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Question Number 38453 by maxmathsup by imad last updated on 25/Jun/18

$$findf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(xt\right)}{t}{e}^{-xt}dtwithx>0$$$$1)findasimpleformoff\left(x\right)$$$$2)calculate{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(\pi t\right)}{t}{e}^{-t}dt$$$$3)calculate{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(3t\right)}{t}{e}^{-2t}dt$$

Commented by math khazana by abdo last updated on 26/Jun/18

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

Commented by math khazana by abdo last updated on 29/Jun/18

$$1)wehavef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(at\right)}{t}{e}^{-xt}dt\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=-{\int }_0^{\mathrm{\infty }}(1-cos\left(at\right))e^{-xt}dt$$$$={\int }_0^{\mathrm{\infty }}cos\left(at\right)e^{-xt}dt-{\int }_0^{\mathrm{\infty }}{e}^{-xt}dtbut$$$${\int }_0^{\mathrm{\infty }}{e}^{-xt}dt={[-\frac{1}{x}{e}^{-xt}]}_0^{+\mathrm{\infty }}=\frac{1}{x}$$$${\int }_0^{\mathrm{\infty }}cos\left(at\right)e^{-xt}dt=Re\left({\int }_0^{\mathrm{\infty }}{e}^{iat-xt}dt\right)$$$${\int }_0^{\mathrm{\infty }}{e}^{(-x+ia)t}dt={\left[\frac{1}{-x+ia}{e}^{(-x+ia)t}\right]}_0^{+\mathrm{\infty }}$$$$=\frac{-1}{-x+ia}=\frac{1}{x-ia}=\frac{x+ia}{x^2+a^2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}cos\left(at\right)e^{-xt}dt=\frac{x}{x^2+a^2}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{x}{x^2+a^2}-\frac{1}{x}\Rightarrow$$$$f\left(x\right)=\frac{1}{2}ln(x^2+a^2)-ln\left(x\right)+cbut\mathrm{\exists }m>0<$$$$\mid {\int }_0^{\mathrm{\infty }}\frac{1-cos\left(at\right)}{t}{e}^{-xt}dt\mid ⩽m{\int }_0^{\mathrm{\infty }}{e}^{-xt}dt=\frac{m}{x}\to 0$$$$whenx\to +\mathrm{\infty }alsowehave$$$$f\left(x\right)=\frac{1}{2}ln\left(\frac{x^2+a^2}{x^2}\right)+c\Rightarrow$$$$c={lim}_{x\to +\mathrm{\infty }}f\left(x\right)-\frac{1}{2}ln\left(\frac{x^2+a^2}{x^2}\right)=0\Rightarrow$$$$f\left(x\right)=\frac{1}{2}ln(x^2+a^2)-ln\left(x\right)withx>0$$

Commented by math khazana by abdo last updated on 29/Jun/18

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

Commented by math khazana by abdo last updated on 29/Jun/18

$$3)?{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(3t\right)}{t}{e}^{-2t}dt=\frac{1}{2}ln(2^2+3^2)-ln\left(2\right)$$$$=\frac{1}{2}ln\left(13\right)-ln\left(2\right).$$

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