Question Number 121869 by mnjuly1970 last updated on 12/Nov/20 | ||
$$\:\:\:\:\:\:\:\:\:...\:{advanced}\:\:{calculus}... \\ $$$$\:\:\:\:{evaluate}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{I}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \left(\frac{{ln}\left({tan}\left({x}\right)\right)}{{sin}\left({x}\right)−{cos}\left({x}\right)}\right)^{\mathrm{2}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.{m}.{n}. \\ $$ | ||
Answered by mindispower last updated on 12/Nov/20 | ||
$${problem}\:{in}\:\pi/\mathrm{4}\:{or}\:{Pv}..? \\ $$ | ||
Commented by mnjuly1970 last updated on 13/Nov/20 | ||
$$\mathrm{0}\:{to}\:\:\frac{\pi}{\mathrm{2}}\:{is}\:{correct}... \\ $$ | ||
Answered by mindispower last updated on 13/Nov/20 | ||
$$\left({sin}\left({x}\right)−{cos}\left({x}\right)\right)^{\mathrm{2}} =\frac{\left({tg}\left({x}\right)−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{1}+{tg}^{\mathrm{2}} \left({x}\right)} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{ln}^{\mathrm{2}} \left({tg}\left({x}\right)\right)}{\left({tg}\left({x}\right)−\mathrm{1}\right)^{\mathrm{2}} }{dtg}\left({x}\right) \\ $$$${tg}\left({x}\right)={t}\Rightarrow{I}=\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({r}\right)}{\left({r}−\mathrm{1}\right)^{\mathrm{2}} }{dr} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{ln}^{\mathrm{2}} \left({r}\right)}{\left({r}−\mathrm{1}\right)^{\mathrm{2}} }.{dr}+\frac{{ln}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{r}}\right)}{\left(\frac{\mathrm{1}}{{r}}−\mathrm{1}\right)^{\mathrm{2}} }.\frac{{dr}}{{r}^{\mathrm{2}} }\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({r}\right){dr}}{\left({r}−\mathrm{1}\right)^{\mathrm{2}} }..{IBP}\Rightarrow \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{−{ln}^{\mathrm{2}} \left({r}\right)}{{r}−\mathrm{1}}\right]_{{t}} ^{\mathrm{1}} +\mathrm{2}\int_{{t}} ^{\mathrm{1}} \frac{{ln}\left({r}\right)}{{r}\left({r}−\mathrm{1}\right)}{dr} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}^{\mathrm{2}} \left({t}\right)}{{t}−\mathrm{1}}+\mathrm{2}\int_{{t}} ^{\mathrm{1}} \frac{{ln}\left({r}\right)}{{r}−\mathrm{1}}{dr}−\mathrm{2}\int_{{t}} ^{\mathrm{1}} \frac{{ln}\left({r}\right)}{{r}}{dr} \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{ln}^{\mathrm{2}} \left({t}\right)}{{t}−\mathrm{1}}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}−{t}} \frac{{ln}\left(\mathrm{1}−{r}\right)}{{r}}{dr}+{ln}^{\mathrm{2}} \left({t}\right)\: \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\left(\frac{{ln}^{\mathrm{2}} \left({t}\right)}{{t}−\mathrm{1}}+{ln}^{\mathrm{2}} \left({t}\right)\right)+\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}−{t}\right)\right] \\ $$$$=\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left[\frac{{tln}^{\mathrm{2}} \left({t}\right)}{{t}−\mathrm{1}}\right]+\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)\:\:{cause}\:{Li}_{\mathrm{2}} {is}\:{continus} \\ $$$$=\mathrm{2}{Li}_{\mathrm{2}} \left(\mathrm{1}\right)=\mathrm{2}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by mnjuly1970 last updated on 13/Nov/20 | ||
$${bravo}\:{bravo} \\ $$$${sir}\:\:{mindspower} \\ $$$$\:{thank}\:{you}\:{so}\:{much}... \\ $$ | ||
Commented by mindispower last updated on 13/Nov/20 | ||
$${thanx}\:{always}\:{a}\:{pleasur} \\ $$ | ||