Step 1

\(f(t)=1 \rightarrow F(s)=\frac{1}{s}\)

Step 2

Given a function f(t), the Laplace transform is a new function F(s) is defined by

\(F(s)=\int_0^\infty e^{-st}f(t)dt\) where we assume s is a positive real number.

For f(t)=1, the Laplace transform is given by

\(F(s)=\int_0^\infty e^{-st}\cdot 1 dt,\)

\(\text{Let } u=-st\Rightarrow du=-s dt \Rightarrow dt=-\frac{du}{s}\)

\(\text{For } t=0, u=-s \cdot 0=0.\)

\(\text{For } t=\infty, u=-s \cdot \infty=-\infty\)

Therefore,

\(F(s)=-\int_0^{-\infty} \frac{(e^u)}{s} du\)

\(=\int_{-\infty}^0 \frac{(e^u)}{s} du [\because \int_a^bf(x)dx=-\int_b^af(x)dx]\)

\(=\frac{1}{s} \int_{-\infty}^0 (e^u) du\)

\(=\frac{1}{s}[e^u]_{-\infty}^0\)

\(=\frac{1}{s}[e^0-e^{-\infty}]\)

\(=\frac{1}{s}[1-0]\)

\(=\frac{1}{s}\)

Hence, verified.

\(f(t)=1 \rightarrow F(s)=\frac{1}{s}\)

Step 2

Given a function f(t), the Laplace transform is a new function F(s) is defined by

\(F(s)=\int_0^\infty e^{-st}f(t)dt\) where we assume s is a positive real number.

For f(t)=1, the Laplace transform is given by

\(F(s)=\int_0^\infty e^{-st}\cdot 1 dt,\)

\(\text{Let } u=-st\Rightarrow du=-s dt \Rightarrow dt=-\frac{du}{s}\)

\(\text{For } t=0, u=-s \cdot 0=0.\)

\(\text{For } t=\infty, u=-s \cdot \infty=-\infty\)

Therefore,

\(F(s)=-\int_0^{-\infty} \frac{(e^u)}{s} du\)

\(=\int_{-\infty}^0 \frac{(e^u)}{s} du [\because \int_a^bf(x)dx=-\int_b^af(x)dx]\)

\(=\frac{1}{s} \int_{-\infty}^0 (e^u) du\)

\(=\frac{1}{s}[e^u]_{-\infty}^0\)

\(=\frac{1}{s}[e^0-e^{-\infty}]\)

\(=\frac{1}{s}[1-0]\)

\(=\frac{1}{s}\)

Hence, verified.