Question Number 108723 by 150505R last updated on 18/Aug/20
Commented by bemath last updated on 19/Aug/20
![I=∫_0 ^(π/2) ln ((√2) cos (x−(π/4)))dx = [ x ln ((√2)) ]_0 ^(π/2) +∫_0 ^(π/2) ln (cos (x−(π/4)))dx = ((π ln (2))/4) + ∫_(−π/4) ^(π/4) ln (cos u) du = ((πln (2))/4)+2∫_0 ^(π/4) ln (cos u) du](https://www.tinkutara.com/question/Q108754.png)
$${I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\:\mathrm{cos}\:\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\right){dx} \\ $$$$\:=\:\left[\:{x}\:\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\right)\:\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} +\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{ln}\:\left(\mathrm{cos}\:\:\left({x}−\frac{\pi}{\mathrm{4}}\right)\right){dx} \\ $$$$\:=\:\frac{\pi\:\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{4}}\:+\:\underset{−\pi/\mathrm{4}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{cos}\:{u}\right)\:{du}\: \\ $$$$\:=\:\frac{\pi\mathrm{ln}\:\left(\mathrm{2}\right)}{\mathrm{4}}+\mathrm{2}\underset{\mathrm{0}} {\overset{\pi/\mathrm{4}} {\int}}\mathrm{ln}\:\left(\mathrm{cos}\:{u}\right)\:{du}\: \\ $$
Answered by mathdave last updated on 19/Aug/20
![solution from sin(x+y)=sinxcosy+cosxsiny let sin(x+(π/4))=sin((π/4))cosx+cos((π/4))sinx=((√2)/2)(sinx+cosx) sinx+cosx=(√2)sin(x+(π/4)) I=∫_0 ^(π/2) ln((√2)sin(x+(π/4)))dx by splitting d integral in half we have I=∫_0 ^(π/4) ln((√2)sin(x+(π/4)))dx+∫_(π/4) ^(π/2) ln((√2)sin(x+(π/4)))dx apply integral reflection identity on integrand with limit (0,(π/4)) by putting y⇒((π/4)−x) and as well putting y⇒(x−(π/4)) on the integrand with limit ((π/4),(π/2)) and after that put back x=y ,so we have ∵I=∫_0 ^(π/4) ln((√2)sin((π/2)−x))dx+∫_0 ^(π/4) ln((√2)sin(x+(π/2)))dx note sin((π/2)−x)=sin(x+(π/2))=cosx I=2∫_0 ^(π/4) ln((√2)cosx)dx=2∫_0 ^(π/4) (1/2)ln2dx+2∫_0 ^(π/4) ln(cosx)dx =((πln2)/4)+2∫_0 ^(π/4) ln(cosx)dx....................(xx) now define two integral I_1 =∫_0 ^(π/4) ln(cosx)dx and I_2 =∫_0 ^(π/4) ln(sinx)dx now adding I_1 and I_(2 ) then we have I_1 +I_2 =∫_0 ^(π/4) ln(cosxsinx)dx but cosxsinx=(1/2)sin2x ∫_0 ^(π/4) ln((1/2)sin2x)dx=∫_0 ^(π/4) ln(sin2x)dx−∫_0 ^(π/4) ln2dx =(1/2)∫_0 ^(π/2) ln(sinx)dx−∫_0 ^(π/4) ln2dx=(1/2)[−(π/2)ln2]−((πln2)/4) I_1 +I_2 =−((πln2)/2).................(1) now subtract I_2 −I_1 =∫_0 ^(π/4) ln(tanx)dx putting u=tanx then dx=(du/((1+x^2 ))) then I_2 −I_1 =∫_0 ^1 ((lnu)/(1+u^2 ))du series of (1/(1+u^2 ))=Σ_(n=0) ^∞ (−1)^n u^(2n) I_2 −I_1 =∫_0 ^1 Σ_(n=0) ^∞ (−1)^n u^(2n) lnudu=Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 u^(2n) lnudu when using I.B.P I_2 −I_1 =−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 ))=−G(called catalan constant).......................(2) solving equation (1) and (2) simultaneously we have I_1 =∫_0 ^(π/4) ln(cosx)dx=(1/2)(G−((πln2)/2)) and I_2 =∫_0 ^(π/4) ln(sinx)dx=−(1/2)(G+((πln2)/2)) now putting back I_1 and I_2 into equation (xx) so that we have I=((πln2)/4)+2I_1 =((πln2)/4)+2[(1/2)(G−((πln2)/4))]=G−((πln2)/4) ∵ ∫_0 ^(π/2) ln(cosx+sinx)dx=G−((πln2)/4) by mathdave (18/08/2020)](https://www.tinkutara.com/question/Q108768.png)
