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0-e-3x-2-x-1-5-dx-




Question Number 1571 by 112358 last updated on 20/Aug/15
∫_0 ^( ∞) (e^(−3x^2 ) /x^(1/5) )dx=?
$$\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−\mathrm{3}{x}^{\mathrm{2}} } }{{x}^{\mathrm{1}/\mathrm{5}} }{dx}=? \\ $$
Answered by prakash jain last updated on 20/Aug/15
3x^2 =u  x=((u/3))^(1/2) ⇒x^(1/5) =((u/3))^(1/10)   6xdx=du⇒dx=(du/(6x))=(du/(u^(1/2) 3^(1/2) 2))  (dx/x^(1/5) )= (du/(2(√(3u)))) ∙ (3^(1/10) /u^(1/10) )=  (du/(2∙3^(2/5) ))∙u^(−3/5)  =(du/3^(2/5) )u^((2/5)−1)   ∫_0 ^( ∞) x^(t−1) e^(−x) dx=Γ(t)  ∫_0 ^∞  u^((2/5) −1) e^(−u) du=Γ((2/5))
$$\mathrm{3}{x}^{\mathrm{2}} ={u} \\ $$$${x}=\left(\frac{{u}}{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{2}} \Rightarrow{x}^{\mathrm{1}/\mathrm{5}} =\left(\frac{{u}}{\mathrm{3}}\right)^{\mathrm{1}/\mathrm{10}} \\ $$$$\mathrm{6}{xdx}={du}\Rightarrow{dx}=\frac{{du}}{\mathrm{6}{x}}=\frac{{du}}{{u}^{\mathrm{1}/\mathrm{2}} \mathrm{3}^{\mathrm{1}/\mathrm{2}} \mathrm{2}} \\ $$$$\frac{{dx}}{{x}^{\mathrm{1}/\mathrm{5}} }=\:\frac{{du}}{\mathrm{2}\sqrt{\mathrm{3}{u}}}\:\centerdot\:\frac{\mathrm{3}^{\mathrm{1}/\mathrm{10}} }{{u}^{\mathrm{1}/\mathrm{10}} }=\:\:\frac{{du}}{\mathrm{2}\centerdot\mathrm{3}^{\mathrm{2}/\mathrm{5}} }\centerdot{u}^{−\mathrm{3}/\mathrm{5}} \:=\frac{{du}}{\mathrm{3}^{\mathrm{2}/\mathrm{5}} }{u}^{\frac{\mathrm{2}}{\mathrm{5}}−\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {x}^{{t}−\mathrm{1}} {e}^{−{x}} {dx}=\Gamma\left({t}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{u}^{\frac{\mathrm{2}}{\mathrm{5}}\:−\mathrm{1}} {e}^{−{u}} {du}=\Gamma\left(\frac{\mathrm{2}}{\mathrm{5}}\right) \\ $$

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