Question Number 141916 by mnjuly1970 last updated on 24/May/21
$$ \\ $$$$\:\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\mathrm{I}:=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {arccosh}\left({sin}\left({x}\right)+{cos}\left({x}\right)\right){dx}=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:.. \\ $$
Answered by mindispower last updated on 25/May/21
![by part withe d(arccoh(x))=(dx/( (√(1−x^2 )))) I=xarccosh(sinx+cos(x))]_0 ^(π/2) −∫_0 ^(π/2) ((x(cos(x)−sin(x)))/( (√(sin(2x)))))dx −∫_0 ^(π/2) ((xcos(x)−xsin(x))/( (√(sin(2x)))))dx =−(1/( (√2)))∫_0 ^(π/2) xsin^(−(1/2)) (x)cos^(1/2) (x)+(1/( (√2)))∫_0 ^(π/2) xsin^(1/2) (x)cos^(−(1/2)) (x) (1/( (√2)))∫_0 ^∞ ((arctg(x)((√x)−(1/( (√x)))))/(1+x^2 ))dx=A ∫_0 ^∞ ((x^s arctan(tx))/(1+x^2 )) dx=f(t),s∈]−1,1[ f′(t)=∫_0 ^∞ (x^(s+1) /((1+x^2 )(1+t^2 x^2 )))dx ∫_(−∞) ^∞ (x^(s+1) /((1+x^2 )(1+t^2 x^2 )))dx=a=2iπ(.(i^(s+1) /(2i(1−t^2 )))+((i^(s+1) t^2 )/(t^(s+1) .(t^2 −1))).(1/(2it))) =((πi^(s+1) )/(1−t^2 ))(1−t^(−s) ) a=f(t)+∫_0 ^∞ (((−x)^(s+1) )/((1+x^2 )(1+t^2 x^2 )))dx =(1+(−1)^(s+1) )f(t) =(1+e^(iπ(s+1)) )f(t) f(t)=(π/(1−t^2 ))(1−t^(−s) ).(e^(i(π/2)(s+1)) /(1+e^(iπ(s+1)) )) =((π(1−t^(−s) ))/(2(1−t^2 )cos((π/2)(s+1))))=f′(t) ∫_0 ^1 f′(t)=((x^s arctan(x))/(1+x^2 )) =(π/(2cos((π/2)(s+1))))∫_0 ^1 ((1−t^(−s) )/(1−t^2 ))dt_(t^2 =x) =−(π/(4sin(((πs)/2))))∫_0 ^1 ((1−x^(−(s/2)) )/(1−x))x^(−(1/2)) =−(π/(4sin(((πs)/2))))∫_0 ^1 ((x^(−(1/2)) −x^(−((s+1)/2)) )/(1−x))dx=g(s) Ψ(s+1)=−γ+∫_0 ^1 ((1−x^s )/(1−x))dx g(s)=−(π/(4sin(((πs)/2)))).(Ψ(((1−s)/2))−Ψ((1/2))) A=−(π/(4sin((π/4))(√2)))(Ψ((1/4))−2Ψ((1/2))+Ψ((3/4))) Ψ((1/2))=−2ln(2)−γ Ψ((1/4))=(−(π/2)−3ln(2)−γ) Ψ((3/4))=Ψ((1/4))+π =−(π/4)(−π−6ln(2)−2γ+π+4ln(2)+2γ) =−(π/4).−2ln(2)=(π/2)ln(2)](https://www.tinkutara.com/question/Q141960.png)
$${by}\:{part}\:{withe}\:{d}\left({arccoh}\left({x}\right)\right)=\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\left.{I}={xarccosh}\left({sinx}+{cos}\left({x}\right)\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}\left({cos}\left({x}\right)−{sin}\left({x}\right)\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{xcos}\left({x}\right)−{xsin}\left({x}\right)}{\:\sqrt{{sin}\left(\mathrm{2}{x}\right)}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xsin}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right){cos}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xsin}^{\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right){cos}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left({x}\right) \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{{arctg}\left({x}\right)\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}={A} \\ $$$$\left.\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}} {arctan}\left({tx}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}={f}\left({t}\right),{s}\in\right]−\mathrm{1},\mathrm{1}\left[\right. \\ $$$${f}'\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{s}+\mathrm{1}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{dx} \\ $$$$\int_{−\infty} ^{\infty} \frac{{x}^{{s}+\mathrm{1}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{dx}={a}=\mathrm{2}{i}\pi\left(.