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Question Number 32994 by abdo imad last updated on 09/Apr/18

$$find{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(\lambda x\right)}{x^2}dxwith\lambda >0.$$

Commented by abdo imad last updated on 15/Apr/18

$$changement\lambda x=ugive$$$$I={\int }_0^{\mathrm{\infty }}\frac{1-cosu}{{\left(\frac{u}{\lambda }\right)}^2}\frac{du}{\lambda }=\lambda {\int }_0^{\mathrm{\infty }}\frac{1-cosu}{u^2}dubut$$$${\int }_0^{\mathrm{\infty }}\frac{1-cosu}{u^2}du=2{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(\frac{u}{2}\right)}{u^2}du$$$${=}_{\frac{u}{2}=t}2{\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{4t^2}2dt={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^2}dt\left(byparts\right)$$$$={[-\frac{1}{t}{sin}^{2}t]}_0^{+\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}-\frac{1}{t}2sintcostdt$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(2t\right)}{t}dt={}_{2t=x}{\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{\frac{x}{2}}\frac{dx}{2}={\int }_0^{+\mathrm{\infty }}\frac{sinx}{x}dx$$$$=\frac{\pi }{2}\Rightarrow I=\frac{\lambda \pi }{2}.$$

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