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Question Number 180350 by mnjuly1970 last updated on 10/Nov/22

	Answered by Ar Brandon last updated on 11/Nov/22
	$L(e^(−x) .erf((√x)))=∫_0 ^∞ e^(−x) erf((√x))e^(−px) dx =∫_0 ^∞ ((2/( (√π)))∫_0 ^(√x) e^(−t^2 ) dt)e^(−(p+1)x) dx=(2/( (√π)))∫_0 ^∞ (∫_0 ^(√x) e^(−t^2 ) dt)e^(−(p+1)x) dx { ((u(x)=∫_0 ^(√x) e^(−t^2 ) dt)),((v′(x)=e^(−(p+1)x) )) :} ⇒ { ((u′(x)=e^(−x) )),((v(x)=−(1/(p+1))e^(−(p+1)x) )) :} =−(2/( (√π)))∙(1/(p+1))[e^(−(p+1)x) ∫_0 ^(√x) e^(−t^2 ) dt]_0 ^∞ +(2/((p+1)(√π)))∫_0 ^∞ e^(−(p+2)x) dx =(2/( (p+1)(√π)))+(2/((p+1)(p+2)(√π)))=((2p+6)/((p^2 +3p+2)(√π)))$
	$L (e^{- x} . e r f (\sqrt{x})) = \int_{0}^{\infty} e^{- x} e r f (\sqrt{x}) e^{- p x} d x$ $= \int_{0}^{\infty} (\frac{2}{\sqrt{π}} \int_{0}^{\sqrt{x}} e^{- t^{2}} d t) e^{- (p + 1) x} d x = \frac{2}{\sqrt{π}} \int_{0}^{\infty} (\int_{0}^{\sqrt{x}} e^{- t^{2}} d t) e^{- (p + 1) x} d x$ ${\begin{cases} u (x) = \int_{0}^{\sqrt{x}} e^{- t^{2}} d t \\ v^{'} (x) = e^{- (p + 1) x} \end{cases} \Rightarrow {\begin{cases} u^{'} (x) = e^{- x} \\ v (x) = - \frac{1}{p + 1} e^{- (p + 1) x} \end{cases}$ $= - \frac{2}{\sqrt{π}} \cdot \frac{1}{p + 1} {[e^{- (p + 1) x} \int_{0}^{\sqrt{x}} e^{- t^{2}} d t]}_{0}^{\infty} + \frac{2}{(p + 1) \sqrt{π}} \int_{0}^{\infty} e^{- (p + 2) x} d x$ $= \frac{2}{(p + 1) \sqrt{π}} + \frac{2}{(p + 1) (p + 2) \sqrt{π}} = \frac{2 p + 6}{(p^{2} + 3 p + 2) \sqrt{π}}$
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