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Question Number 83367 by M±th+et£s last updated on 01/Mar/20

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{sin}^7\left(x\right)}{x^7}dx$$

Commented by mathmax by abdo last updated on 02/Mar/20

$$thisintegralissolvablebutifindalotsofcalculus$$$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{sin}^{7}x}{x^7}dx=2{\int }_0^{\mathrm{\infty }}\frac{{sin}^{7}x}{x^7}dxandbypsrts{u}^{{}^{\prime }}=x^{-7}andv={sin}^{7}x$$$$\frac{1}{2}I={[-\frac{1}{6}{x}^{-6}{sin}^{7}x]}_0^{+\mathrm{\infty }}+\frac{7}{6}{\int }_0^{\mathrm{\infty }}\frac{1}{x^6}cosx{sin}^{6}xdx$$$$=\frac{7}{12}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right){sin}^{4}x}{x^6}dx=\frac{7}{12}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)}{x^6}{\left(\frac{1-cos\left(2x\right)}{2}\right)}^{2}dx$$$$=\frac{7}{48}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)(1-2cos\left(2x\right)+{cos}^2\left(2x\right))}{x^6}dx$$$$=\frac{7}{48}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)}{x^6}dx-\frac{14}{96}{\int }_0^{\mathrm{\infty }}\frac{sin\left(4x\right)}{x^6}+\frac{7}{48}{\int }_0^{\mathrm{\infty }}\frac{sin\left(2x\right)\left(\frac{1+cos\left(2x\right)}{2}\right)}{x^6}dx$$$$.....becontinued....$$

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