Question Number 68100 by ~ À ® @ 237 ~ last updated on 04/Sep/19 | ||
$${Find}\:\:{K}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}−{t}+{t}^{\mathrm{2}} \right)}{{t}}\:{dt}\:\:\:\:\: \\ $$ | ||
Commented by mind is power last updated on 05/Sep/19 | ||
$${y},{re}\:{welcom} \\ $$ | ||
Commented by mind is power last updated on 05/Sep/19 | ||
$${f}\left({s}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right){s}\right)}{{t}} \\ $$$${s}\in\left[\mathrm{0},\mathrm{1}\right] \\ $$$${f}'\left({s}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−\left({t}−{t}^{\mathrm{2}} \right)}{{t}\left(\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right){s}\right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\mathrm{1}}{\mathrm{1}−\left({t}−{t}^{\mathrm{2}} \right){s}}{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\mathrm{1}}{{t}^{\mathrm{2}} {s}−{ts}+\mathrm{1}}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} {s}−{ts}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{t}^{\mathrm{2}} {s}−{ts}+\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{s}}\left[{ln}\left({t}^{\mathrm{2}} {s}−{ts}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{s}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{{t}^{\mathrm{2}} −{t}+\frac{\mathrm{1}}{{s}}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}{s}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{s}}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$−\frac{\mathrm{2}}{\mathrm{4}−{s}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left\{\left(\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right).\sqrt{\frac{\mathrm{4}{s}}{\mathrm{4}−{s}}}\right)^{\mathrm{2}} +\mathrm{1}\right\}.} \\ $$$$=−\frac{\mathrm{1}}{\sqrt{{s}\left(\mathrm{4}−{s}\right)}}\left[{arctan}\left(\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{4}{s}}{\mathrm{4}−{s}}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=−\frac{\mathrm{2}}{\sqrt{{s}\left(\mathrm{4}−{s}\right)}}\left[{arctg}\left(\sqrt{\frac{{s}}{\mathrm{4}−{s}}}\right)\right] \\ $$$${f}\left({s}\right)=\int−\frac{\mathrm{2}}{\sqrt{{s}\left(\mathrm{4}−{s}\right)}}{arctan}\left(\sqrt{\frac{{s}}{\mathrm{4}−{s}}}\right){ds} \\ $$$${let}\:{u}=\sqrt{\frac{{s}}{\mathrm{4}−{s}}}\Rightarrow{s}=\frac{\mathrm{4}{u}^{\mathrm{2}} }{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$${ds}=\frac{\mathrm{8}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du} \\ $$$${s}\left(\mathrm{4}−{s}\right)=\frac{\mathrm{16}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int\frac{−\mathrm{2}}{\sqrt{{s}\left(\mathrm{4}−{s}\right)}}\left[{arctan}\left(\sqrt{\frac{\mathrm{4}{s}}{\mathrm{4}−{s}}}\right)\right]{ds} \\ $$$$=\int\frac{−\mathrm{2}}{\sqrt{\frac{\mathrm{16}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }}}.{arctan}\left({u}\right).\frac{\mathrm{8}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\int\frac{−\mathrm{2}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}{u}}{arctan}\left({u}\right).\frac{\mathrm{8}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\int\frac{−\mathrm{4}}{\mathrm{1}+{u}^{\mathrm{2}} }{arctan}\left({u}\right){du} \\ $$$$=−\mathrm{2}\left({arctan}\left({u}\right)\right)^{\mathrm{2}} +{c} \\ $$$$=−\mathrm{2}\left({arctan}\left(\sqrt{\frac{{s}}{\mathrm{4}−{s}}}\right)\right)^{\mathrm{2}} +{c}={f}\left({s}\right) \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}=>{c}=\mathrm{0} \\ $$$${f}\left({s}\right)=−\mathrm{2}\left({arctan}\left(\sqrt{\frac{{s}}{\mathrm{4}−{s}}}\right)\right)^{\mathrm{2}} \\ $$$${k}={f}\left(\mathrm{1}\right)=−\mathrm{2}\left({arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right)^{\mathrm{2}} =−\mathrm{2}\left(\frac{\pi}{\mathrm{6}}\right)^{\mathrm{2}} =\frac{−\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$$$ \\ $$$$ \\ $$ | ||
Commented by ~ À ® @ 237 ~ last updated on 05/Sep/19 | ||
