# Laplace Transform of Heaviside Step Function times Function

## Theorem

Let $\map f t: \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.

Let $f$ be piecewise continuous with one-sided limits on any closed interval of the form $\closedint 0 b$ where $b > 0$.

Let $\map {u_c} t$ be the Heaviside step function.

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of $f$.

Then:

$\laptrans {\map {u_c} t \, \map f {t - c} } = e^{-s c} \map F s$

for $\map \Re s > a$.

## Proof

 $\ds \laptrans {\map {u_c} t \, \map f {t - c} }$ $=$ $\ds \int_0^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t$ Definition of Laplace Transform $\ds$ $=$ $\ds \int_0^{\to c^-} \map {u_c} t e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} \map {u_c} t e^{-s t} \map f {t - c} \rd t$ Sum of Complex Integrals on Adjacent Intervals $\ds$ $=$ $\ds \int_0^{\to c^-} 0 \times e^{-s t} \map f {t - c} \rd t + \int_{\to c^+}^{\to +\infty} 1 \times e^{-s t} \map f {t - c} \rd t$ Definition of Heaviside Step Function $\ds$ $=$ $\ds \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t$

Let $u = t - c$.

Then $\dfrac {\rd u} {\rd t} = 1$.

Then $u \to 0^-$ as $t \to c^+$.

Also, $u \to +\infty$ as $t \to +\infty$.

So:

 $\ds \int_{\to c^+}^{\to +\infty} e^{-s t} \map f {t - c} \rd t$ $=$ $\ds \int_{\to c^+}^{\to +\infty} e^{-s \paren {u + c} } \map f u \frac {\rd u} {\rd t} \rd t$ $\ds$ $=$ $\ds \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u$ Integration by Substitution $\ds$ $=$ $\ds \int_0^{\to 0^+} e^{-s \paren {u + c} } \map f u \rd u + \int_{\to 0^-}^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u$ Integral on Zero Interval $\ds$ $=$ $\ds \int_0^{\to +\infty} e^{-s \paren {u + c} } \map f u \rd u$ Sum of Complex Integrals on Adjacent Intervals $\ds$ $=$ $\ds \int_0^{\to +\infty} e^{-s u - s c} \map f u \rd u$ $\ds$ $=$ $\ds e^{-s c} \int_0^{\to +\infty} e^{-s u} \map f u \rd u$ Exponent Combination Laws $\ds$ $=$ $\ds e^{-s c} \int_0^{\to +\infty} e^{-s t} \map f t \rd t$ renaming the variable of integration $\ds$ $=$ $\ds e^{-s c} \laptrans {\map f t}$ Definition of Laplace Transform

$\blacksquare$