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Question Number 152947 by mnjuly1970 last updated on 03/Sep/21

     prove ::     𝛗=∫_(−∞) ^( ∞)  (( e^( −(1/x^( 2) )) )/x^( 4) ) dx =^?  (1/2) Γ ((1/2) )

$$ \\ $$$$\:\:\:{prove}\::: \\ $$$$\:\:\:\boldsymbol{\phi}=\int_{−\infty} ^{\:\infty} \:\frac{\:{e}^{\:−\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }} }{{x}^{\:\mathrm{4}} }\:{dx}\:\overset{?} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\:\Gamma\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:\right) \\ $$$$ \\ $$

Answered by mindispower last updated on 03/Sep/21

=2∫_0 ^∞ (1/x)e^(−(1/x^2 )) .−(1/2)d((1/x^2 )),(1/x^2 )=t  =∫_0 ^∞ (√t)e^(−t) dt=∫_0 ^∞ t^((3/2)−t) e^(−t) =Γ((3/2))=Γ(1+(1/2))  =(1/2)Γ((1/2))

$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}{e}^{−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }} .−\frac{\mathrm{1}}{\mathrm{2}}{d}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right),\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={t} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \sqrt{{t}}{e}^{−{t}} {dt}=\int_{\mathrm{0}} ^{\infty} {t}^{\frac{\mathrm{3}}{\mathrm{2}}−{t}} {e}^{−{t}} =\Gamma\left(\frac{\mathrm{3}}{\mathrm{2}}\right)=\Gamma\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

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