I have to prove the Laplace transform equality
$${\cal L}^{-1}\left(\sqrt{s-\alpha}-\sqrt{s-\beta}\right)=\frac{1}{2t\sqrt{\pi t}}[e^{(\beta t)} - e^{(\alpha t)}]$$
Help.
Prove Laplace equality ${\cal L}^{-1}\left(\sqrt{s-\alpha}-\sqrt{s-\beta}\right)=\frac{1}{2t\sqrt{\pi t}}[e^{(\beta t)} - e^{(\alpha t)}]$.
0
$\begingroup$
complex-analysis
laplace-transform
gamma-function
-
2You should add some details about what you have done so far, and what your ideas are to compute the laplace transform. – 2017-02-12
-
0A Laplace transform ($\mathcal{L}$) is actually missing in your equality. As written, it is not an equality. – 2017-02-12
1 Answers
0
With $\displaystyle\Gamma(\frac12)=\sqrt{\pi}$, $\Gamma(p+1)=p\Gamma(p)$ and $$\color{blue}{{\cal L}(t^p)=\frac{\Gamma(p+1)}{s^{p+1}}}$$ write $${\cal L}\left(\frac{1}{t\sqrt{t}}\right)={\cal L}(t^\frac{-3}{2})=\frac{\Gamma(\frac{-3}{2}+1)}{s^{\frac{-3}{2}+1}}=\Gamma(\frac{-1}{2})\sqrt{s}=\frac{\Gamma(\frac{1}{2})}{\frac{-1}{2}}\sqrt{s}=-2\sqrt{\pi}\sqrt{s}$$ and by shifting property $$\color{blue}{{\cal L}(e^{ct}f(t))=F(s-c)}$$ so $${\cal L}\left(e^{\alpha t}\frac{1}{t\sqrt{t}}\right)=-2\sqrt{\pi}\sqrt{s-\alpha}$$