\frac{{i}^{{s}+\mathrm{1}} }{\mathrm{2}{i}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}+\frac{{i}^{{s}+\mathrm{1}} {t}^{\mathrm{2}} }{{t}^{{s}+\mathrm{1}} .\left({t}^{\mathrm{2}} −\mathrm{1}\right)}.\frac{\mathrm{1}}{\mathrm{2}{it}}\right) \\ $$$$=\frac{\pi{i}^{{s}+\mathrm{1}} }{\mathrm{1}−{t}^{\mathrm{2}} }\left(\mathrm{1}−{t}^{−{s}} \right) \\ $$$${a}={f}\left({t}\right)+\int_{\mathrm{0}} ^{\infty} \frac{\left(−{x}\right)^{{s}+\mathrm{1}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\mathrm{1}+{t}^{\mathrm{2}} {x}^{\mathrm{2}} \right)}{dx} \\ $$$$=\left(\mathrm{1}+\left(−\mathrm{1}\right)^{{s}+\mathrm{1}} \right){f}\left({t}\right) \\ $$$$=\left(\mathrm{1}+{e}^{{i}\pi\left({s}+\mathrm{1}\right)} \right){f}\left({t}\right) \\ $$$${f}\left({t}\right)=\frac{\pi}{\mathrm{1}−{t}^{\mathrm{2}} }\left(\mathrm{1}−{t}^{−{s}} \right).\frac{{e}^{{i}\frac{\pi}{\mathrm{2}}\left({s}+\mathrm{1}\right)} }{\mathrm{1}+{e}^{{i}\pi\left({s}+\mathrm{1}\right)} } \\ $$$$=\frac{\pi\left(\mathrm{1}−{t}^{−{s}} \right)}{\mathrm{2}\left(\mathrm{1}−{t}^{\mathrm{2}} \right){cos}\left(\frac{\pi}{\mathrm{2}}\left({s}+\mathrm{1}\right)\right)}={f}'\left({t}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}'\left({t}\right)=\frac{{x}^{{s}} {arctan}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}\left({s}+\mathrm{1}\right)\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{−{s}} }{\mathrm{1}−{t}^{\mathrm{2}} }{dt}_{{t}^{\mathrm{2}} ={x}} \\ $$$$=−\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{−\frac{{s}}{\mathrm{2}}} }{\mathrm{1}−{x}}{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=−\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} −{x}^{−\frac{{s}+\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}−{x}}{dx}={g}\left({s}\right) \\ $$$$\Psi\left({s}+\mathrm{1}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{s}} }{\mathrm{1}−{x}}{dx} \\ $$$${g}\left({s}\right)=−\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi{s}}{\mathrm{2}}\right)}.\left(\Psi\left(\frac{\mathrm{1}−{s}}{\mathrm{2}}\right)−\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${A}=−\frac{\pi}{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)\sqrt{\mathrm{2}}}\left(\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\mathrm{2}\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{2}{ln}\left(\mathrm{2}\right)−\gamma \\ $$$$\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)=\left(−\frac{\pi}{\mathrm{2}}−\mathrm{3}{ln}\left(\mathrm{2}\right)−\gamma\right) \\ $$$$\Psi\left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\Psi\left(\frac{\mathrm{1}}{\mathrm{4}}\right)+\pi \\ $$$$=−\frac{\pi}{\mathrm{4}}\left(−\pi−\mathrm{6}{ln}\left(\mathrm{2}\right)−\mathrm{2}\gamma+\pi+\mathrm{4}{ln}\left(\mathrm{2}\right)+\mathrm{2}\gamma\right) \\ $$$$=−\frac{\pi}{\mathrm{4}}.−\mathrm{2}{ln}\left(\mathrm{2}\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 25/May/21
$${very}\:{nice}\:.{thanks}\:{alot}… \\ $$
Commented by mindispower last updated on 25/May/21
$${withe}\:{pleasur}\:{sir} \\ $$