my problem in inverse laplace $frac1sarctanfrac1s$

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2
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I want to find inverse laplace
$$dfrac1sarctandfrac1s$$
My approach:
$$F(s)=dfrac1sarctandfrac1s$$
$$(sF)'=dfrac1s^2+1$$
$$L(-ty')=dfrac1s^2+1$$
$$-ty'=sin t+C$$
$$y=-intdfracsin t+Ctdt$$
how can i continue from here?




laplace-transform




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  • Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
    – Davide Morgante
    Jul 21 at 11:58










  • @DavideMorgante I wanted to accept but it didn't allow before 5 mins.
    – Lolita
    Jul 21 at 12:00










  • Yeah, that's true. Thank you!
    – Davide Morgante
    Jul 21 at 12:00










  • No dfrac in titles, unless they would be absolutely necessary.
    – Did
    Jul 21 at 12:32










  • @Did Sure thnx.
    – Lolita
    Jul 21 at 12:33














up vote
2
down vote

favorite












I want to find inverse laplace
$$dfrac1sarctandfrac1s$$
My approach:
$$F(s)=dfrac1sarctandfrac1s$$
$$(sF)'=dfrac1s^2+1$$
$$L(-ty')=dfrac1s^2+1$$
$$-ty'=sin t+C$$
$$y=-intdfracsin t+Ctdt$$
how can i continue from here?




laplace-transform




share|cite|improve this question





















  • Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
    – Davide Morgante
    Jul 21 at 11:58










  • @DavideMorgante I wanted to accept but it didn't allow before 5 mins.
    – Lolita
    Jul 21 at 12:00










  • Yeah, that's true. Thank you!
    – Davide Morgante
    Jul 21 at 12:00










  • No dfrac in titles, unless they would be absolutely necessary.
    – Did
    Jul 21 at 12:32










  • @Did Sure thnx.
    – Lolita
    Jul 21 at 12:33












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I want to find inverse laplace
$$dfrac1sarctandfrac1s$$
My approach:
$$F(s)=dfrac1sarctandfrac1s$$
$$(sF)'=dfrac1s^2+1$$
$$L(-ty')=dfrac1s^2+1$$
$$-ty'=sin t+C$$
$$y=-intdfracsin t+Ctdt$$
how can i continue from here?




laplace-transform




share|cite|improve this question













I want to find inverse laplace
$$dfrac1sarctandfrac1s$$
My approach:
$$F(s)=dfrac1sarctandfrac1s$$
$$(sF)'=dfrac1s^2+1$$
$$L(-ty')=dfrac1s^2+1$$
$$-ty'=sin t+C$$
$$y=-intdfracsin t+Ctdt$$
how can i continue from here?





laplace-transform





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 21 at 12:31









Did

242k23208443




242k23208443









asked Jul 21 at 11:43









Lolita

52318




52318











  • Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
    – Davide Morgante
    Jul 21 at 11:58










  • @DavideMorgante I wanted to accept but it didn't allow before 5 mins.
    – Lolita
    Jul 21 at 12:00










  • Yeah, that's true. Thank you!
    – Davide Morgante
    Jul 21 at 12:00










  • No dfrac in titles, unless they would be absolutely necessary.
    – Did
    Jul 21 at 12:32










  • @Did Sure thnx.
    – Lolita
    Jul 21 at 12:33
















  • Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
    – Davide Morgante
    Jul 21 at 11:58










  • @DavideMorgante I wanted to accept but it didn't allow before 5 mins.
    – Lolita
    Jul 21 at 12:00










  • Yeah, that's true. Thank you!
    – Davide Morgante
    Jul 21 at 12:00










  • No dfrac in titles, unless they would be absolutely necessary.
    – Did
    Jul 21 at 12:32










  • @Did Sure thnx.
    – Lolita
    Jul 21 at 12:33















Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
– Davide Morgante
Jul 21 at 11:58




Remember that on MSE you should mark an answer as accepted if you think that the answer was what you were looking for! This, besides giving points to the users, let's all other users know that this forum is closed
– Davide Morgante
Jul 21 at 11:58












@DavideMorgante I wanted to accept but it didn't allow before 5 mins.
– Lolita
Jul 21 at 12:00




@DavideMorgante I wanted to accept but it didn't allow before 5 mins.
– Lolita
Jul 21 at 12:00












