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Question Number 126465 by mnjuly1970 last updated on 20/Dec/20

$$...nicecalculus...$$$$\mathcal{E}valuate...$$$$\varphi ={\int }_0^1\frac{sin\left(ln\left(x\right)\right)-ln\left(x\right)}{{ln}^2\left(x\right)}dx$$$$\mathcal{A}ns::ln\left(\sqrt{2}\right)+\frac{\pi }{4}-1...$$

Commented by talminator2856791 last updated on 22/Dec/20

$$\mathrm{do}\mathrm{you}\mathrm{have}\mathrm{more}\mathrm{difficult}\mathrm{question}\mathrm{than}\mathrm{this}$$

Answered by mindispower last updated on 23/Dec/20

$$-ln\left(x\right)=t$$$$={\int }_0^{\mathrm{\infty }}\frac{t-sin\left(t\right)}{t^2}{e}^{-t}$$$$sin\left(t\right)-t=\sum _{k⩾1}\frac{{(-1)}^k{t}^{2k+1}}{(2k+1)!}$$$$=\sum _{k⩾1}{\int }_0^{\mathrm{\infty }}\frac{{(-1)}^k}{(2k+1)!}{t}^{2k-1}{e}^{-t}dt$$$$=\sum _{k⩾1}\frac{{(-1)}^k}{(2k+1)!}\mathrm{\Gamma }\left(2k\right)$$$$=\sum _{k⩾1}\frac{{(-1)}^k}{2k(2k+1)}=\sum _{k⩾1}\frac{{(-1)}^k}{2k}-\sum _{k⩾1}\frac{{(-1)}^k}{2k+1}$$$$=\frac{1}{2}\sum _{k⩾1}{(-1)}^k{\int }_0^1{x}^{k-1}$$$$=\frac{1}{2}{\int }_0^1\frac{1}{x}\sum _{k⩾1}{(-x)}^{k}dx-\sum _{k⩾1}{\int }_0^1{(-1)}^k{x}^{2k}dx$$$$=-\frac{1}{2}{\int }_0^1\frac{1}{1+x}dx-{\int }_0^1\frac{-x^2}{1+x^2}dx$$$$=-\frac{1}{2}ln\left(2\right)+{\int }_0^{1}dx+{\int }_0^1\frac{dx}{1+x^2}$$$$=1+\frac{\pi }{4}-ln\left(\sqrt{2}\right)$$

Commented by mnjuly1970 last updated on 24/Dec/20

$$thanksalot..$$$$$$

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