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Question Number 59175 by maxmathsup by imad last updated on 05/May/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{1-cos\left(x\right)}{x^2}dx$$

Commented by maxmathsup by imad last updated on 07/May/19

$$letI={\int }_0^{\mathrm{\infty }}\frac{1-cos\left(x\right)}{x^2}dx\Rightarrow I={\int }_0^{\mathrm{\infty }}\frac{2{sin}^2\left(\frac{x}{2}\right)}{x^2}dx{=}_{\frac{x}{2}=t}2{\int }_0^{\mathrm{\infty }}\frac{{sin}^2\left(t\right)}{4t^2}\left(2\right)dt$$$$={\int }_0^{\mathrm{\infty }}\frac{{sin}^{2}t}{t^2}dtbyparts{u}^{{}^{\prime }}=\frac{1}{t^2}andv={sin}^{2}t\Rightarrow$$$$I={[-\frac{1}{t}{sin}^2\left(t\right)]}_0^{+\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}-\frac{1}{t}\left(2sintcost\right)dt$$$$={\int }_0^{\mathrm{\infty }}\frac{sin\left(2t\right)}{t}dt{=}_{2t=u}{\int }_0^{\mathrm{\infty }}\frac{sin\left(u\right)}{\frac{u}{2}}\frac{du}{2}={\int }_0^{\mathrm{\infty }}\frac{sinu}{u}du=\frac{\pi }{2}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{1-cosx}{x^2}dx=\frac{\pi }{2}.$$

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