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Question Number 132729 by mnjuly1970 last updated on 16/Feb/21

$$....advancedcalculus...$$$$evaluate:$$$$\mathit{\varphi }={\int }_0^{\mathrm{\infty }}{xe}^{-2x}ln\left(x\right)dx=???$$$$$$

Answered by Olaf last updated on 16/Feb/21

$$\varphi ={\int }_0^{\mathrm{\infty }}{xe}^{-2x}\ln xdx$$$$\varphi ={\int }_0^{\mathrm{\infty }}(x\ln x-x)e^{-2x}dx+{\int }_0^{\mathrm{\infty }}{xe}^{-2x}dx$$$$\varphi ={[-\frac{e^{-2x}}{2}(x\ln x-x)]}_0^{\mathrm{\infty }}+\frac{1}{2}{\int }_0^{\mathrm{\infty }}\ln {xe}^{-2x}dx$$$$+{[-(\frac{1}{2}x+\frac{1}{4})e^{-2x}]}_0^{\mathrm{\infty }}$$$$\varphi =\frac{1}{2}{\int }_0^{\mathrm{\infty }}\ln {xe}^{-2x}dx+\frac{1}{4}$$$$\varphi =\frac{1}{2}(-\frac{\gamma }{2}-\frac{\ln 2}{2})+\frac{1}{4}$$$$\varphi =\frac{1}{4}(1-\gamma -\ln 2)$$

Commented by mnjuly1970 last updated on 16/Feb/21

$$tayeballahmrolaf...$$

Answered by Dwaipayan Shikari last updated on 16/Feb/21

$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{x}^{a-1}{e}^{-2x}dx=\frac{1}{2^a}{\int }_0^{\mathrm{\infty }}{u}^{a-1}{e}^{-u}du=\frac{\mathrm{\Gamma }\left(a\right)}{2^a}$$$$I^{\prime }\left(a\right)=\frac{{\mathrm{\Gamma }}^{\prime }\left(a\right)}{2^a}-\frac{\mathrm{\Gamma }\left(a\right)}{2^a}log\left(2\right)$$$$I^{\prime }\left(2\right)={\int }_0^{\mathrm{\infty }}{xe}^{-2x}log\left(x\right)dx=\frac{\mathrm{\Gamma }\left(2\right)\psi \left(2\right)}{4}-\frac{\mathrm{\Gamma }\left(2\right)}{4}log\left(2\right)=\frac{1}{4}(1-\gamma -log\left(2\right))$$

Answered by mnjuly1970 last updated on 16/Feb/21

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