Question Number 218582 by Nicholas666 last updated on 12/Apr/25 | ||
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$$ \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({ax}\right)\:−\:{cos}\left({bx}\right)}{{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}\:\mid{b}−{a}\mid\: \\ $$$$ \\ $$ | ||
Answered by aleks041103 last updated on 12/Apr/25 | ||
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$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}−{cos}\left({ax}\right)}{{x}^{\mathrm{2}} }{dx}\:=\:\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}−{cos}\left({ax}\right)\right){d}\left(−\frac{\mathrm{1}}{{x}}\right)\:= \\ $$$$=\:\left(\frac{{cos}\left({ax}\right)−\mathrm{1}}{{x}}\right)_{\mathrm{0}} ^{\infty} +\int_{\mathrm{0}} ^{\infty} \frac{{d}\left(\mathrm{1}−{cos}\left({ax}\right)\right)}{{x}}= \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{asin}\left({ax}\right)}{{x}}{dx}={a}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({ax}\right)}{\left({ax}\right)}{d}\left({ax}\right)= \\ $$$$={a}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}={aI} \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({ax}\right)\:−\:{cos}\left({bx}\right)}{{x}^{\mathrm{2}} }\:{dx}\:= \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}−{cos}\left({bx}\right)}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}−{cos}\left({ax}\right)}{{x}^{\mathrm{2}} }\right){dx}= \\ $$$$={bI}−{aI}={I}\left({b}−{a}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}\right)}{{x}}{dx}\:=\:{I}\left(\mathrm{0}\right) \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{tx}} {sin}\left({x}\right)}{{x}}{dx} \\ $$$${I}\:'\left({t}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \frac{\left(−{x}\right){e}^{−{tx}} {sin}\left({x}\right)}{{x}}{dx}= \\ $$$$=−\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {sin}\left({x}\right){dx}= \\ $$$$={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−{tx}} {e}^{−{ix}} {dx}\right)= \\ $$$$={Im}\left(\int_{\mathrm{0}} ^{\infty} {e}^{−\left({t}+{i}\right){x}} {dx}\right)= \\ $$$$={Im}\left(\frac{−\mathrm{1}}{{t}+{i}}\left({e}^{−\left({t}+{i}\right){x}} \right)_{\mathrm{0}} ^{\infty} \right)= \\ $$$$={Im}\left(\frac{\mathrm{1}}{{t}+{i}}\left(\mathrm{1}−\mathrm{0}\right)\right)={Im}\left(\frac{{t}−{i}}{{t}^{\mathrm{2}} +\mathrm{1}}\right)=−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$\Rightarrow{I}\left({t}\right)−{I}\left(\mathrm{0}\right)=\int_{\mathrm{0}} ^{\:{t}} \frac{−{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={arctan}\left(\mathrm{0}\right)−{arctan}\left({t}\right) \\ $$$${I}\left(\infty\right)=\mathrm{0} \\ $$$$\Rightarrow−{I}\left(\mathrm{0}\right)=−{arctan}\left(\infty\right)=−\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow{I}={I}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({ax}\right)\:−\:{cos}\left({bx}\right)}{{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}\:\left({b}−{a}\right) \\ $$ | ||
Commented by Nicholas666 last updated on 12/Apr/25 | ||
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$${thanks} \\ $$ | ||