# If L\left\{f(t)\right\}=F(s) , then L{f(4t)} (a)frac(1)(4) F(s) (b)F(frac(s)(4)) (c)4F(s) (d)frac(1)(4) F(frac(s)(4))

If $$\displaystyle{L{f{{\left({t}\right)}}={F}{\left({s}\right)},\ \text{ then }\ {L}{f{{\left({4}{t}\right)}}}}}$$
a) $$\displaystyle{\frac{{{1}}}{{{4}}}}{F}{\left({s}\right)}$$
b) $$\displaystyle{F}{\left({\frac{{{s}}}{{{4}}}}\right)}$$
c) $$\displaystyle{4}{F}{\left({s}\right)}$$
d) $$\displaystyle{\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}$$

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Step 1
Here we solve this from the definition of Laplace transformation.
step 2
$$\displaystyle{L}{\left({f{{\left({t}\right)}}}\right)}={F}{\left({s}\right)}={\int_{{0}}^{\infty}}{f{{\left({t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle\rightarrow{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{f{{\left({4}{t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}$$
Step 3
Let us take transformation $$\displaystyle{z}={4}{t}\rightarrow{\left.{d}{t}\right.}={\frac{{{\left.{d}{z}\right.}}}{{{4}}}}$$
Then the integration becomes
$$\displaystyle{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\int_{{0}}^{\infty}}{f{{\left({4}{t}\right)}}}{e}^{{-{s}{t}}}{\left.{d}{t}\right.}={\int_{{0}}^{\infty}}{f{{\left({z}\right)}}}{e}^{{-{\frac{{{s}{z}}}{{{4}}}}}}{\frac{{{\left.{d}{z}\right.}}}{{{4}}}}$$
$$\displaystyle={\frac{{{1}}}{{{4}}}}{\int_{{0}}^{\infty}}{f{{\left({z}\right)}}}{e}^{{{\frac{{{s}}}{{{4}}}}{z}}}{\left.{d}{z}\right.}={\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}$$
Step 4
The final answer is $$\displaystyle{L}{\left({f{{\left({4}{t}\right)}}}\right)}={\frac{{{1}}}{{{4}}}}{F}{\left({\frac{{{s}}}{{{4}}}}\right)}$$ so 4th option is correct.