Question Number 190371 by TUN last updated on 02/Apr/23
Answered by qaz last updated on 02/Apr/23
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\:\left({ln}\frac{\mathrm{1}}{{x}}\right)\frac{{x}^{{b}} −{x}^{{a}} }{{lnx}}{dx}=\int_{−\infty} ^{\mathrm{0}} \mathrm{sin}\:\left(−{u}\right)\centerdot\frac{{e}^{{ub}} −{e}^{{ua}} }{{u}}{e}^{{u}} {du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{u}}{{u}}\left[{e}^{−{u}\left({a}+\mathrm{1}\right)} −{e}^{−{u}\left({b}+\mathrm{1}\right)} \right]{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left\{{e}^{−{u}\left({a}+\mathrm{1}\right)} \mathrm{sin}\:{u}−{e}^{−{u}\left({b}+\mathrm{1}\right)} \mathrm{sin}\:{u}\right\}\centerdot\mathscr{L}^{−\mathrm{1}} \left\{{u}^{−\mathrm{1}} \right\}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\left({u}+{a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\left({u}+{b}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right]{du} \\ $$$$=\left[\mathrm{arctan}\:\left({u}+{a}+\mathrm{1}\right)−\mathrm{arctan}\:\left({u}+{b}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{arctan}\:\frac{{a}−{b}}{\mathrm{1}+\left({u}+{a}+\mathrm{1}\right)\left({u}+{b}+\mathrm{1}\right)}\mid_{\mathrm{0}} ^{\infty} =\mathrm{arctan}\:\frac{{b}−{a}}{\mathrm{1}+\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)} \\ $$