Question Number 192388 by sudipmoi last updated on 16/May/23
Answered by aleks041103 last updated on 22/May/23
![p>0 ∫_1 ^( ∞) ((sin(x)dx)/x^p )=[−((cos(x))/x^p )]_1 ^∞ −∫_1 ^( ∞) (−cos(x))d((1/x^p ))= =cos(1)−p∫_1 ^∞ ((cos(x))/x^(p+1) )dx ∫_1 ^∞ ∣((cos(x))/x^(p+1) )∣dx<∫_1 ^( ∞) (dx/x^(p+1) )=−[(1/(px^p ))]_1 ^∞ =(1/p) ⇒the integral is absolutely convergent ⇒∫_1 ^∞ ((cos(x))/x^(p+1) )dx converges ⇒∫_1 ^( ∞) ((sin(x)dx)/x^p ) converges for p>0](https://www.tinkutara.com/question/Q192611.png)
$${p}>\mathrm{0} \\ $$$$\int_{\mathrm{1}} ^{\:\infty} \frac{{sin}\left({x}\right){dx}}{{x}^{{p}} }=\left[−\frac{{cos}\left({x}\right)}{{x}^{{p}} }\right]_{\mathrm{1}} ^{\infty} −\int_{\mathrm{1}} ^{\:\infty} \left(−{cos}\left({x}\right)\right){d}\left(\frac{\mathrm{1}}{{x}^{{p}} }\right)= \\ $$$$={cos}\left(\mathrm{1}\right)−{p}\int_{\mathrm{1}} ^{\infty} \frac{{cos}\left({x}\right)}{{x}^{{p}+\mathrm{1}} }{dx} \\ $$$$\int_{\mathrm{1}} ^{\infty} \mid\frac{{cos}\left({x}\right)}{{x}^{{p}+\mathrm{1}} }\mid{dx}<\int_{\mathrm{1}} ^{\:\infty} \frac{{dx}}{{x}^{{p}+\mathrm{1}} }=−\left[\frac{\mathrm{1}}{{px}^{{p}} }\right]_{\mathrm{1}} ^{\infty} =\frac{\mathrm{1}}{{p}} \\ $$$$\Rightarrow{the}\:{integral}\:{is}\:{absolutely}\:{convergent} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\infty} \frac{{cos}\left({x}\right)}{{x}^{{p}+\mathrm{1}} }{dx}\:{converges} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{\:\infty} \frac{{sin}\left({x}\right){dx}}{{x}^{{p}} }\:{converges}\:{for}\:{p}>\mathrm{0} \\ $$