Laplace transform of complementary error function $operatorname{erfc}(1/sqrt{t})$- using infinite series












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$begingroup$


The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



$$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



That is



$$begin{aligned}
Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
&=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
&=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
&=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
&=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
end{aligned}$$



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







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    0












    $begingroup$


    The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



    $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



    That is



    $$begin{aligned}
    Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
    &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
    &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
    &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
    &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
    end{aligned}$$



    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











    $endgroup$















      0












      0








      0





      $begingroup$


      The Laplace transform of the complementary error function $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



      $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



      That is



      $$begin{aligned}
      Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
      &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
      &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
      end{aligned}$$



      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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$ is



      $$Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}=Lleft{1-operatorname{erf}left(frac{1}{sqrt{t}}right) right}$$



      That is



      $$begin{aligned}
      Lleft{operatorname{erfc}left(frac{1}{sqrt{t}}right)right}
      &=Lleft{1-frac{2}{sqrt{pi}} int_{0}^{1/sqrt{t}} e^{-x^2} dx right}\
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!} int_{0}^{1/sqrt{t}} x^{2n} dx right} \
      &=Lleft{1-frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} frac{1}{t^{n+1/2}} right} \
      &=frac{1}{p} - frac{2}{sqrt{pi}} sum_{n=0}^{infty} frac{(-1)^n}{n!(2n+1)} Lleft{frac{1}{t^{n+1/2}} right} \
      end{aligned}$$



      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!






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      edited Aug 22 '16 at 1:29









      Bungo

      13.7k22148




      13.7k22148










      asked Aug 22 '16 at 1:08









      Venkatesan MurugesanVenkatesan Murugesan

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          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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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          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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
















          0












          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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














          0












          0








          0





          $begingroup$

          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$






          share|cite|improve this answer









          $endgroup$



          It looks fine. Now you just have to exploit
          $$mathcal{L}(1)=frac{1}{s},qquad mathcal{L}left(frac{1}{t^{n+1/2}}right) = s^{n-frac{1}{2}}Gammaleft(frac{1}{2}-nright)=s^{n-frac{1}{2}}(-1)^nfrac{2^nsqrt{pi}}{(2n-1)!!}. $$
          Anyway, It would have been faster to exploit the properties of the Laplace transform. $text{Erfc}$ is defined by an integral, and through a change of variable it is not difficult to check that
          $$ mathcal{L}left(text{Erfc}left(frac{1}{sqrt{t}}right)right) = color{red}{frac{e^{-2sqrt{s}}}{s}}. $$







          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

          290k33282662




          290k33282662








          • 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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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 $operatorname{erfc}left(frac{1}{sqrt{t}}right)$. 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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