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Question Number 130011 by mnjuly1970 last updated on 21/Jan/21

$$...advancedcalculus...$$$$provethat::$$$$\mathrm{\Phi }=\underset{\mathbb{R}}{\int }{e}^{(-e^x+2x)}{x}^{2}dx={(1-\gamma )}^2+\frac{{\pi }^2-6}{6}$$$$$$

Answered by mindispower last updated on 21/Jan/21

$$notintegrablover\mathbb{R}$$$$$$

Commented by mnjuly1970 last updated on 21/Jan/21

$$thankyouforyourmention$$$$icorrectedit...$$

Answered by mindispower last updated on 22/Jan/21

$$e^x=t$$$$\iff {\int }_0^{\mathrm{\infty }}{e}^{-t}.t.{\left(ln\left(t\right)\right)}^{2}dt=\mathrm{\Phi }$$$$\mathrm{\Gamma }\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}dt\Rightarrow {\mathrm{\Gamma }}^{″}\left(x\right)={\int }_0^{\mathrm{\infty }}{t}^{x-1}{e}^{-t}{ln}^2\left(t\right)dt$$$$\mathrm{\Phi }={\mathrm{\Gamma }}^{″}\left(2\right)$$$${\mathrm{\Gamma }}^{\prime }\left(x\right)=\mathrm{\Gamma }\left(x\right)\mathrm{\Psi }\left(x\right)\Rightarrow {\mathrm{\Gamma }}^{″}\left(x\right)=\mathrm{\Gamma }\left(x\right){\mathrm{\Psi }}^{\prime }\left(x\right)+{\mathrm{\Psi }}^2\left(x\right)$$$$\mathrm{\Phi }=\mathrm{\Gamma }\left(2\right){\mathrm{\Psi }}^{\prime }\left(2\right)+\mathrm{\Psi }{\left(2\right)}^2$$$$=1.\sum _{n⩾1}\frac{1}{{(n+1)}^2}+{(\mathrm{\Psi }\left(1\right)+1)}^2$$$$=\zeta \left(2\right)-1+{(1-\gamma )}^2=\frac{{\pi }^2-6}{6}+{(1-\gamma )}^2$$

Commented by mnjuly1970 last updated on 22/Jan/21

$$thanksalotsirpower$$$$grateful...$$

Commented by mindispower last updated on 23/Jan/21

$$pleasursir$$

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