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Question Number 131227 by mnjuly1970 last updated on 02/Feb/21

$$\dots realanalysis\dots$$$$prove::$$$$\mathbf{\Omega }={\int }_0^{1}ln\left(ln\left(\frac{1}{x}\right)\right){ln}^2\left(x\right)dx=3-2\gamma$$$$$$

Answered by Dwaipayan Shikari last updated on 02/Feb/21

$$I\left(a\right)={\int }_0^1{x}^{a}log(-log\left(x\right))dxlogx=u$$$$I\left(a\right)={\int }_{-\mathrm{\infty }}^0{e}^{(a+1)u}log(-u)du$$$$I\left(a\right)={\int }_0^{\mathrm{\infty }}{e}^{-(a+1)u}log\left(u\right)du=\frac{1}{a+1}{\int }_0^{\mathrm{\infty }}{e}^{-t}log\left(t\right)dt-\frac{1}{a+1}{\int }_0^{\mathrm{\infty }}{e}^{-t}log(a+1)$$$$=-\frac{\gamma }{a+1}-\frac{1}{a+1}log(a+1)$$$$I^{\prime }\left(a\right)=\frac{\gamma }{{(a+1)}^2}+\frac{log(a+1)}{{(a+1)}^2}-\frac{1}{{(a+1)}^2}$$$$I^{″}\left(a\right)=\frac{-2\gamma }{{(a+1)}^3}+\frac{1}{{(a+1)}^3}-2\frac{log(a+1)}{{(a+1)}^3}+\frac{2}{{(a+1)}^3}$$$$I^{″}\left(0\right)=3-2\gamma ={\int }_0^{\mathrm{\infty }}log\left(log\left(\frac{1}{x}\right)\right){log}^2\left(x\right)dx$$

Commented by mnjuly1970 last updated on 02/Feb/21

$$thankyoumrdwaipayan\dots$$

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