Yeah, that's true. Thank you!
– Davide Morgante
Jul 21 at 12:00




Yeah, that's true. Thank you!
– Davide Morgante
Jul 21 at 12:00












No dfrac in titles, unless they would be absolutely necessary.
– Did
Jul 21 at 12:32




No dfrac in titles, unless they would be absolutely necessary.
– Did
Jul 21 at 12:32












@Did Sure thnx.
– Lolita
Jul 21 at 12:33




@Did Sure thnx.
– Lolita
Jul 21 at 12:33










1 Answer
1






active

oldest

votes

















up vote
2
down vote



accepted










At this point, the integral $$int fracsin(t)tdt$$ has no closed form: this is called the sine integral






share|cite|improve this answer





















  • thnx. so is my answer correct?
    – Lolita
    Jul 21 at 11:51










  • Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
    – Davide Morgante
    Jul 21 at 11:53











  • why, integration const.
    – Lolita
    Jul 21 at 11:53






  • 2




    The laplace transform is a definite integral.
    – Botond
    Jul 21 at 11:55










  • The integral for the Laplace transform is not an indefinite integral
    – Davide Morgante
    Jul 21 at 11:55










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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










At this point, the integral $$int fracsin(t)tdt$$ has no closed form: this is called the sine integral






share|cite|improve this answer





















  • thnx. so is my answer correct?
    – Lolita
    Jul 21 at 11:51










  • Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
    – Davide Morgante
    Jul 21 at 11:53











  • why, integration const.
    – Lolita
    Jul 21 at 11:53






  • 2




    The laplace transform is a definite integral.
    – Botond
    Jul 21 at 11:55










  • The integral for the Laplace transform is not an indefinite integral
    – Davide Morgante
    Jul 21 at 11:55














up vote
2
down vote



accepted










At this point, the integral $$int fracsin(t)tdt$$ has no closed form: this is called the sine integral






share|cite|improve this answer





















  • thnx. so is my answer correct?
    – Lolita
    Jul 21 at 11:51










  • Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
    – Davide Morgante
    Jul 21 at 11:53











  • why, integration const.
    – Lolita
    Jul 21 at 11:53






  • 2




    The laplace transform is a definite integral.
    – Botond
    Jul 21 at 11:55










  • The integral for the Laplace transform is not an indefinite integral
    – Davide Morgante
    Jul 21 at 11:55












up vote
2
down vote



accepted







up vote
2
down vote



accepted






At this point, the integral $$int fracsin(t)tdt$$ has no closed form: this is called the sine integral






share|cite|improve this answer













At this point, the integral $$int fracsin(t)tdt$$ has no closed form: this is called the sine integral







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 21 at 11:50









Davide Morgante

1,812220




1,812220











  • thnx. so is my answer correct?
    – Lolita
    Jul 21 at 11:51










  • Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
    – Davide Morgante
    Jul 21 at 11:53











  • why, integration const.
    – Lolita
    Jul 21 at 11:53






  • 2




    The laplace transform is a definite integral.
    – Botond
    Jul 21 at 11:55










  • The integral for the Laplace transform is not an indefinite integral
    – Davide Morgante
    Jul 21 at 11:55
















  • thnx. so is my answer correct?
    – Lolita
    Jul 21 at 11:51










  • Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
    – Davide Morgante
    Jul 21 at 11:53











  • why, integration const.
    – Lolita
    Jul 21 at 11:53






  • 2




    The laplace transform is a definite integral.
    – Botond
    Jul 21 at 11:55










  • The integral for the Laplace transform is not an indefinite integral
    – Davide Morgante
    Jul 21 at 11:55















thnx. so is my answer correct?
– Lolita
Jul 21 at 11:51




thnx. so is my answer correct?
– Lolita
Jul 21 at 11:51












Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
– Davide Morgante
Jul 21 at 11:53





Just a correction $$mathcalL(sin(t))= frac1s^2+1$$ no need for the $+C$
– Davide Morgante
Jul 21 at 11:53













why, integration const.
– Lolita
Jul 21 at 11:53




why, integration const.
– Lolita
Jul 21 at 11:53




2




2




The laplace transform is a definite integral.
– Botond
Jul 21 at 11:55




The laplace transform is a definite integral.
– Botond
Jul 21 at 11:55












The integral for the Laplace transform is not an indefinite integral
– Davide Morgante
Jul 21 at 11:55




The integral for the Laplace transform is not an indefinite integral
– Davide Morgante
Jul 21 at 11:55












 

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