Question Number 121862 by Bird last updated on 12/Nov/20 | ||
$$\left.\mathrm{1}\right){explicite}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}−\mathrm{1}} {lnt}}{\mathrm{1}+{t}}{dt} \\ $$$${with}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnt}}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt} \\ $$ | ||
Answered by mnjuly1970 last updated on 12/Nov/20 | ||
$${solution}:\mathrm{1}\:::\:{g}\left({b}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{{t}^{{a}+{b}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({a}\right)={g}'\left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:{g}\left({b}\right)=\Gamma\left({a}+{b}\right)\Gamma\left(\mathrm{1}−{a}−{b}\right)=\frac{\pi}{{sin}\left(\pi\left({a}+{b}\right)\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{g}'\left({b}\right)=\frac{−\pi^{\mathrm{2}} {cos}\left(\pi\left({a}+{b}\right)\right)}{{sin}^{\mathrm{2}} \left(\pi\left({a}+{b}\right)\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{g}'\left(\mathrm{0}\right)=\frac{−\pi^{\mathrm{2}} {cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}={f}\left({a}\overset{} {\right)} \\ $$$$\:\:\:\:\:\:{f}\left({a}\right)=−\pi^{\mathrm{2}} {cot}\left(\pi{a}\right){csc}\left(\pi{a}\right)\:... \\ $$$$\:\:\:\:\:\mathrm{2}::\:\:\:{f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{we}\:{know}\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\overset{{easy}} {=}\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\: \\ $$ | ||