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- Homework Statement
- Suppose ##F(s)=\mathcal{L}\{f(t)\}## exists for ##s> a\geq 0##.

Show that if c is a positive constant:

$$\mathcal{L} \{f(ct)\}=\frac{1}{c}F(\frac{s}{c}), \ s> ca$$

- Relevant Equations
- Laplace transform

Could someone check whether my proof for this simple theorem is correct? I get to the result, but with the feeling of having done something very wrong :)

$$\mathcal{L} \{f(ct)\}=\int_{0}^{\infty}e^{-st}f(ct)dt \ \rightarrow ct=u, \ dt=\frac{1}{c}du, \

\mathcal{L} \{f(ct)\}=\frac{1}{c}\int_{0}^{\infty}e^{\frac{-s}{c}u}f(u)du=\frac{1}{c}F(\frac{s}{c})$$

felt very straightforward, but looking back at it the very last step seems weird: in spite of a variable substitution, does it make sense to still get the same transform(only with the argument changed) of ##f(t)##?

$$\mathcal{L} \{f(ct)\}=\int_{0}^{\infty}e^{-st}f(ct)dt \ \rightarrow ct=u, \ dt=\frac{1}{c}du, \

\mathcal{L} \{f(ct)\}=\frac{1}{c}\int_{0}^{\infty}e^{\frac{-s}{c}u}f(u)du=\frac{1}{c}F(\frac{s}{c})$$

felt very straightforward, but looking back at it the very last step seems weird: in spite of a variable substitution, does it make sense to still get the same transform(only with the argument changed) of ##f(t)##?