$${Thanks}\:\:{you}\:{sir} \\ $$$${let}\:{consider}\:\:{g}\:{defined}\:{by}\:\:{g}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\:\:{ln}\left(\mathrm{1}+{at}+{t}^{\mathrm{2}} \right)}{{t}}{dt}\:\:\:,\:\:{we}\:{have}\:\:{D}_{{g}} =\left[−\mathrm{2};\mathrm{2}\right] \\ $$$${when}\:{stating}\:\:\:{t}=\frac{\mathrm{1}}{{u}}\:\:\:\:\:\:\:{we}\:{get}\:\:\:{g}\left({a}\right)=\int_{\mathrm{1}} ^{\infty} \:\left[\frac{−\mathrm{2}{lnt}}{{t}}\:+\:\frac{{ln}\left(\mathrm{1}+{at}+{t}^{\mathrm{2}} \right)}{{t}}\:\right]{dt} \\ $$$${Now}\:{g}'\left({a}\right)=\:\int_{\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{t}}\:.\frac{{t}}{\mathrm{1}+{at}+{t}^{\mathrm{2}} }\:{dt} \\ $$$${using}\:\:{t}^{\mathrm{2}} +{at}+\mathrm{1}=\left[\left({t}+\frac{{a}}{\mathrm{2}}\right)+\frac{\mathrm{4}−{a}^{\mathrm{2}} }{\mathrm{4}}\right]=\left(\frac{\mathrm{4}−{a}^{\mathrm{2}} }{\mathrm{4}}\right)\left[\left(\frac{\mathrm{2}{t}+{a}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right)^{\mathrm{2}} +\mathrm{1}\right]\:\:\:\:\:\:{cause}\:\:{a}\in{D}_{{g}} \: \\ $$$${So}\:\:{g}'\left({a}\right)=\frac{\mathrm{4}}{\mathrm{4}−{a}^{\mathrm{2}} }\:\int_{\mathrm{1}} ^{\infty} \frac{\:\:\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{2}{t}+{a}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:}\right)^{\mathrm{2}} }\:{dt}\:=\frac{\mathrm{2}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:}\:\left[{arctan}\left(\frac{\mathrm{2}{t}+{a}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:}\right)\right]_{\mathrm{1}} ^{\infty} \\ $$$${we}\:{get}\:\:{g}'\left({a}\right)=\:\frac{\pi−\mathrm{2}{arctan}\left(\frac{\mathrm{2}+{a}}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}\right)}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:}\: \\ $$$${Now}\:\:{g}\left({a}\right)=\int\:\frac{\pi}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:}{da}\:−\mathrm{2}\int\:\frac{{arctan}\left(\sqrt{\frac{\mathrm{2}+{a}}{\mathrm{2}−{a}}\:}\:\right)}{\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }}{da} \\ $$$${for}\:{the}\:{second}\:{whole}\:{let}\:{state}\:\:\:{u}=\sqrt{\frac{\mathrm{2}+{a}}{\mathrm{2}−{a}}}\:\:\:\Rightarrow{a}=\frac{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:{and}\:\:\sqrt{\mathrm{4}−{a}^{\mathrm{2}} }\:=\frac{\mathrm{4}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:,\:{da}=\:\frac{\mathrm{8}{udu}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\: \\ $$$${g}\left({a}\right)=\:\int\:\pi\frac{\frac{\mathrm{1}}{\mathrm{2}}}{\sqrt{\mathrm{1}−\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} }}\:{da}\:−\:\mathrm{2}\int\:\:\:\frac{\mathrm{2}{arctanu}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du} \\ $$$$\:\:\:\:{then}\:\:{g}\left({a}\right)=\:\pi{arcsin}\left(\frac{{a}}{\mathrm{2}}\right)\:−\mathrm{2}\left({arctan}\sqrt{\frac{\mathrm{2}+{a}}{\mathrm{2}−{a}}}\:\right)^{\mathrm{2}} +{c} \\ $$$${firstly}\:\:{g}\left(−\mathrm{2}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−\mathrm{2}{t}+{t}^{\mathrm{2}} \right)}{{t}}{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}{dt} \\ $$$${g}\left(−\mathrm{2}\right)=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{t}^{{n}−\mathrm{1}} }{{n}}\:=−\mathrm{2}\underset{{n}=\mathrm{1}_{} } {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\: \\ $$$${secondly}\:\:{g}\left(−\mathrm{2}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$${then}\:{c}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\: \\ $$$${Finally}\:\:{we}\:{were}\:{searching}\:{about} \\ $$$${g}\left(−\mathrm{1}\right)=\:\pi{arcsin}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\:−\mathrm{2}\left({arctan}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\right)\right)^{\mathrm{2}} +\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\: \\ $$$$\:\:{g}\left(−\mathrm{1}\right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{6}}\:−\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{36}}\:+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{18}}\: \\ $$$$ \\ $$ | ||
Commented by mathmax by abdo last updated on 06/Sep/19 | ||
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{xt}\:+{t}^{\mathrm{2}} \right)}{{t}}{dt}\:\:\:\:{we}\:{must}\:{have}\:{t}^{\mathrm{2}} \:+{xt}\:+\mathrm{1}>\mathrm{0}\:{for}\:{all}\:{t}\Rightarrow \\ $$$${x}^{\mathrm{2}} −\mathrm{4}<\mathrm{0}\:\Rightarrow−\mathrm{2}<{x}<\mathrm{2}\:\:{we}\:{have}\:{f}^{'} \left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}}{{t}\left(\mathrm{1}+{xt}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+{xt}\:+\mathrm{1}}\:=_{{t}=\frac{\mathrm{1}}{{u}}} \:\:\:\:−\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{−{du}}{{u}^{\mathrm{2}} \left\{\frac{\mathrm{1}}{{u}^{\mathrm{2}} }\:+\frac{{x}}{{u}}+\mathrm{1}\right\}} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{du}}{\mathrm{1}+{xu}+{u}^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}\frac{{x}}{\mathrm{2}}{u}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{du}}{\left({u}+\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{4}−{x}^{\mathrm{2}} }{\mathrm{4}}}\:{changement}\:\:{u}+\frac{{x}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}{z}\:{givez}=\frac{\mathrm{2}{u}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }} \\ $$$${f}^{'} \left({x}\right)\:=\frac{\mathrm{4}}{\mathrm{4}−{x}^{\mathrm{2}} }\:\:\int_{\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}} ^{+\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+{z}^{\mathrm{2}} }×\frac{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{\mathrm{2}}{dz} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\int_{\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}} ^{+\infty} \:\:\frac{{dz}}{\mathrm{1}+{z}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\left[{arctanz}\right]_{\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}} ^{+\infty} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\left\{\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right)\right\}=\frac{\pi}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}−\frac{\mathrm{2}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:{arctan}\left(\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{f}\left({x}\right)\:=\int\:\frac{\pi{dx}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:−\mathrm{2}\:\int\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:{arctan}\left(\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right)\:+{c} \\ $$$${changement}\:\:{x}\:=\mathrm{2}{cos}\theta\:{give} \\ $$$$\int\:\frac{\mathrm{1}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:{arctan}\left(\frac{\mathrm{2}+{x}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\right)\:=\int\:\frac{\mathrm{1}}{\mathrm{2}{sin}\theta}\:{arctan}\left(\frac{\mathrm{2}+\mathrm{2}{cos}\theta}{\mathrm{2}{sin}\theta}\right)\left(−\mathrm{2}{sin}\theta\right){cos}\theta \\ $$$$=−\int\:\:{arctan}\left(\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)}{\mathrm{2}{cos}\left(\frac{\theta}{\mathrm{2}}\right){sin}\left(\frac{\theta}{\mathrm{2}}\right)}\right)\:=−\int\:\:{arctan}\left(\frac{\mathrm{1}}{{tan}\theta}\right) \\ $$$$=−\left(\frac{\pi}{\mathrm{2}}−\theta\right)\:=\theta−\frac{\pi}{\mathrm{2}}\:={arcos}\left(\frac{{x}}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}} \\ $$$$\int\:\frac{\pi{dx}}{\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}\:=_{{x}\:=\mathrm{2}{cos}\theta} \:\:\:\:\int\:\:\frac{−\mathrm{2}\pi{sin}\theta\:{d}\theta}{\mathrm{2}{sin}\theta}\:=−\pi\theta\:=−\pi{arccos}\left(\frac{{x}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${f}\left({x}\right)=−\pi\:{arcos}\left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{2}\:{arcos}\left(\frac{{x}}{\mathrm{2}}\right)+\pi\:+{c} \\ $$$$=\pi−\left(\pi+\mathrm{2}\right)\:{arcos}\left(\frac{{x}}{\mathrm{2}}\right)\:+{c} \\ $$$${f}\left(\mathrm{0}\right)\:=\pi−\left(\pi+\mathrm{2}\right)\frac{\pi}{\mathrm{2}}\:+{c}\:=\pi−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi\:+{c}\:={c}−\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow{c}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}+{f}\left(\mathrm{0}\right) \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\:+\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}{dt} \\ $$$${we}\:{have}\:{ln}^{,} \left(\mathrm{1}+{u}\right)\:=\frac{\mathrm{1}}{\mathrm{1}+{u}}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} \:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{u}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {u}^{{n}} }{{n}+\mathrm{1}}\:+{c}\left({c}=\mathrm{0}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{u}^{{n}} }{{n}}\:\:\Rightarrow \\ $$$${ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {t}^{\mathrm{2}{n}} }{{n}}\:\Rightarrow\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {t}^{\mathrm{2}{n}−\mathrm{1}} }{{n}} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{{t}}{dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{t}^{\mathrm{2}{n}−\mathrm{1}} {dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)=−\frac{\mathrm{1}}{\mathrm{4}}\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:\Rightarrow \\ $$$${c}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{24}}\:\:\:\:{be}\:{continued}... \\ $$ | ||