$${solution} \\ $$$${from}\:\mathrm{sin}\left({x}+{y}\right)=\mathrm{sin}{x}\mathrm{cos}{y}+\mathrm{cos}{x}\mathrm{sin}{y} \\ $$$${let} \\ $$$$\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)=\mathrm{s}{in}\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{cos}{x}+\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}\right)\mathrm{sin}{x}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{sin}{x}+\mathrm{cos}{x}\right) \\ $$$$\mathrm{sin}{x}+\mathrm{cos}{x}=\sqrt{\mathrm{2}}\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx} \\ $$$${by}\:{splitting}\:{d}\:{integral}\:{in}\:{half}\:{we}\:{have} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx}+\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)\right){dx} \\ $$$${apply}\:{integral}\:{reflection}\:{identity}\:{on}\:{integrand}\:{with}\:{limit} \\ $$$$\left(\mathrm{0},\frac{\pi}{\mathrm{4}}\right)\:\:{by}\:{putting}\:\:{y}\Rightarrow\left(\frac{\pi}{\mathrm{4}}−{x}\right)\:{and}\:{as}\:{well}\:{putting}\:\:{y}\Rightarrow\left({x}−\frac{\pi}{\mathrm{4}}\right)\:{on}\:{the}\:{integrand}\:{with}\:{limit} \\ $$$$\left(\frac{\pi}{\mathrm{4}},\frac{\pi}{\mathrm{2}}\right)\:\:{and}\:{after}\:{that}\:\:{put}\:{back}\:{x}={y}\:,{so}\:{we}\:{have} \\ $$$$\because{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)\right){dx}+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{2}}\right)\right){dx} \\ $$$${note}\:\mathrm{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)=\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{2}}\right)=\mathrm{cos}{x} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\sqrt{\mathrm{2}}\mathrm{cos}{x}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln2}{dx}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx} \\ $$$$=\frac{\pi\mathrm{ln2}}{\mathrm{4}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}………………..\left({xx}\right) \\ $$$${now}\:{define}\:{two}\:{integral}\:\:{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}\:\:{and} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sin}{x}\right){dx}\:\:\:{now}\:{adding}\:\:\:{I}_{\mathrm{1}} \:{and}\:{I}_{\mathrm{2}\:} \:{then}\:{we} \\ $$$${have}\:\:{I}_{\mathrm{1}} +{I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\mathrm{sin}{x}\right){dx}\:\:{but}\: \\ $$$$\mathrm{cos}{x}\mathrm{sin}{x}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin2}{x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sin2}{x}\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln2}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{sin}{x}\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln2}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[−\frac{\pi}{\mathrm{2}}\mathrm{ln2}\right]−\frac{\pi\mathrm{ln2}}{\mathrm{4}} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{2}} =−\frac{\pi\mathrm{ln2}}{\mathrm{2}}……………..\left(\mathrm{1}\right) \\ $$$${now}\:{subtract}\:{I}_{\mathrm{2}} −{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{tan}{x}\right){dx}\:\:\:{putting} \\ $$$${u}=\mathrm{tan}{x}\:\:\:{then}\:\:{dx}=\frac{{du}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:\:{then}\:{I}_{\mathrm{2}} −{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$${series}\:{of}\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}} \: \\ $$$${I}_{\mathrm{2}} −{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} {u}^{\mathrm{2}{n}} \mathrm{ln}{udu}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\mathrm{2}{n}} \mathrm{ln}{udu} \\ $$$${when}\:{using}\:{I}.{B}.{P} \\ $$$${I}_{\mathrm{2}} −{I}_{\mathrm{1}} =−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=−{G}\left({called}\:{catalan}\right. \\ $$$$\left.{constant}\right)…………………..