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Question Number 140401 by mnjuly1970 last updated on 07/May/21

evaluate :: Φ:=∫_0 ^( ∞) xe^(−(x^2 /4)) ln(x)dx = m.( π γ) find ” m ” ......

$$\:\:\: \\ $$$$\:\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\Phi:=\int_{\mathrm{0}} ^{\:\infty} {xe}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} {ln}\left({x}\right){dx}\:=\:{m}.\left(\:\pi\:\gamma\right) \\ $$$$\:\:\:\:\:\:{find}\:\:\:''\:\:{m}\:\:''\:...... \\ $$$$ \\ $$

Answered by qaz last updated on 07/May/21

φ(a)=∫_0 ^∞ x^a e^(−(x^2 /4)) dx=((Γ(((a+1)/2)))/(2((1/4))^((a+1)/2) ))=2^a Γ(((a+1)/2)) Φ=φ(1)′={2^a ln2Γ(((a+1)/2))+2^a Γ(((a+1)/2))ψ(((a+1)/2))∙(1/2)}_(a=1) =2ln2+ψ(1) ⇒m=((2ln2−γ)/(πγ))

$$\phi\left({a}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} {dx}=\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\frac{{a}+\mathrm{1}}{\mathrm{2}}} }=\mathrm{2}^{{a}} \Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Phi=\phi\left(\mathrm{1}\right)'=\left\{\mathrm{2}^{{a}} {ln}\mathrm{2}\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)+\mathrm{2}^{{a}} \Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\psi\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\centerdot\frac{\mathrm{1}}{\mathrm{2}}\right\}_{{a}=\mathrm{1}} \\ $$$$=\mathrm{2}{ln}\mathrm{2}+\psi\left(\mathrm{1}\right) \\ $$$$\Rightarrow{m}=\frac{\mathrm{2}{ln}\mathrm{2}−\gamma}{\pi\gamma} \\ $$

Commented by mnjuly1970 last updated on 07/May/21

thanks alot mr qaz..

$${thanks}\:{alot}\:{mr}\:{qaz}.. \\ $$

Answered by mnjuly1970 last updated on 07/May/21

((x/2))^2 =y ⇒x=2(√y) Φ=∫_0 ^( ∞) 2(√y) e^(−y) ln(2(√y) )(dy/( (√y))) =2∫_0 ^( ∞) e^(−y) ln(2)dy+∫_0 ^( ∞) e^(−y) ln(y)dy = ln(4)−γ ln(4)−γ=m(πγ) m=((ln(4)−γ)/(πγ))

$$\:\:\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} ={y}\:\Rightarrow{x}=\mathrm{2}\sqrt{{y}} \\ $$$$\:\:\:\:\Phi=\int_{\mathrm{0}} ^{\:\infty} \mathrm{2}\sqrt{{y}}\:{e}^{−{y}} {ln}\left(\mathrm{2}\sqrt{{y}}\:\right)\frac{{dy}}{\:\sqrt{{y}}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{y}} {ln}\left(\mathrm{2}\right){dy}+\int_{\mathrm{0}} ^{\:\infty} {e}^{−{y}} {ln}\left({y}\right){dy} \\ $$$$\:\:=\:{ln}\left(\mathrm{4}\right)−\gamma \\ $$$$\:\:\:\:\:{ln}\left(\mathrm{4}\right)−\gamma={m}\left(\pi\gamma\right) \\ $$$$\:\:\:\:\:\:{m}=\frac{{ln}\left(\mathrm{4}\right)−\gamma}{\pi\gamma} \\ $$

Commented by qaz last updated on 07/May/21

i edited again....

$${i}\:{edited}\:\:{again}.... \\ $$

Commented by mnjuly1970 last updated on 07/May/21

grateful ....

$$\:{grateful}\:.... \\ $$

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