Question Number 79480 by TawaTawa last updated on 25/Jan/20
Commented by mathmax by abdo last updated on 25/Jan/20
![let I =∫_0 ^(π/4) ((ln(tanx))/(cos(2x)))dx changement tanx =t give I =∫_0 ^1 ((ln(t))/((1−t^2 )/(1+t^2 ))) (dt/(1+t^2 )) =∫_0 ^1 ((ln(t))/(1−t^2 ))dt =∫_0 ^1 ln(t)(Σ_(n=0) ^∞ t^(2n) )dt =Σ_(n=0) ^∞ ∫_0 ^1 t^(2n) ln(t)dt we have by parts ∫_0 ^1 t^(2n) ln(t)dt =[(1/(2n+1))t^(2n+1) ln(t)]_0 ^1 −∫_0 ^1 (t^(2n) /(2n+1))dt =−(1/((2n+1))) ∫_0 ^1 t^(2n) dt =−(1/((2n+1)^2 ))[t^(2n+1) ]_0 ^1 =−(1/((2n+1)^2 )) ⇒ I =−Σ_(n=0) ^∞ (1/((2n+1)^2 )) we have Σ_(n=1) ^∞ (1/n^2 ) =(1/4)Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1−(1/4))Σ_(n=1) ^∞ (1/n^2 ) =(3/4)×(π^2 /6) =(π^2 /8) ⇒I =−(π^2 /8) the value is proved.](Q79507.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{ln}\left({tanx}\right)}{{cos}\left(\mathrm{2}{x}\right)}{dx}\:\:{changement}\:{tanx}\:={t}\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left({t}\right)}{\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({t}\right)\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{t}^{\mathrm{2}{n}} \right){dt} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:\:\:\:{we}\:{have}\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {ln}\left({t}\right){dt}\:=\left[\frac{\mathrm{1}}{\mathrm{2}{n}+\mathrm{1}}{t}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\mathrm{2}{n}} }{\mathrm{2}{n}+\mathrm{1}}{dt} \\ $$$$=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} \:{dt}\:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\left[{t}^{\mathrm{2}{n}+\mathrm{1}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:{I}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:{we}\:{have} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{4}}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\mathrm{3}}{\mathrm{4}}×\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:\:\Rightarrow{I}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$$${the}\:{value}\:{is}\:{proved}. \\ $$
Commented by TawaTawa last updated on 25/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by abdomathmax last updated on 25/Jan/20
$${you}\:\:{are}\:{welcome}\:. \\ $$