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I was looking at this answer to the question asked and I am curious about the

$$\int_0^\infty F(u)g(u) du = \int_0^\infty f(u)G(u) du $$

relationship being used. I referred to the link provided in the answer, but it discussed the Mellin transform, not the Laplace transform, and I am not making the connection.

I was wondering if someone could explain where this relationship comes from or where I could learn more about it, and if it is true for laplace transforms in general, or only for that specific problem.

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user103104
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  • It is almost obvious if you just substitute $F(u)=\int_0^\infty f(x)e^{-xu}dx$ and $G(u)=\int_0^\infty g(x)e^{-xu}dx$. – Andrey Sokolov Oct 25 '13 at 04:32
  • I have not done any multivariable calculus in awhile so this is probably a stupid question... When you substitute in you get

    $\int_0^\infty \int_0^\infty e^{-ut}f(t)g(u) ,dt ,du = \int_0^\infty \int_0^\infty e^{-ut}g(t)f(u) ,dt ,du$

    Do the different variables for $f$ and $g$ in the integrals not make any difference to its value?

     – user103104 Oct 25 '13 at 04:43
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    Notice that $u$ and $t$ play interchangeable roles in the above integrals, and that by having them both range from $0$ to $\infty$ you are essentially sweeping the upper right plane region, by interchanging the order of integration the area you integrate over remains the same. – alonso s Oct 25 '13 at 04:46

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