Question Number 143783 by mnjuly1970 last updated on 18/Jun/21
$$ \\ $$$$\:\:\:\:\:\:\:\:\Omega\::=\int_{−\infty} ^{\:\infty} \frac{{log}\left(\mathrm{2}+{x}^{\:\mathrm{2}} \right)}{\mathrm{4}+{x}^{\:\mathrm{2}} }{dx}=? \\ $$$$ \\ $$
Answered by mindispower last updated on 18/Jun/21
![f(a)=∫_(−∞) ^∞ ((ln(2+2ax^2 ))/(4+x^2 )),a≥0 f′(a)=∫_(−∞) ^∞ ((2x^2 )/((4+x^2 )(2+2ax^2 ))) by residue theorem f′(a)=2iπ.(((2(2i)^2 )/(2.2i(2+2a.(2i)^2 )))+((2.(((−1)/a)))/((4−(1/a)).4a.(i/( (√a))))) =2iπ(.((−8)/(4i(2−8a)))−(2/((4a−1).4i(√a)))) =((2π)/(4a−1))−(π/((4a−1)(√a))) f(a)=(π/2)ln∣4a−1∣+π∫((dx/(2x+1))−(1/(2x−1))) =(π/2)ln∣4a−1∣+(π/2)ln∣((2(√a)+1)/(2(√a)−1))∣+c f(0)=c=∫((ln(2))/(4+x^2 ))dx= ((ln(2))/2) arctan(x)]_(−∞) ^∞ =((ln(2)π)/2) (π/2)ln∣4a+4(√a)+1∣+((ln(2)π)/2)=f(a) Ω=f((1/2))=(π/2)ln(6+4(√2))≃3,85](https://www.tinkutara.com/question/Q143834.png)
$${f}\left({a}\right)=\int_{−\infty} ^{\infty} \frac{{ln}\left(\mathrm{2}+\mathrm{2}{ax}^{\mathrm{2}} \right)}{\mathrm{4}+{x}^{\mathrm{2}} },{a}\geqslant\mathrm{0} \\ $$$${f}'\left({a}\right)=\int_{−\infty} ^{\infty} \frac{\mathrm{2}{x}^{\mathrm{2}} }{\left(\mathrm{4}+{x}^{\mathrm{2}} \right)\left(\mathrm{2}+\mathrm{2}{ax}^{\mathrm{2}} \right)} \\ $$$${by}\:{residue}\:{theorem} \\ $$$${f}'\left({a}\right)=\mathrm{2}{i}\pi.\left(\frac{\mathrm{2}\left(\mathrm{2}{i}\right)^{\mathrm{2}} }{\mathrm{2}.\mathrm{2}{i}\left(\mathrm{2}+\mathrm{2}{a}.\left(\mathrm{2}{i}\right)^{\mathrm{2}} \right)}+\frac{\mathrm{2}.\left(\frac{−\mathrm{1}}{{a}}\right)}{\left(\mathrm{4}−\frac{\mathrm{1}}{{a}}\right).\mathrm{4}{a}.\frac{{i}}{\:\sqrt{{a}}}}\right. \\ $$$$=\mathrm{2}{i}\pi\left(.\frac{−\mathrm{8}}{\mathrm{4}{i}\left(\mathrm{2}−\mathrm{8}{a}\right)}−\frac{\mathrm{2}}{\left(\mathrm{4}{a}−\mathrm{1}\right).\mathrm{4}{i}\sqrt{{a}}}\right) \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{4}{a}−\mathrm{1}}−\frac{\pi}{\left(\mathrm{4}{a}−\mathrm{1}\right)\sqrt{{a}}} \\ $$$${f}\left({a}\right)=\frac{\pi}{\mathrm{2}}{ln}\mid\mathrm{4}{a}−\mathrm{1}\mid+\pi\int\left(\frac{{dx}}{\mathrm{2}{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}{ln}\mid\mathrm{4}{a}−\mathrm{1}\mid+\frac{\pi}{\mathrm{2}}{ln}\mid\frac{\mathrm{2}\sqrt{{a}}+\mathrm{1}}{\mathrm{2}\sqrt{{a}}−\mathrm{1}}\mid+{c} \\ $$$$\left.