ mathscrL^-1mathscrLt^2 f(t)= mathscrL^-1 -frac4(s^2-1)^2 $$
That is equal to:
$$ t^2 f(t) = e^-t(-1-e^2t(t-1)-t)$$
Finally:
$$ f(t) = frac(e^-t(-1-e^2t(t-1)-t))t^2$$

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
  – Oscar Lanzi
  Jul 21 at 9:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks. see my problem math.stackexchange.com/questions/2858434/…
  – Lolita
  Jul 21 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
  – Maxim
  Jul 21 at 13:08
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You can use the fact that:
$$ mathscrLt^2 f(t)=F''(s)$$

So $$mathscrLt^2 f(t)= F''(s) = -frac4(s^2-1)^2$$

Now you can antitrasform both sides and then divide by $t^2$

So
$$ mathscrL^-1mathscrLt^2 f(t)= mathscrL^-1 -frac4(s^2-1)^2 $$
That is equal to:
$$ t^2 f(t) = e^-t(-1-e^2t(t-1)-t)$$
Finally:
$$ f(t) = frac(e^-t(-1-e^2t(t-1)-t))t^2$$

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
  – Oscar Lanzi
  Jul 21 at 9:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks. see my problem math.stackexchange.com/questions/2858434/…
  – Lolita
  Jul 21 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
  – Maxim
  Jul 21 at 13:08
  
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You can use the fact that:
$$ mathscrLt^2 f(t)=F''(s)$$

So $$mathscrLt^2 f(t)= F''(s) = -frac4(s^2-1)^2$$

Now you can antitrasform both sides and then divide by $t^2$

So
$$ mathscrL^-1mathscrLt^2 f(t)= mathscrL^-1 -frac4(s^2-1)^2 $$
That is equal to:
$$ t^2 f(t) = e^-t(-1-e^2t(t-1)-t)$$
Finally:
$$ f(t) = frac(e^-t(-1-e^2t(t-1)-t))t^2$$

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

You can use the fact that:
$$ mathscrLt^2 f(t)=F''(s)$$

So $$mathscrLt^2 f(t)= F''(s) = -frac4(s^2-1)^2$$

Now you can antitrasform both sides and then divide by $t^2$

So
$$ mathscrL^-1mathscrLt^2 f(t)= mathscrL^-1 -frac4(s^2-1)^2 $$
That is equal to:
$$ t^2 f(t) = e^-t(-1-e^2t(t-1)-t)$$
Finally:
$$ f(t) = frac(e^-t(-1-e^2t(t-1)-t))t^2$$

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 9:07

Luca Savant

763

answered Jul 21 at 9:07

Luca Savant

763

answered Jul 21 at 9:07

Luca Savant

763

763

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
  – Oscar Lanzi
  Jul 21 at 9:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks. see my problem math.stackexchange.com/questions/2858434/…
  – Lolita
  Jul 21 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
  – Maxim
  Jul 21 at 13:08
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
  – Oscar Lanzi
  Jul 21 at 9:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks. see my problem math.stackexchange.com/questions/2858434/…
  – Lolita
  Jul 21 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
  – Maxim
  Jul 21 at 13:08
  
  

  
  
  
  

Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
– Oscar Lanzi
Jul 21 at 9:21

Don't render the second derivative that way. Leave the separate terms obtained by differentiation uncombined. They become the partial fractions used for the inverse Laplace transformation.
– Oscar Lanzi
Jul 21 at 9:21

thanks. see my problem math.stackexchange.com/questions/2858434/…
– Lolita
Jul 21 at 11:44

thanks. see my problem math.stackexchange.com/questions/2858434/…
– Lolita
Jul 21 at 11:44

Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
– Maxim
Jul 21 at 13:08

Note that $s ln((s-1) / (s+1))$ approaches $-2$ on a vertical line, the Bromwich integral doesn't exist. In terms of distributions, the inverse transform is your $f(t)$ minus $2delta(t)$.
– Maxim
Jul 21 at 13:08

add a comment |