L(((1+2bt)/( (√t)))e^(bt) )(s)=?


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Question Number 142285 by qaz last updated on 29/May/21

$L (\frac{1 + 2 bt}{\sqrt{t}} e^{bt}) (s) = ?$
	Answered by Dwaipayan Shikari last updated on 29/May/21

	$L (f (t)) (s) = \int_{0}^{\infty} f (t) e^{- s t} d t$ $L (\frac{1 + 2 b t}{\sqrt{t}} e^{b t}) (s) = \int_{0}^{\infty} \frac{1 + 2 b t}{\sqrt{t}} e^{- s t + b t} d t$ $= \int_{0}^{\infty} \frac{e^{- s t + b t}}{\sqrt{t}} d t + 2 b \int_{0}^{\infty} \sqrt{t} e^{- s t + b t} d t (s - b) t = u$ $= \frac{1}{(s - b)} \int_{0}^{\infty} {(\frac{u}{s - b})}^{- 1 / 2} e^{- u} d u + 2 b {(\frac{1}{s - b})}^{3 / 2} \int_{0}^{\infty} u^{1 / 2} e^{- u} d u$ $= \frac{\sqrt{π}}{\sqrt{s - b}} + \frac{b \sqrt{π}}{{(s - b)}^{3 / 2}} = \sqrt{\frac{π}{(s - b)}} (1 + \frac{b}{s - b}) = \frac{\sqrt{π} s}{{(s - b)}^{3 / 2}}$
	Answered by mathmax by abdo last updated on 29/May/21
	$L(((1+2bx)/( (√x)))e^(bx) )(s)=∫_0 ^∞ ((1+2bx)/( (√x)))e^(bx) e^(−sx) dx =∫_0 ^∞ ((1+2bx)/( (√x)))e^(−(s−b)x) dx =_((√x)=t) ∫_0 ^∞ ((1+2bt^2 )/t)e^(−(s−b)t^2 ) (2t)dt =2∫_0 ^∞ (1+2bt^2 )e^(−(s−b)t^2 ) dt =2∫_0 ^∞ e^(−(s−b)t^2 ) dt+4b∫_0 ^∞ t^2 e^(−(s−b)t^2 ) dt we have ∫_0 ^∞ e^(−(s−b)t^2 ) dt =_((√(s−b))t=z) ∫_0 ^∞ e^(−z^2 ) (dz/( (√(s−b))))=(1/( (√(s−b)))).((√π)/2) =((√π)/(2(√(s−b)))) ∫_0 ^∞ t^2 e^(−(s−b)t^2 ) dt =_((√(s−b))t=z) ∫_0 ^∞ (z^2 /(s−b))e^(−z^2 ) (dz/( (√(s−b))))=(1/((s−b)^(3/2) ))∫_0 ^∞ z^2 e^(−z^2 ) dz by parts ∫_0 ^∞ z^2 e^(−z^2 ) dz =−(1/2)∫_0 ^∞ z(−2z)e^(−z^2 ) dz =−(1/2){[ze^(−z^2 ) ]_0 ^∞ −∫_0 ^∞ e^(−z^2 ) dz} =(1/2).((√π)/2)=((√π)/4) ⇒ L(((1+2bx)/( (√x)))e^(bx) ) =((√π)/( (√(s−b))))+4b.(1/((s−b)^(3/2) )).((√π)/4) =((√π)/( (√(s−b))))+((b(√π))/((s−b)^(3/2) ))$
	$L (\frac{1 + 2 bx}{\sqrt{x}} e^{bx}) (s) = \int_{0}^{\infty} \frac{1 + 2 bx}{\sqrt{x}} e^{bx} e^{- sx} dx$ $= \int_{0}^{\infty} \frac{1 + 2 bx}{\sqrt{x}} e^{- (s - b) x} dx =_{\sqrt{x} = t} \int_{0}^{\infty} \frac{1 + {2 bt}^{2}}{t} e^{- (s - b) t^{2}} (2 t) dt$ $= 2 \int_{0}^{\infty} (1 + {2 bt}^{2}) e^{- (s - b) t^{2}} dt = 2 \int_{0}^{\infty} e^{- (s - b) t^{2}} dt + 4 b \int_{0}^{\infty} t^{2} e^{- (s - b) t^{2}} dt$ $we have \int_{0}^{\infty} e^{- (s - b) t^{2}} dt =_{\sqrt{s - b} t = z} \int_{0}^{\infty} e^{- z^{2}} \frac{dz}{\sqrt{s - b}} = \frac{1}{\sqrt{s - b}} . \frac{\sqrt{π}}{2}$ $= \frac{\sqrt{π}}{2 \sqrt{s - b}}$ $\int_{0}^{\infty} t^{2} e^{- (s - b) t^{2}} dt =_{\sqrt{s - b} t = z} \int_{0}^{\infty} \frac{z^{2}}{s - b} e^{- z^{2}} \frac{dz}{\sqrt{s - b}} = \frac{1}{{(s - b)}^{\frac{3}{2}}} \int_{0}^{\infty} z^{2} e^{- z^{2}} dz$ $by parts \int_{0}^{\infty} z^{2} e^{- z^{2}} dz = - \frac{1}{2} \int_{0}^{\infty} z (- 2 z) e^{- z^{2}} dz$ $= - \frac{1}{2} {{[{ze}^{- z^{2}}]}_{0}^{\infty} - \int_{0}^{\infty} e^{- z^{2}} dz} = \frac{1}{2} . \frac{\sqrt{π}}{2} = \frac{\sqrt{π}}{4} \Rightarrow$ $L (\frac{1 + 2 bx}{\sqrt{x}} e^{bx}) = \frac{\sqrt{π}}{\sqrt{s - b}} + 4 b . \frac{1}{{(s - b)}^{\frac{3}{2}}} . \frac{\sqrt{π}}{4}$ $= \frac{\sqrt{π}}{\sqrt{s - b}} + \frac{b \sqrt{π}}{{(s - b)}^{\frac{3}{2}}}$
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