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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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