determine-L-e-x-2-x-with-L-laplace-transform-

Question Number 96658 by mathmax by abdo last updated on 03/Jun/20

determine L (e^{- x^{2} - x}) with L laplace transform

Answered by mathmax by abdo last updated on 04/Jun/20

L (e^{- x^{2} - x}) = \int_{0}^{\infty} e^{- t^{2} - t} e^{- xt} dt = \int_{0}^{\infty} e^{- t^{2} - (x + 1) t} dt

= \int_{0}^{\infty} e^{- (t^{2} + 2 t \times \frac{x + 1}{2} + \frac{{(x + 1)}^{2}}{4} - \frac{{(x + 1)}^{2}}{4})} = e^{\frac{{(x + 1)}^{2}}{4}} \int_{0}^{\infty} e^{- {t + \frac{x + 1}{2}}^{2}} dt

=_{t + \frac{x + 1}{2} = u} e^{\frac{{(x + 1)}^{2}}{4}} \int_{\frac{x + 1}{2}}^{\infty} e^{- u^{2}} du = e^{\frac{{(x + 1)}^{2}}{4}} {\int_{\frac{x + 1}{2}}^{0} e^{- u^{2}} du + \int_{0}^{\infty} e^{- u^{2}} du}

= e^{\frac{{(x + 1)}^{2}}{4}} {\frac{π}{2} - \int_{0}^{\frac{x + 1}{2}} e^{- u^{2}} du} \Rightarrow

L (e^{- x^{2} - x}) = \frac{π}{2} e^{\frac{{(x + 1)}^{2}}{4}} - e^{\frac{{(x + 1)}^{2}}{4}} \int_{0}^{\frac{x + 1}{2}} e^{- u^{2}} du

Commented by mathmax by abdo last updated on 04/Jun/20

sorry L (e^{- x^{2} - x}) = e^{\frac{{(x + 1)}^{2}}{2}} {\frac{\sqrt{π}}{2} - \int_{0}^{\frac{x + 1}{2}} e^{- u^{2}} du} \Rightarrow

L (e^{- x^{2} - x}) = \frac{\sqrt{π}}{2} e^{\frac{{(x + 1)}^{2}}{4}} - e^{\frac{{(x + 1)}^{2}}{4}} \int_{0}^{\frac{x + 1}{2}} e^{- u^{2}} du