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Question Number 101073 by Dwaipayan Shikari last updated on 30/Jun/20
∫_0 ^∞ ((sin(logx))/(logx))dx
$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({logx}\right)}{{logx}}{dx} \\ $$
Answered by mathmax by abdo last updated on 30/Jun/20
A =∫_0 ^∞ ((sin(lnx))/(lnx))dx changement lnx =−t give x =e^(−t) A =∫_(+∞) ^(−∞) ((sin(−t))/(−t)) (−e^(−t) )dt =∫_(−∞) ^(+∞) ((sint)/t) e^(−t) dt =∫_(−∞) ^0 ((sint)/t) e^(−t) dt(→t =−u) +∫_0 ^∞ ((sint)/t) e^(−t) dt =∫_(+∞) ^0 ((−sinu)/(−u)) e^u (−du) +∫_0 ^∞ ((sint)/t) e^(−t) dt =∫_0 ^(+∞ ) ((sint)/t) e^t dt +∫_0 ^∞ ((sint)/t) e^(−t) [dt =∫_0 ^∞ ((sint)/t)(e^t +e^(−t) )dt =∫_0 ^∞ ((sint )/t) e^t dt (diverrgent) +∫_0 ^∞ ((sint)/t) e^(−t) dt (convervent) ⇒∫_0 ^∞ ((sin(lnx))/(lnx))dx is divergent
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{lnx}\right)}{\mathrm{lnx}}\mathrm{dx}\:\:\:\mathrm{changement}\:\mathrm{lnx}\:=−\mathrm{t}\:\mathrm{give}\:\:\mathrm{x}\:=\mathrm{e}^{−\mathrm{t}} \\ $$$$\mathrm{A}\:=\int_{+\infty} ^{−\infty} \:\frac{\mathrm{sin}\left(−\mathrm{t}\right)}{−\mathrm{t}}\:\left(−\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\int_{−\infty} ^{\mathrm{0}} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\left(\rightarrow\mathrm{t}\:=−\mathrm{u}\right)\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt} \\ $$$$=\int_{+\infty} ^{\mathrm{0}} \:\frac{−\mathrm{sinu}}{−\mathrm{u}}\:\mathrm{e}^{\mathrm{u}} \:\left(−\mathrm{du}\right)\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:=\int_{\mathrm{0}} ^{+\infty\:} \frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \left[\mathrm{dt}\right. \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sint}}{\mathrm{t}}\left(\mathrm{e}^{\mathrm{t}} \:+\mathrm{e}^{−\mathrm{t}} \right)\mathrm{dt}\:\:\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}\:}{\mathrm{t}}\:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:\left(\mathrm{diverrgent}\right)\:+\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sint}}{\mathrm{t}}\:\mathrm{e}^{−\mathrm{t}} \:\mathrm{dt}\:\left(\mathrm{convervent}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{sin}\left(\mathrm{lnx}\right)}{\mathrm{lnx}}\mathrm{dx}\:\mathrm{is}\:\mathrm{divergent} \\ $$

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