Question Number 190371 by TUN last updated on 02/Apr/23
Answered by qaz last updated on 02/Apr/23
![∫_0 ^1 sin (ln(1/x))((x^b −x^a )/(lnx))dx=∫_(−∞) ^0 sin (−u)∙((e^(ub) −e^(ua) )/u)e^u du =∫_0 ^∞ ((sin u)/u)[e^(−u(a+1)) −e^(−u(b+1)) ]du =∫_0 ^∞ L{e^(−u(a+1)) sin u−e^(−u(b+1)) sin u}∙L^(−1) {u^(−1) }du =∫_0 ^∞ [(1/((u+a+1)^2 +1))−(1/((u+b+1)^2 +1))]du =[arctan (u+a+1)−arctan (u+b+1)]_0 ^∞ =arctan ((a−b)/(1+(u+a+1)(u+b+1)))∣_0 ^∞ =arctan ((b−a)/(1+(a+1)(b+1)))](https://www.tinkutara.com/question/Q190390.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{sin}\:\left({ln}\frac{\mathrm{1}}{{x}}\right)\frac{{x}^{{b}} −{x}^{{a}} }{{lnx}}{dx}=\int_{−\infty} ^{\mathrm{0}} \mathrm{sin}\:\left(−{u}\right)\centerdot\frac{{e}^{{ub}} −{e}^{{ua}} }{{u}}{e}^{{u}} {du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\:{u}}{{u}}\left[{e}^{−{u}\left({a}+\mathrm{1}\right)} −{e}^{−{u}\left({b}+\mathrm{1}\right)} \right]{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \mathscr{L}\left\{{e}^{−{u}\left({a}+\mathrm{1}\right)} \mathrm{sin}\:{u}−{e}^{−{u}\left({b}+\mathrm{1}\right)} \mathrm{sin}\:{u}\right\}\centerdot\mathscr{L}^{−\mathrm{1}} \left\{{u}^{−\mathrm{1}} \right\}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\left({u}+{a}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\left({u}+{b}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}\right]{du} \\ $$$$=\left[\mathrm{arctan}\:\left({u}+{a}+\mathrm{1}\right)−\mathrm{arctan}\:\left({u}+{b}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\infty} \\ $$$$=\mathrm{arctan}\:\frac{{a}−{b}}{\mathrm{1}+\left({u}+{a}+\mathrm{1}\right)\left({u}+{b}+\mathrm{1}\right)}\mid_{\mathrm{0}} ^{\infty} =\mathrm{arctan}\:\frac{{b}−{a}}{\mathrm{1}+\left({a}+\mathrm{1}\right)\left({b}+\mathrm{1}\right)} \\ $$