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Question Number 125173 by mohammad17 last updated on 08/Dec/20

Commented by mohammad17 last updated on 08/Dec/20

$p l e a s e h e l p m e$
	Answered by Dwaipayan Shikari last updated on 08/Dec/20

	$\int_{- \infty}^{\infty} e^{- z^{2} - \frac{{(x - z)}^{2}}{4 α^{2} t}} d z$ $= \int_{- \infty}^{\infty} e^{- z^{2} (\frac{1}{4 α^{2} t} + 1) + z (\frac{x}{2 α^{2} t}) - \frac{x^{2}}{4 α^{2} t}} d z$ $= \int_{- \infty}^{\infty} e^{- (\frac{1}{4 α^{2} t} + 1) ({(z - \frac{x}{(1 + 4 α^{2} t)})}^{2} + \frac{x^{2}}{4 α^{2} t} - \frac{x^{2}}{{(1 + 4 α t)}^{2}})} d z$ $= \int_{- \infty}^{\infty} e^{- {(\sqrt{\frac{4 α^{2} t + 1}{4 α^{2} t}} (z - \frac{x}{(1 + 4 α^{2} t)}))}^{2}} e^{(- \frac{x^{2}}{4 α^{2} t} + \frac{x^{2}}{{(1 + 4 α t)}^{2}}) (\frac{1}{4 α^{2} t} + 1)} d z$ $= 2 α \sqrt{\frac{π t}{4 α^{2} t + 1}} e^{(\frac{x}{(1 + 4 α^{2} t)} - \frac{x^{2}}{4 α^{2} t}) (\frac{4 α^{2} t + 1}{4 α^{2} t})}$ $= 2 α \sqrt{\frac{π t}{4 α^{2} t + 1}} e^{\frac{- x^{2}}{4 α^{2} t + 1}}$
	Commented by mohammad17 last updated on 08/Dec/20

	$t h a n k y o u s i r$
	Answered by mathmax by abdo last updated on 08/Dec/20
	$i suppose that we have 3 parametr in this integral x ,α ant I(xαt) =∫_(−∞) ^(+∞) e^(−(z^2 +(((x−z)^2 )/(4α^2 t)))) dz =∫_(−∞) ^(+∞) e^(−{((4α^2 tz^2 +z^2 −2xz +x^2 )/(4α^2 t))}) dz =∫_(−∞) ^(+∞) e^(−{ (((4α^2 t+1)z^2 −2xz+x^2 )/(4α^2 t))}) dz but (4α^2 t +1)z^2 −2xz +x^2 =(4α^2 t+1){z^2 −((2x)/(4α^2 t +1))z +(x^2 /(4α^2 t +1))} =_(m=4α^2 t+1) m{z^2 −((2x)/m)z +(x^2 /m)} =m{z^2 −2(x/m)z +(x^2 /m^2 )+(x^2 /m)−(x^2 /m^2 )} =m{(z−(x/m))^2 +(((m−1)x^2 )/m^2 )} ⇒ I(x,α,t) =∫_(−∞) ^(+∞) e^(−((1m)/(4α^2 t)){(z−(x/m))^2 +(((m−1)x^2 )/m^2 )}) dz =∫_(−∞) ^(+∞) e^(−((√(m/(4α^2 t)))(z−(x/m)))^2 ) ×e^(−(((m−1)x^2 )/(4mα^2 t))) dz =_((√(m/(4α^2 t)))(z−(x/m))=w) e^(−(((m−1)x^2 )/(4mα^2 t))) ∫_(−∞) ^(+∞) e^(−w^2 ) (dw/( (√(m/(4α^2 t))))) =((m/(4α^2 t)))^(−(1/2)) (√π)× e^(−(((m−1)x^2 )/(4mα^2 t))) =(√π)(((4α^2 t+1)/(4α^2 t)))^(−(1/2)) ×e^(−(((4α^2 t+1−1)x^2 )/(4α^2 t(4α^2 t +1)))) =(√π)(1+(1/(4α^2 t)))^(−(1/2)) . e^(−(x^2 /(4α^2 t+1)))$
	$i suppose that we have 3 parametr in this integral x, α ant$ $I (x α t) = \int_{- \infty}^{+ \infty} e^{- (z^{2} + \frac{{(x - z)}^{2}}{4 α^{2} t})} dz = \int_{- \infty}^{+ \infty} e^{- {\frac{4 α^{2} {tz}^{2} + z^{2} - 2 xz + x^{2}}{4 α^{2} t}}} dz$ $= \int_{- \infty}^{+ \infty} e^{- {\frac{(4 α^{2} t + 1) z^{2} - 2 xz + x^{2}}{4 α^{2} t}}} dz but$ $(4 α^{2} t + 1) z^{2} - 2 xz + x^{2} = (4 α^{2} t + 1) {z^{2} - \frac{2 x}{4 α^{2} t + 1} z + \frac{x^{2}}{4 α^{2} t + 1}}$ $=_{m = 4 α^{2} t + 1} m {z^{2} - \frac{2 x}{m} z + \frac{x^{2}}{m}} = m {z^{2} - 2 \frac{x}{m} z + \frac{x^{2}}{m^{2}} + \frac{x^{2}}{m} - \frac{x^{2}}{m^{2}}}$ $= m {{(z - \frac{x}{m})}^{2} + \frac{(m - 1) x^{2}}{m^{2}}} \Rightarrow$ $I (x, α, t) = \int_{- \infty}^{+ \infty} e^{- \frac{1 m}{4 α^{2} t} {{(z - \frac{x}{m})}^{2} + \frac{(m - 1) x^{2}}{m^{2}}}} dz$ $= \int_{- \infty}^{+ \infty} e^{- {(\sqrt{\frac{m}{4 α^{2} t}} (z - \frac{x}{m}))}^{2}} \times e^{- \frac{(m - 1) x^{2}}{4 m α^{2} t}} dz$ $=_{\sqrt{\frac{m}{4 α^{2} t}} (z - \frac{x}{m}) = w} e^{- \frac{(m - 1) x^{2}}{4 m α^{2} t}} \int_{- \infty}^{+ \infty} e^{- w^{2}} \frac{dw}{\sqrt{\frac{m}{4 α^{2} t}}}$ $= {(\frac{m}{4 α^{2} t})}^{- \frac{1}{2}} \sqrt{π} \times e^{- \frac{(m - 1) x^{2}}{4 m α^{2} t}}$ $= \sqrt{π} {(\frac{4 α^{2} t + 1}{4 α^{2} t})}^{- \frac{1}{2}} \times e^{- \frac{(4 α^{2} t + 1 - 1) x^{2}}{4 α^{2} t (4 α^{2} t + 1)}}$ $= \sqrt{π} {(1 + \frac{1}{4 α^{2} t})}^{- \frac{1}{2}} . e^{- \frac{x^{2}}{4 α^{2} t + 1}}$
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