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$$F(s)=L[f(t)]$$ exists for $$s>a \geq 0$$ and if $$c$$ is a positive constant then
$$L\left[U_{c}(t) f(t-c)\right]=e^{-c s} L[f(t)]=e^{-c s} f(s)$$ and $$u_{c}(t) f(t-c)=L^{-1}\left[e^{-c s} f(s)\right]$$

$$L\left[U_{c}(t) f\left(t-c \right)\right]=\int_{0}^{\infty} e^{-s t} v_{c}(t) f(t-c) d t$$

$$=\int_{c}^{\infty} e^{-s t} f(t-c) d t$$

Let $$t-c=w \rightarrow d w=1 d t$$

$$L\left[U_{c}(t) f(t-c)\right]=\int_{0}^{\infty} e^{-s(w+c)} f(w) d w=e^{-c s} \int_{0}^{\infty} e^{-s w} f(w) d w$$

$$L\left[u_{c}(t) f(t-c)\right]=e^{-c s} f(s) \text { Take } L^{-1}$$

$$L^{-1} L\left[u_{c}(t) f(t-c)\right]=L^{-1} e^{-c s} f(s)$$

$$U_{c}(t) f(t-c)=L^{-1}\left[e^{-c s} f(s)\right]$$

$$\text { Show that } L\left\{t^{-1 / 2}\right\}=\sqrt{\frac{\pi}{s}} $$

$$\mathcal{L}\left\{t^{-1 / 2}\right\}=\int_{0}^{\infty} e^{-s t} t^{-1 / 2} d t$$

let $$u^{2}=s t \rightarrow t=\frac{u^{2}}{s} \rightarrow d t=\frac{1}{s} 2udu$$

$$\mathcal{L}\left\{t^{-1 / 2}\right\}=\int_{0}^{\infty} e^{-s \cdot \frac{u^{2}}{s}}\left(\frac{u^{2}}{s}\right)^{-\frac{1}{2}} \frac{1}{s} 2 u d u$$

$$=\int_{0}^{\infty} e^{-u^{2}} \cdot \frac{u^{-1}}{s^{-1 / 2}} \cdot \frac{1}{s} \cdot 2 u d u$$

$$=\frac{2}{s^{1 / 2}} \int_{0}^{\infty} e^{-u^{2}} \cdot d u \quad, s>0$$

$$=\frac{2}{\sqrt{s}} \cdot \frac{\sqrt{\pi}}{2}=\sqrt{\frac{\pi}{s}} $$

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