Convolution/Laplace T., Verifying step of proof: $int_0^Tleft(int_0^tf(u)g(t-u)duright)e^-sudt=int_0^Tdtint_0^tf(u)e^-sug(t-u)e^-s(t-u)du$

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While reading about convolution and Laplace Transform (Fourier Analysis and Its Applications, Anders Vretblad), I came across the following theorem and proof (picture at the bottom).



There is one step that I don't understand fully:$$int_0^Tleft(int_0^tf(u)g(t-u)duright)e^-sudt=int_0^Tdtint_0^tf(u)e^-sug(t-u)e^-s(t-u)du$$
What makes it true?



enter image description here




proof-explanation laplace-transform convolution




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    I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
    – M4g1ch
    Jul 22 at 10:59














up vote
0
down vote

favorite












While reading about convolution and Laplace Transform (Fourier Analysis and Its Applications, Anders Vretblad), I came across the following theorem and proof (picture at the bottom).



There is one step that I don't understand fully:$$int_0^Tleft(int_0^tf(u)g(t-u)duright)e^-sudt=int_0^Tdtint_0^tf(u)e^-sug(t-u)e^-s(t-u)du$$
What makes it true?



enter image description here




proof-explanation laplace-transform convolution




share|cite|improve this question















  • 1




    I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
    – M4g1ch
    Jul 22 at 10:59












up vote
0
down vote

favorite









up vote
0
down vote

favorite











While reading about convolution and Laplace Transform (Fourier Analysis and Its Applications, Anders Vretblad), I came across the following theorem and proof (picture at the bottom).



There is one step that I don't understand fully:$$int_0^Tleft(int_0^tf(u)g(t-u)duright)e^-sudt=int_0^Tdtint_0^tf(u)e^-sug(t-u)e^-s(t-u)du$$
What makes it true?



enter image description here




proof-explanation laplace-transform convolution




share|cite|improve this question











While reading about convolution and Laplace Transform (Fourier Analysis and Its Applications, Anders Vretblad), I came across the following theorem and proof (picture at the bottom).



There is one step that I don't understand fully:$$int_0^Tleft(int_0^tf(u)g(t-u)duright)e^-sudt=int_0^Tdtint_0^tf(u)e^-sug(t-u)e^-s(t-u)du$$
What makes it true?



enter image description here





proof-explanation laplace-transform convolution





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 22 at 7:14









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  • 1




    I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
    – M4g1ch
    Jul 22 at 10:59












  • 1




    I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
    – M4g1ch
    Jul 22 at 10:59







1




1




I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
– M4g1ch
Jul 22 at 10:59




I'd say that the RHS is some kind of notation to denote he is reversing the order of integration. Note also that in the LHS there's an error, it should be $e^-st$ instead of $e^-su$.
– M4g1ch
Jul 22 at 10:59















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