Question Number 83367 by M±th+et£s last updated on 01/Mar/20
$$\int_{−\infty} ^{\infty} \frac{{sin}^{\mathrm{7}} \left({x}\right)}{{x}^{\mathrm{7}} }{dx} \\ $$
Commented by mathmax by abdo last updated on 02/Mar/20
![this integral is solvable but i find a lots of calculus I =∫_(−∞) ^(+∞) ((sin^7 x)/x^7 )dx =2∫_0 ^∞ ((sin^7 x)/x^7 )dx and by psrts u^′ =x^(−7) and v=sin^7 x (1/2)I =[−(1/6)x^(−6) sin^7 x]_0 ^(+∞) +(7/6)∫_0 ^∞ (1/x^6 )cosx sin^6 x dx =(7/(12))∫_0 ^∞ ((sin(2x)sin^4 x)/x^6 )dx =(7/(12))∫_0 ^∞ ((sin(2x))/x^6 )(((1−cos(2x))/2))^2 dx =(7/(48))∫_0 ^∞ ((sin(2x)(1−2cos(2x)+cos^2 (2x)))/x^6 )dx =(7/(48))∫_0 ^∞ ((sin(2x))/x^6 ) dx −((14)/(96))∫_0 ^∞ ((sin(4x))/x^6 ) +(7/(48))∫_0 ^∞ ((sin(2x)(((1+cos(2x))/2)))/x^6 )dx .....be continued....](https://www.tinkutara.com/question/Q83446.png)
$${this}\:{integral}\:{is}\:{solvable}\:{but}\:{i}\:{find}\:{a}\:{lots}\:{of}\:{calculus}\: \\ $$$${I}\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}^{\mathrm{7}} {x}}{{x}^{\mathrm{7}} }{dx}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}^{\mathrm{7}} {x}}{{x}^{\mathrm{7}} }{dx}\:\:{and}\:{by}\:{psrts}\:{u}^{'} ={x}^{−\mathrm{7}} \:{and}\:{v}={sin}^{\mathrm{7}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{I}\:=\left[−\frac{\mathrm{1}}{\mathrm{6}}{x}^{−\mathrm{6}} \:{sin}^{\mathrm{7}} {x}\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\mathrm{7}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{{x}^{\mathrm{6}} }{cosx}\:{sin}^{\mathrm{6}} {x}\:{dx} \\ $$$$=\frac{\mathrm{7}}{\mathrm{12}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{2}{x}\right){sin}^{\mathrm{4}} {x}}{{x}^{\mathrm{6}} }{dx}\:=\frac{\mathrm{7}}{\mathrm{12}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{6}} }\left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \:{dx} \\ $$$$=\frac{\mathrm{7}}{\mathrm{48}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)\left(\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)}{{x}^{\mathrm{6}} }{dx} \\ $$$$=\frac{\mathrm{7}}{\mathrm{48}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}^{\mathrm{6}} }\:{dx}\:−\frac{\mathrm{14}}{\mathrm{96}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left(\mathrm{4}{x}\right)}{{x}^{\mathrm{6}} }\:+\frac{\mathrm{7}}{\mathrm{48}}\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{x}\right)\left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)}{{x}^{\mathrm{6}} }{dx} \\ $$$$…..{be}\:{continued}…. \\ $$