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Question Number 101959 by MathQoutes last updated on 05/Jul/20

    Starting from          y=(4/(√π))t^((3/2) ) ∫_0 ^∞ x^3 e^(−tx^2 ) dx  find   (π/8)=?

$$ \\ $$$$ \\ $$$${Starting}\:{from} \\ $$$$\:\:\:\:\:\:\:\:{y}=\frac{\mathrm{4}}{\sqrt{\pi}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}\:} \int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{3}} {e}^{−{tx}^{\mathrm{2}} } {dx} \\ $$$${find}\:\:\:\frac{\pi}{\mathrm{8}}=? \\ $$

Answered by mathmax by abdo last updated on 06/Jul/20

the question is not clear  let find y(t) =(4/(√π))t^(3/2)  ∫_0 ^∞  x^3  e^(−tx^2 )  dx for t>0  we do the changement x(√t)=u ⇒  y(t) =(4/(√π))t^(3/2)  ∫_0 ^∞  ((u/(√t)))^(3 )  e^(−u^2 )  (du/(√t))  =(4/(√(πt))) ∫_0 ^∞   u^3  e^(−u^2 ) du  =_(u = (√z))   (4/(√(πt))) ∫_0 ^∞  z^(3/2)  e^(−z)  (dz/(2(√z)))  =(2/(√(πt)))∫_0 ^∞  z e^(−z)  dz  we know  Γ(z) =∫_0 ^∞  z^(x−1)  e^(−z)  dt  ⇒∫_0 ^∞  z e^(−z ) dz =Γ(2) =1!=1  ⇒y(t) =(2/(√(πt)))

$$\mathrm{the}\:\mathrm{question}\:\mathrm{is}\:\mathrm{not}\:\mathrm{clear}\:\:\mathrm{let}\:\mathrm{find}\:\mathrm{y}\left(\mathrm{t}\right)\:=\frac{\mathrm{4}}{\sqrt{\pi}}\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{tx}^{\mathrm{2}} } \:\mathrm{dx}\:\mathrm{for}\:\mathrm{t}>\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}\sqrt{\mathrm{t}}=\mathrm{u}\:\Rightarrow \\ $$$$\mathrm{y}\left(\mathrm{t}\right)\:=\frac{\mathrm{4}}{\sqrt{\pi}}\mathrm{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\int_{\mathrm{0}} ^{\infty} \:\left(\frac{\mathrm{u}}{\sqrt{\mathrm{t}}}\right)^{\mathrm{3}\:} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \:\frac{\mathrm{du}}{\sqrt{\mathrm{t}}} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\pi\mathrm{t}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{u}^{\mathrm{3}} \:\mathrm{e}^{−\mathrm{u}^{\mathrm{2}} } \mathrm{du}\:\:=_{\mathrm{u}\:=\:\sqrt{\mathrm{z}}} \:\:\frac{\mathrm{4}}{\sqrt{\pi\mathrm{t}}}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}^{\frac{\mathrm{3}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{z}} \:\frac{\mathrm{dz}}{\mathrm{2}\sqrt{\mathrm{z}}} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\pi\mathrm{t}}}\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}\:\mathrm{e}^{−\mathrm{z}} \:\mathrm{dz}\:\:\mathrm{we}\:\mathrm{know}\:\:\Gamma\left(\mathrm{z}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}^{\mathrm{x}−\mathrm{1}} \:\mathrm{e}^{−\mathrm{z}} \:\mathrm{dt}\:\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\mathrm{z}\:\mathrm{e}^{−\mathrm{z}\:} \mathrm{dz}\:=\Gamma\left(\mathrm{2}\right)\:=\mathrm{1}!=\mathrm{1} \\ $$$$\Rightarrow\mathrm{y}\left(\mathrm{t}\right)\:=\frac{\mathrm{2}}{\sqrt{\pi\mathrm{t}}} \\ $$

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