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Laplace transform of complementary error function $operatornameerfc(1/sqrtt)$- using infinite series



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Finding Laplace Transform of a FunctionSolve transport equations by using Laplace transformLaplace transform of complementary error function using infinite seriesInverse Laplace Transform to arrive at Error FunctionLaplace transforms and error functionLaplace Transform $sinh(sqrtt)$Laplace transform $mathcalL_xleft[frac1(x+9) sqrtx+8right](s)$How to get the Laplace transformation of function $fraca2sqrtπtoperatornameerf(frac-a4t)$Formula for complementary cumulative distribution function (CCDF) using Laplace and inverse Laplace transformLaplace Transform of Complementary Error Function










1












$begingroup$


The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is



$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$



That is



$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$



After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






laplace-transform







share|cite|improve this question











$endgroup$











  • $begingroup$
    There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
    $endgroup$
    – Aron OME
    Feb 28 at 16:07















1












$begingroup$


The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is



$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$



That is



$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$



After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






laplace-transform







share|cite|improve this question











$endgroup$











  • $begingroup$
    There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
    $endgroup$
    – Aron OME
    Feb 28 at 16:07













1












1








1





$begingroup$


The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is



$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$



That is



$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$



After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






laplace-transform







share|cite|improve this question











$endgroup$




The Laplace transform of the complementary error function $operatornameerfcleft(frac1sqrttright)$ is



$$Lleftoperatornameerfcleft(frac1sqrttright)right=Lleft1-operatornameerfleft(frac1sqrttright) right$$



That is



$$beginaligned
Lleftoperatornameerfcleft(frac1sqrttright)right
&=Lleft1-frac2sqrtpi int_0^1/sqrtt e^-x^2 dx right\
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn! int_0^1/sqrtt x^2n dx right \
&=Lleft1-frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) frac1t^n+1/2 right \
&=frac1p - frac2sqrtpi sum_n=0^infty frac(-1)^nn!(2n+1) Lleftfrac1t^n+1/2 right \
endaligned$$



After this step, I do not know how to proceed. Please help, if you know the procedure or if there is any mistake please point out. Thank you!






laplace-transform




laplace-transform






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 22 '16 at 1:29









Bungo

13.7k22148




13.7k22148










asked Aug 22 '16 at 1:08









Venkatesan MurugesanVenkatesan Murugesan

1413




1413











  • $begingroup$
    There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
    $endgroup$
    – Aron OME
    Feb 28 at 16:07
















  • $begingroup$
    There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
    $endgroup$
    – Aron OME
    Feb 28 at 16:07















$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07




$begingroup$
There is a problem because $displaystyle mathcalLleft(frac1t^n+1/2right)$ doesn't exist when $n>1/2$ !
$endgroup$
– Aron OME
Feb 28 at 16:07










1 Answer
1






active

oldest

votes


















1












$begingroup$

It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
    $endgroup$
    – Venkatesan Murugesan
    Aug 22 '16 at 15:52











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









1












$begingroup$

It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
    $endgroup$
    – Venkatesan Murugesan
    Aug 22 '16 at 15:52















1












$begingroup$

It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$






share|cite|improve this answer









$endgroup$








  • 1




    $begingroup$
    Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
    $endgroup$
    – Venkatesan Murugesan
    Aug 22 '16 at 15:52













1












1








1





$begingroup$

It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$






share|cite|improve this answer









$endgroup$



It looks fine. Now you just have to exploit
$$mathcalL(1)=frac1s,qquad mathcalLleft(frac1t^n+1/2right) = s^n-frac12Gammaleft(frac12-nright)=s^n-frac12(-1)^nfrac2^nsqrtpi(2n-1)!!. $$
Anyway, It would have been faster to exploit the properties of the Laplace transform. $textErfc$ is defined by an integral, and through a change of variable it is not difficult to check that
$$ mathcalLleft(textErfcleft(frac1sqrttright)right) = colorredfrace^-2sqrtss. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 22 '16 at 1:26









Jack D'AurizioJack D'Aurizio

292k33284674




292k33284674







  • 1




    $begingroup$
    Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
    $endgroup$
    – Venkatesan Murugesan
    Aug 22 '16 at 15:52












  • 1




    $begingroup$
    Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
    $endgroup$
    – Venkatesan Murugesan
    Aug 22 '16 at 15:52







1




1




$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52




$begingroup$
Sir, I have obtained the result by exploiting the properties of the Laplace transform. However, my interest is to get the Laplace transform using infinite series of $operatornameerfcleft(frac1sqrttright)$. I have used the result as suggested by you. But I didn't get the required result. Please help if possible. –
$endgroup$
– Venkatesan Murugesan
Aug 22 '16 at 15:52

















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