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Question Number 35630 by abdo mathsup 649 cc last updated on 21/May/18

$$1)findthevalueoff\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t^2}{e}^{-t}dt$$$$2)calculate{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(t\right)}{t^2}{e}^{-t}dt.$$

Commented by abdo mathsup 649 cc last updated on 24/May/18

$$byparts{u}^{{}^{\prime }}=\frac{1}{t^2}andv=(1-cos\left(xt\right))e^{-t}\Rightarrow$$$$v^{{}^{\prime }}=xsin\left(xt\right)e^{-t}-(1-cos\left(xt\right))e^{-t}$$$$=(xsin\left(xt\right)+cos\left(xt\right)-1)e^{-t}$$$$f\left(x\right)={[-\frac{1}{t}(1-cos\left(xt\right))e^{-t}]}_0^{+\mathrm{\infty }}$$$$-{\int }_0^{\mathrm{\infty }}\frac{-1}{t}(xsin\left(xt\right)+cos\left(xt\right)-1)e^{-t}dtbut$$$$1-cos\left(xt\right)\sim \frac{x^2{t}^2}{2}\Rightarrow \frac{1-cos\left(xt\right)}{t}\sim \frac{{tx}^2}{2}\Rightarrow$$$${lim}_{t\to 0}\frac{(1-cos\left(xt\right))e^{-t}}{t}=0\Rightarrow$$$$f\left(x\right)=x{\int }_0^{\mathrm{\infty }}\frac{sin\left(xt\right)}{t}{e}^{-t}dt+{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t}{e}^{-t}dt$$$$letcalculateh\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{sin\left(xt\right)}{t}{e}^{-t}dt$$$$h^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}cos\left(xt\right)e^{-t}dt=Re\left({\int }_0^{\mathrm{\infty }}{e}^{ixt-t}dt\right)$$$${\int }_0^{\mathrm{\infty }}{e}^{(-1+ix)t}dt={\left[\frac{1}{-1+ix}{e}^{(-1+\mathit{i}\mathit{x})t}\right]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{-1+ix}=\frac{-1}{1-ix}=\frac{-1-ix}{1+x^2}\Rightarrow h^{{}^{\prime }}\left(x\right)=\frac{-1}{1+x^2}$$$$\Rightarrow h\left(x\right)=\lambda -arctan\left(x\right)but\lambda =h\left(0\right)=0\Rightarrow$$$$h\left(x\right)=-arctan\left(x\right)$$

Commented by abdo mathsup 649 cc last updated on 24/May/18

$$letcalculateK\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t}{e}^{-t}dt$$$$K^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}sin\left(xt\right)e^{-t}dt=Im\left({\int }_0^{\mathrm{\infty }}{e}^{ixt-t}dt\right)$$$$=Im\left({\int }_0^{\mathrm{\infty }}{e}^{(-1+ix)t}dt\right)=\frac{-x}{1+x^2}\Rightarrow$$$$K\left(x\right)=\lambda -\frac{1}{2}ln(1+x^2)but\lambda =K\left(0\right)=0\Rightarrow$$$$K\left(x\right)=-\frac{1}{2}ln(1+x^2)\Rightarrow$$$$f\left(x\right)=-xarctan\left(x\right)-\frac{1}{2}ln(1+x^2)$$

Commented by abdo mathsup 649 cc last updated on 24/May/18

$$errorinthefinalline$$$$f\left(x\right)=-xarctan\left(x\right)+\frac{1}{2}ln(1+x^2).$$

Commented by abdo mathsup 649 cc last updated on 24/May/18

$$2)wehaveprovedthat$$$${\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t^2}{e}^{-t}dt=\frac{1}{2}ln(1+x^2)-xarctan\left(x\right)$$$$lettakex=1\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{1-cos\left(t\right)}{t^2}{e}^{-t}=\frac{1}{2}ln\left(2\right)-\frac{\pi }{4}.$$

Commented by abdo mathsup 649 cc last updated on 24/May/18

$$errorinline8$$$$f\left(x\right)=x{\int }_0^{\mathrm{\infty }}\frac{sin\left(xt\right)}{t}{e}^{-t}dt-{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t}{e}^{-t}dt$$

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