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Question Number 127446 by mnjuly1970 last updated on 29/Dec/20

$$...challangingintegral...$$$$provethat::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}(cos\left(x\right)-\frac{1}{1+x^2})\frac{dx}{x}=-\gamma$$$$$$

Answered by mindispower last updated on 31/Dec/20

$$Ci\left(x\right)=-{\int }_x^{\mathrm{\infty }}\frac{cos\left(t\right)}{t}dt=\gamma +ln\left(x\right)-{\int }_0^x\frac{1-cos\left(t\right)}{t}dt$$$$\mathrm{\Omega }=\underset{t\to 0}{\lim }{\int }_t^{\mathrm{\infty }}(cos\left(x\right)-\frac{1}{1+x^2}).\frac{dx}{x}$$$$=\underset{t\to 0}{\lim }(-\gamma -ln\left(t\right)+{\int }_0^t\frac{1-cos\left(x\right)}{x}dx-{\int }_t^{\mathrm{\infty }}\frac{dx}{x(1+x^2)})$$$$=\underset{t\to 0}{\lim }(-\gamma -ln\left(t\right)+{\int }_0^t\frac{1-cos\left(x\right)}{x}dx-B)$$$$B={\int }_t^{\mathrm{\infty }}\frac{1}{x(1+x^2)}=\frac{1+x^2-x.x}{x(1+x^2)}={\int }_t^{\mathrm{\infty }}\frac{1}{x}-\frac{x}{1+x^2}dx$$$$={[ln\left(x\right)-\frac{ln(1+x^2)}{2}]}_t^{\mathrm{\infty }}=-ln\left(t\right)+\frac{ln(1+t^2)}{2}$$$$\mathrm{\Omega }=\underset{t\to 0}{\lim }(-\gamma -ln\left(t\right)+{\int }_0^t\frac{1-cos\left(x\right)}{x}dx+ln\left(t\right)-\frac{ln(1+t^2)}{2})$$$$\mathrm{\Omega }=\underset{t\to 0}{\lim }(-\gamma +{\int }_0^t\frac{1-cos\left(x\right)}{x}dx)$$$$\underset{t\to 0}{\lim }{\int }_0^t\frac{1-cos\left(x\right)}{x}dx=0$$$$\frac{1-cos\left(x\right)}{x}=g\left(x\right)$$$$0⩽\frac{1-cos\left(x\right)}{x}\underset{x\in [0,t]}{}=\sum _{k⩾1}\frac{{(-1)}^{k+1}}{(2k!)}{x}^{2k-1}⩽\frac{x}{2}$$$$\Rightarrow 0⩽{\int }_0^t\frac{1-cos\left(x\right)}{x}dx⩽\frac{t^2}{4}\to 0$$$$\mathrm{\Omega }=\underset{t\to 0}{\lim }(-\gamma +{\int }_0^t\frac{1-cos\left(x\right)}{x}dx)=-\gamma +0=-\gamma$$$$\mathrm{\Omega }=-\gamma$$$$$$$$$$

Commented by mnjuly1970 last updated on 31/Dec/20

$$verynice.thanksalot..$$$$$$

Commented by mindispower last updated on 31/Dec/20

$$pleasursirhappynewyearsallgoodthingsfor$$$$you$$

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