Laplace transform restriction and differentiation












1












$begingroup$


every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.



1



The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.



Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?



The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
L(cosat) = 2



the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.



Thank you so much for helping! Greatly appreciated.






laplace-transform







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$endgroup$

















    1












    $begingroup$


    every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.



    1



    The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
    But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.



    Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?



    The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
    L(cosat) = 2



    the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.



    Thank you so much for helping! Greatly appreciated.






    laplace-transform







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.



      1



      The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
      But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.



      Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?



      The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
      L(cosat) = 2



      the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.



      Thank you so much for helping! Greatly appreciated.






      laplace-transform







      share|cite|improve this question











      $endgroup$




      every one.I have just started learning Laplace transform.However, there are two main conceptual problems I can't convince myself.



      1



      The first problem is about the restriction of this integral, I understand that for a real variable s, it needs to be positive so that the integral converges to a value, e.g. f(t)=1 then L(f(t)) = 1/s . s>0
      But,if the s is a complex variable, I don't understand why the book says , e.g. for f(t) = exp(-at) ,then Re(s+a)>0.



      Why did they simply ignore the imaginary part of variable s, or are they just considering the real part of the function?(which they didn't mention)?



      The second question is when I was asked to verify L(tsinat) =3 http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table_files/eq0021M.gif by using
      L(cosat) = 2



      the problem is why do we have to differentiate this formula with respect to a? I thought a was a constant value which can't be differentiated in many problems I encountered while solving ODEs before.



      Thank you so much for helping! Greatly appreciated.






      laplace-transform




      laplace-transform






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      share|cite|improve this question













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      edited Jan 22 at 8:13









      Glorfindel

      3,41981830




      3,41981830










      asked Sep 9 '13 at 7:39









      el psy Congrooel psy Congroo

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      168111






















          2 Answers
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          $begingroup$

          For the first question: If $s=sigma + iomega$ and $t$ is real, then
          $$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
          since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Hint for the second part: Use this fact that:



            $$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              - needs a TU! +1
              $endgroup$
              – amWhy
              Sep 9 '13 at 10:36











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            2 Answers
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            active

            oldest

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            2 Answers
            2






            active

            oldest

            votes









            active

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            active

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            1












            $begingroup$

            For the first question: If $s=sigma + iomega$ and $t$ is real, then
            $$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
            since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              For the first question: If $s=sigma + iomega$ and $t$ is real, then
              $$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
              since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                For the first question: If $s=sigma + iomega$ and $t$ is real, then
                $$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
                since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.






                share|cite|improve this answer









                $endgroup$



                For the first question: If $s=sigma + iomega$ and $t$ is real, then
                $$|e^{-st}|=|e^{-sigma t}e^{-iomega t}| = e^{-sigma t}$$
                since $|e^{iomega t}|=1$. In other words it's only the real part of $s$ that determines where the ontegral converges.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 9 '13 at 8:00









                mrfmrf

                37.5k54685




                37.5k54685























                    1












                    $begingroup$

                    Hint for the second part: Use this fact that:



                    $$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      - needs a TU! +1
                      $endgroup$
                      – amWhy
                      Sep 9 '13 at 10:36
















                    1












                    $begingroup$

                    Hint for the second part: Use this fact that:



                    $$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      - needs a TU! +1
                      $endgroup$
                      – amWhy
                      Sep 9 '13 at 10:36














                    1












                    1








                    1





                    $begingroup$

                    Hint for the second part: Use this fact that:



                    $$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$






                    share|cite|improve this answer









                    $endgroup$



                    Hint for the second part: Use this fact that:



                    $$mathcal{L}(t^nf(t))=(-1)^nfrac{d^n F(s)}{ds^n}$$ wherein $mathcal{L}(f(t))=F(s).$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Sep 9 '13 at 7:46









                    mrsmrs

                    1




                    1












                    • $begingroup$
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                      $endgroup$
                      – amWhy
                      Sep 9 '13 at 10:36


















                    • $begingroup$
                      - needs a TU! +1
                      $endgroup$
                      – amWhy
                      Sep 9 '13 at 10:36
















                    $begingroup$
                    - needs a TU! +1
                    $endgroup$
                    – amWhy
                    Sep 9 '13 at 10:36




                    $begingroup$
                    - needs a TU! +1
                    $endgroup$
                    – amWhy
                    Sep 9 '13 at 10:36


















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