\left(\mathrm{2}\right) \\ $$$${solving}\:{equation}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{simultaneously}\:{we} \\ $$$${have}\:\:\:{I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({G}−\frac{\pi\mathrm{ln2}}{\mathrm{2}}\right)\:\:\:{and} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{sin}{x}\right){dx}=−\frac{\mathrm{1}}{\mathrm{2}}\left({G}+\frac{\pi\mathrm{ln2}}{\mathrm{2}}\right)\:\:{now}\:{putting} \\ $$$${back}\:{I}_{\mathrm{1}} \:{and}\:{I}_{\mathrm{2}} \:{into}\:\:{equation}\:\left({xx}\right)\:{so}\:{that}\:{we}\:{have} \\ $$$${I}=\frac{\pi\mathrm{ln2}}{\mathrm{4}}+\mathrm{2}{I}_{\mathrm{1}} =\frac{\pi\mathrm{ln2}}{\mathrm{4}}+\mathrm{2}\left[\frac{\mathrm{1}}{\mathrm{2}}\left({G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}}\right)\right]={G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}} \\ $$$$\because\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}\right){dx}={G}−\frac{\pi\mathrm{ln2}}{\mathrm{4}} \\ $$$${by}\:{mathdave}\:\left(\mathrm{18}/\mathrm{08}/\mathrm{2020}\right) \\ $$
Commented by mathmax by abdo last updated on 19/Aug/20
$$\mathrm{you}\:\mathrm{have}\:\mathrm{done}\:\mathrm{a}\:\mathrm{good}\:\mathrm{work}\:\mathrm{your}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{correct}\:\mathrm{thanks} \\ $$
Answered by mathmax by abdo last updated on 19/Aug/20
![let I =∫_0 ^(π/2) ln(sinx +cosx)dx ⇒I =∫_0 ^(π/2) ln{cosx(1+tanx)}dx =∫_0 ^(π/2) ln(cosx)dx +∫_0 ^(π/2) ln(1+tanx)dx =−(π/2)ln(2)+J J =∫_0 ^(π/2) ln(1+tanx)dx =_(tanx=t) ∫_0 ^∞ ((ln(1+t))/(1+t^2 ))dt let f(a) =∫_0 ^∞ ((ln(a+t))/(t^2 +1))dt (a>0) ⇒f^′ (a) =∫_0 ^∞ (dt/((t+a)(t^2 +1))) let decompose F(t) =(1/((t+a)(t^2 +1))) ⇒F(t) =(α/(t+a)) +((et +f)/(t^2 +1)) α =(1/(a^2 +1)) ,lim_(t→+∞) tF(t) =0 =α+e ⇒e=−(1/(a^2 +1)) F(o) =(1/a) =(α/a) +f ⇒f =(1/a)−(1/(a(a^2 +1))) =(1/a)(1−(1/(a^2 +1))) =(a/((a^2 +1))) ⇒F(t)=(1/((a^2 +1)(t+a)))+((−(1/(a^2 +1))t+(a/((a^2 +1))))/(t^(2 ) +1)) ⇒ f^′ (a) =(1/(a^2 +1)){ ∫_0 ^∞ (dt/(t+a))−(1/2)∫_0 ^∞ ((2t)/(t^2 +1))dt+a∫_0 ^∞ (dt/(t^2 +1))} =(1/(a^2 +1)){[ln∣((t+a)/( (√(t^2 +1))))∣]_0 ^∞ +((πa)/2)} =−((lna)/(a^2 +1)) +((πa)/(2(a^2 +1))) ⇒ f(a) =−∫_0 ^a ((lnx)/(1+x^2 ))dx +(π/2)∫_0 ^(a ) ((xdx)/((x^2 +1))) +c =−∫_0 ^a ((lnx)/(1+x^2 ))dx +(π/4)ln(1+a^2 ) +c we have c=f(o)=∫_0 ^∞ ((lnt)/(1+t^2 ))dt=0 ⇒ f(a) =(π/4)ln(1+a^2 )−∫_0 ^a ((lnx)/(1+x^2 ))dx and ∫_0 ^∞ ((ln(1+t))/(1+t^2 ))dt =f(1) =(π/4)ln(2)−∫_0 ^1 ((lnx)/(1+x^2 ))dx but ∫_0 ^1 ((lnx)/(1+x^2 ))dx =∫_0 ^1 ln(x)(Σ_(n=0) ^∞ (−1)^n x^(2n) )dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n) ln(x)dx =Σ_(n=0) ^∞ (−1)^n {[(x^(2n+1) /(2n+1)) lnx]_0 ^1 −∫_0 ^1 (x^(2n) /(2n+1))dx} =−Σ_(n=0) ^∞ (((−1)^n )/(2n+1)) ∫_0 ^1 x^(2n ) dx=−Σ_(n=0) ^∞ (((−1)^n )/((2n+1)^2 )) =−k( catalan constant) ∫_0 ^∞ ((ln(1+t))/(1+t^2 ))dt =(π/4)ln(2) +k ⇒ ∫_0 ^(π/2) ln(cosx +sinx)dx =−(π/2)ln2 +(π/4)ln(2)+k =k−(π/4)ln(2)](https://www.tinkutara.com/question/Q108849.png)
$$\mathrm{let}\:\:\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{ln}\left(\mathrm{sinx}\:+\mathrm{cosx}\right)\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left\{\mathrm{cosx}\left(\mathrm{1}+\mathrm{tanx}\right)\right\}\mathrm{dx} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{J} \\ $$$$\mathrm{J}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{tanx}\right)\mathrm{dx}\:=_{\mathrm{tanx}=\mathrm{t}} \:\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\mathrm{let} \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{a}+\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}+\mathrm{a}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$$\mathrm{let}\:\mathrm{decompose}\:\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\mathrm{1}}{\left(\mathrm{t}+\mathrm{a}\right)\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)\:=\frac{\alpha}{\mathrm{t}+\mathrm{a}}\:+\frac{\mathrm{et}\:+\mathrm{f}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\alpha\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:\:,\mathrm{lim}_{\mathrm{t}\rightarrow+\infty} \mathrm{tF}\left(\mathrm{t}\right)\:=\mathrm{0}\:=\alpha+\mathrm{e}\:\Rightarrow\mathrm{e}=−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\mathrm{F}\left(\mathrm{o}\right)\:=\frac{\mathrm{1}}{\mathrm{a}}\:=\frac{\alpha}{\mathrm{a}}\:+\mathrm{f}\:\Rightarrow\mathrm{f}\:=\frac{\mathrm{1}}{\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{a}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{a}}{\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{1}}{\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}+\mathrm{a}\right)}+\frac{−\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} +\mathrm{1}}\mathrm{t}+\frac{\mathrm{a}}{\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}}{\mathrm{t}^{\mathrm{2}\:} +\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\left\{\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{t}+\mathrm{a}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2t}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}+\mathrm{a}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\left\{\left[\mathrm{ln}\mid\frac{\mathrm{t}+\mathrm{a}}{\:\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}}\mid\right]_{\mathrm{0}} ^{\infty} \:+\frac{\pi\mathrm{a}}{\mathrm{2}}\right\}\:=−\frac{\mathrm{lna}}{\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}}\:+\frac{\pi\mathrm{a}}{\mathrm{2}\left(\mathrm{a}^{\mathrm{2}} \:+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{a}} \frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:+\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{a}\:} \:\frac{\mathrm{xdx}}{\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\:+\mathrm{c} \\ $$$$=−\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:+\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)\:+\mathrm{c}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{c}=\mathrm{f}\left(\mathrm{o}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{1}+\mathrm{a}^{\mathrm{2}} \right)−\int_{\mathrm{0}} ^{\mathrm{a}} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{and}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\mathrm{f}\left(\mathrm{1}\right)\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)−\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\left(\mathrm{x}\right)\left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \right)\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}} \:\mathrm{ln}\left(\mathrm{x}\right)\mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \left\{\left[\frac{\mathrm{x}^{\mathrm{2n}+\mathrm{1}} }{\mathrm{2n}+\mathrm{1}}\:\mathrm{lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}} }{\mathrm{2n}+\mathrm{1}}\mathrm{dx}\right\} \\ $$$$=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}\:} \mathrm{dx}=−\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{2}} }\:=−\mathrm{k}\left(\:\mathrm{catalan}\:\mathrm{constant}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)\:+\mathrm{k}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cosx}\:+\mathrm{sinx}\right)\mathrm{dx}\:=−\frac{\pi}{\mathrm{2}}\mathrm{ln2}\:+\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{k} \\ $$$$=\mathrm{k}−\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$