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Question Number 102058 by Dwaipayan Shikari last updated on 06/Jul/20

∫_0 ^1 ((sin(logx))/(logx))dx

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({logx}\right)}{{logx}}{dx} \\ $$

Answered by prakash jain last updated on 06/Jul/20

x=e^u   dx=e^u du  I=∫_(−∞) ^0 ((sin u)/u)e^u du=−∫_0 ^∞ ((e^(−u) sin u)/u)du  J(t)=−∫_0 ^∞  ((e^(−tu) sin u)/u)du  J′(t)=−∫_0 ^∞ e^(−tu) sin udu=−(1/(1+t^2 ))  J(t)=−tan^(−1) t+C  I=J(1)  Will continue.

$${x}={e}^{{u}} \\ $$$${dx}={e}^{{u}} {du} \\ $$$${I}=\int_{−\infty} ^{\mathrm{0}} \frac{\mathrm{sin}\:{u}}{{u}}{e}^{{u}} {du}=−\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−{u}} \mathrm{sin}\:{u}}{{u}}{du} \\ $$$${J}\left({t}\right)=−\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{tu}} \mathrm{sin}\:{u}}{{u}}{du} \\ $$$${J}'\left({t}\right)=−\int_{\mathrm{0}} ^{\infty} {e}^{−{tu}} \mathrm{sin}\:{udu}=−\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${J}\left({t}\right)=−\mathrm{tan}^{−\mathrm{1}} {t}+{C} \\ $$$${I}={J}\left(\mathrm{1}\right) \\ $$$$\mathrm{Will}\:\mathrm{continue}. \\ $$

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