find-a-simple-form-of-L-e-x-L-is-laplace-transform-

Question Number 38197 by maxmathsup by imad last updated on 22/Jun/18

f i n d a s i m p l e f o r m o f L (e^{- \sqrt{x}}) L i s l a p l a c e t r a n s f o r m

Commented by math khazana by abdo last updated on 25/Jun/18

L (e^{- \sqrt{x}}) = \int_{0}^{\infty} f (t) e^{- x t} d t = \int_{0}^{\infty} e^{- \sqrt{t} - x t} d t

c h a n g e m e n t \sqrt{t} = u g i v e

L (e^{- \sqrt{x}}) = \int_{0}^{\infty} e^{- u - {x u}^{2}} 2 u d u

= - \frac{1}{x} \int_{0}^{\infty} - 2 x u e^{- {x u}^{2}} e^{- u} d u

- \frac{1}{x} {{[e^{- {x u}^{2}} e^{- u}]}_{0}^{+ \infty} + \int_{0}^{\infty} e^{- u} e^{- {x u}^{2}}}

= \frac{1}{x} - \frac{1}{x} \int_{0}^{\infty} e^{- ({x u}^{2} + u)} d u b u t

\int_{0}^{\infty} e^{- ({x u}^{2} + u)} d u = \int_{0}^{\infty} e^{- {{(\sqrt{x} u)}^{2} + 2 \frac{\sqrt{x} u}{\sqrt{x}} + \frac{1}{x} - \frac{1}{x}}} d u

= \int_{0}^{\infty} e^{- {{(\sqrt{x} u + \frac{1}{\sqrt{x}})}^{2}} + \frac{1}{x}} d u

= e^{\frac{1}{x}} \int_{0}^{\infty} e^{- {{(\sqrt{x} u + \frac{1}{\sqrt{x}})}^{2}}} d u

= e^{\frac{1}{x}} \int_{\frac{1}{\sqrt{x}}}^{+ \infty} e^{- t^{2}} \frac{d t}{\sqrt{x}} (c h a n g . \sqrt{x} u + \frac{1}{\sqrt{x}} = t)

= \frac{e^{\frac{1}{x}}}{\sqrt{x}} \int_{\frac{1}{\sqrt{x}}}^{+ \infty} e^{- t^{2}} d t \Rightarrow

L (e^{- \sqrt{x}}) = \frac{1}{x} - \frac{e^{\frac{1}{\sqrt{x}}}}{x \sqrt{x}} \int_{\frac{1}{\sqrt{x}}}^{+ \infty} e^{- t^{2}} d t .