Question Number 167510 by Mathspace last updated on 18/Mar/22
$${explicite}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({a}+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$${a}\geqslant\mathrm{2} \\ $$
Answered by ArielVyny last updated on 20/Mar/22
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({a}+{tan}^{\mathrm{2}} {x}\right){dx} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{{a}+{tan}^{\mathrm{2}} {x}}{dx} \\ $$$${t}={tanx}\rightarrow{dt}=\left(\mathrm{1}+{t}^{\mathrm{2}} \right){dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{a}+{t}^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{a}+{t}^{\mathrm{2}} }.\left[\mathrm{1}−\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right]{dt} \\ $$$$\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{t}}{\:\sqrt{{a}}}\right)^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{t}^{\mathrm{2}} \right)} \\ $$$$\frac{\sqrt{{a}}}{{a}}\left[{arctg}\left(\frac{\mathrm{1}}{\:\sqrt{{a}}}\right)\right]−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{t}^{\mathrm{2}} \right)}=\frac{\alpha}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\beta}{{a}+{t}^{\mathrm{2}} }=\frac{\alpha\left({a}+{t}^{\mathrm{2}} \right)+\beta\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{{t}^{\mathrm{2}} \left(\alpha+\beta\right)+\alpha{a}+\beta}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({a}+{t}^{\mathrm{2}} \right)} \\ $$$$\alpha+\beta=\mathrm{1}\:{et}\:\alpha{a}+\beta=\mathrm{0} \\ $$$$\alpha−\alpha{a}=\mathrm{1}\rightarrow\alpha\left(\mathrm{1}−{a}\right)=\mathrm{1}\rightarrow\alpha=−\frac{\mathrm{1}}{{a}−\mathrm{1}} \\ $$$$\beta=\frac{{a}}{{a}−\mathrm{1}} \\ $$$${f}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{a}+{t}^{\mathrm{2}} }−\left[−\frac{\mathrm{1}}{{a}−\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{{a}}{{a}−\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{a}+{t}^{\mathrm{2}} }{dt}\right] \\ $$$$\:\:=\left(\mathrm{1}+\frac{{a}}{{a}−\mathrm{1}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{a}+{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}−\mathrm{1}}.\frac{\pi}{\mathrm{4}} \\ $$$$\:\:=\frac{\pi}{\mathrm{4}\left({a}−\mathrm{1}\right)}+\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{a}−\mathrm{1}}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{a}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{{a}}}=\underset{{n}\geqslant\mathrm{0}} {\sum}\int_{\mathrm{0}} ^{\mathrm{1}} \left(−\mathrm{1}\right)^{{n}} \left(\frac{{t}^{\mathrm{2}} }{{a}}\right)^{{n}} =\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{a}^{{n}} } \\ $$$${f}'\left({a}\right)=\frac{\pi}{\mathrm{4}\left({a}−\mathrm{1}\right)}+\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{a}−\mathrm{1}}\right)\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} }{{a}^{{n}} } \\ $$$${f}\left({a}\right)=\frac{\pi}{\mathrm{4}}{ln}\left({a}−\mathrm{1}\right)+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{na}^{{n}} }+\underset{{n}\geqslant\mathrm{0}} {\sum}\int\frac{\left(−\mathrm{1}\right)^{{n}} }{{a}^{{n}+\mathrm{1}} −{a}^{{n}} }{da} \\ $$$$ \\ $$