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Question Number 129418 by mnjuly1970 last updated on 15/Jan/21

$$...advsncedcalculus....$$$$$$$$calculate:\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{x}}ln(1+\frac{1}{\sqrt{x}})dx$$$$$$

Answered by Dwaipayan Shikari last updated on 15/Jan/21

$${\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{x}}log(1+\sqrt{x})-{\int }_0^{\mathrm{\infty }}{e}^{-\sqrt{x}}log\left(\sqrt{x}\right)dx$$$$=2{\int }_0^{\mathrm{\infty }}{te}^{-t}log(1+t)-2{\int }_0^{\mathrm{\infty }}{te}^{-t}log\left(t\right)$$$$=2-2\left({\mathrm{\Gamma }}^{\prime }\left(2\right)\right)=2-2(-\gamma +1)=2\gamma$$

Commented by mnjuly1970 last updated on 15/Jan/21

$$excellent...grateful..$$

Answered by mindispower last updated on 15/Jan/21

$$\sqrt{x}=t$$$$\Rightarrow {\int }_0^{\mathrm{\infty }}{e}^{-t}ln(1+\frac{1}{t}).2tdt$$$$\int ln(1+t){te}^{-t}dtbypart$$$$\int {te}^{-t}dt=-(t+1)e^{-t}dt$$$${\int }_0^{\mathrm{\infty }}ln(1+t){te}^{-t}dt={[-(t+1)e^{-t}ln(1+t)]}_0^{\mathrm{\infty }}+\int \frac{(t+1)e^{-t}}{t+1}dt$$$$={\int }_0^{\mathrm{\infty }}{e}^{-t}dt=\mathrm{\Gamma }\left(1\right)=1$$$${\int }_0^{\mathrm{\infty }}{te}^{-t}ln\left(t\right)dt={\partial }_x{\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt{\mid }_{x=2}={\mathrm{\Gamma }}^{\prime }\left(2\right)=\mathrm{\Gamma }\left(2\right)\mathrm{\Psi }\left(2\right)$$$$=\mathrm{\Psi }\left(2\right)=1+\mathrm{\Psi }\left(1\right)=1-\gamma$$$${\int }_0^{\mathrm{\infty }}{e}^{-t}ln(1+\frac{1}{t}).2tdt=2{\int }_0^{\mathrm{\infty }}{e}^{-t}ln(1+t)tdt-2{\int }_0^{\mathrm{\infty }}{e}^{-t}tln\left(t\right)dt$$$$=2.1-2(1-\gamma )=2\gamma$$

Commented by mnjuly1970 last updated on 16/Jan/21

$$veryniceasalways...$$

Answered by Lordose last updated on 16/Jan/21

$$$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\sqrt{\mathrm{x}}}\ln (1+\frac{1}{\sqrt{\mathrm{x}}})\mathrm{dx}$$$$\mathrm{u}=\sqrt{\mathrm{x}}\Rightarrow \mathrm{dx}=2\mathrm{udu}$$$$\mathrm{\Omega }=2{\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln (1+\frac{1}{\mathrm{u}})\mathrm{du}=2{\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln \left(\frac{1+\mathrm{u}}{\mathrm{u}}\right)\mathrm{du}$$$$\mathrm{\Omega }=2({\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln (1+\mathrm{u})\mathrm{du}-{\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln \left(\mathrm{u}\right)\mathrm{du})$$$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln (1+\mathrm{u})\mathrm{du}\stackrel{\mathrm{IBP}}{=}\mid -{\mathrm{e}}^{-\mathrm{u}}(1+\mathrm{u})\ln (1+\mathrm{u}){\mid }_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}$$$$\mathrm{\Phi }=1$$$$\mathrm{\Omega }=2(\mathrm{\Phi }-{\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln \left(\mathrm{u}\right)\mathrm{du})$$$$\mathrm{\Omega }=2(1-{\int }_0^{\mathrm{\infty }}{\mathrm{ue}}^{-\mathrm{u}}\ln \left(\mathrm{u}\right)\mathrm{du})=2-\frac{\partial }{\partial \mathrm{a}}{\mid }_{\mathrm{a}=0}2{\int }_0^{\mathrm{\infty }}{\mathrm{u}}^{1+\mathrm{a}-1}{\mathrm{e}}^{-\mathrm{u}}\mathrm{du}$$$$\mathrm{\Omega }=2-\frac{\partial }{\partial \mathrm{a}}{\mid }_{\mathrm{a}=0}2\mathrm{\Gamma }(1+\mathrm{a})=2-2\mathrm{\Gamma }(1+\mathrm{a}){\psi }^0(1+\mathrm{a}){\mid }_{\mathrm{a}=0}=2-2(1-\gamma )$$$$\mathrm{\Omega }=2\gamma$$$$\mathrm{Where}\mathit{\gamma }=\mathrm{Euler}\mathrm{mascheroni}\mathrm{constant}$$$$★\mathbf{L}\mathit{\varphi }\mathbf{rD}\mathit{\varnothing }\mathbf{sE}$$

Commented by mnjuly1970 last updated on 16/Jan/21

$$thankyoumrlordos...$$

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