{f}\left(\mathrm{0}\right)={c}=\int\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{4}+{x}^{\mathrm{2}} }{dx}=\:\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}\:{arctan}\left({x}\right)\overset{\infty} {\right]}_{−\infty} =\frac{{ln}\left(\mathrm{2}\right)\pi}{\mathrm{2}} \\ $$$$\frac{\pi}{\mathrm{2}}{ln}\mid\mathrm{4}{a}+\mathrm{4}\sqrt{{a}}+\mathrm{1}\mid+\frac{{ln}\left(\mathrm{2}\right)\pi}{\mathrm{2}}={f}\left({a}\right) \\ $$$$\Omega={f}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{6}+\mathrm{4}\sqrt{\mathrm{2}}\right)\simeq\mathrm{3},\mathrm{85} \\ $$
Commented by mnjuly1970 last updated on 18/Jun/21
$${thanks}\:{alot}… \\ $$
Commented by mindispower last updated on 18/Jun/21
$${withe}\:{pleasur} \\ $$
Answered by mnjuly1970 last updated on 18/Jun/21
Answered by mnjuly1970 last updated on 18/Jun/21
Answered by mathmax by abdo last updated on 18/Jun/21
![Φ=∫_(−∞) ^(+∞) ((log(2+x^2 ))/(4+x^2 ))dx ⇒Φ=_(x=2t) ∫_(−∞) ^(+∞) ((log(2+4t^2 ))/(4(1+t^2 )))(2dt) =(1/2) ∫_(−∞) ^(+∞) ((log2 +log(1+2t^2 ))/(1+t^2 ))dt =(1/2)log2 [arctant]_(−∞) ^(+∞) +(1/2)∫_(−∞) ^(+∞) ((log(2t^2 +1))/(t^2 +1))dt =πlog2 +(1/2)∫_(−∞) ^(+∞) ((log(2t^2 +1))/(t^2 +1))dt ϕ(a)=∫_(−∞) ^(+∞) ((log(2t^2 +a))/(t^2 +1))dt (a>0) ⇒ ϕ^′ (a)=∫_(−∞) ^(+∞) (1/((t^2 +1)(2t^2 +a)))dt =(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)(t^2 +((√(a/2)))^2 ))) =(1/2)2iπ{Res(ϕ^∼ ,i) +Res(ϕ^∼ ,i(√(a/2)))} ϕ^∼ (z)=(1/((z−i)(z+i)(z−i(√(a/2)))(z+i(√(a/2))))) Res(ϕ,i)=(1/(2i(−1+(a/2)))) Res(ϕ^∼ ,i(√(a/2)))=(1/(((a/2)+1)2i(√(a/2)))) ⇒ ϕ^′ (a)=iπ{(1/(2i((a/2)−1)))+(1/(2i((a/2)+1)(√(a/2))))} =(π/(a−2)) +(π/((a+2)(√(a/2)))) ⇒ ϕ(a)=πlog∣a−2∣ +π∫ (da/((a+2)(√(a/2)))) +K ∫ (da/((a+2)(√(a/2)))) =_((√(a/2))=t) ∫ ((4tdt)/((2t^2 +2)t)) (a=2t^2 ) =4∫ (dt/(2t^2 +2))=2 ∫ (dt/(t^2 +1))=2arctan((√(a/2))) ⇒ ϕ(a)=πlog∣a−2∣+2π arctan((√(a/2))) +K ϕ(o)=∫_(−∞) ^(+∞) ((log(2t^2 ))/(t^2 +1))dt =2∫_0 ^∞ ((log2+2logt)/(t^2 +1))dt =2log2 ×(π/2)+4∫_0 ^∞ ((logt)/(1+t^2 ))dt=πlog2 +0=πlog2+K ⇒K=0 ϕ(a)=πlog∣a−2∣+2πarctan((√(a/2))) ϕ(1)=2πarctan((1/( (√2)))) ⇒ Φ=πlog2 +π arctan((1/( (√2)))) =πlog2+π((π/2)−arctan(√2)) =πlog2+(π^2 /2)−π arctan((√2))](https://www.tinkutara.com/question/Q143848.png)
$$\Phi=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2}+\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{4}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\Phi=_{\mathrm{x}=\mathrm{2t}} \:\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2}+\mathrm{4t}^{\mathrm{2}} \right)}{\mathrm{4}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\left(\mathrm{2dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log2}\:+\mathrm{log}\left(\mathrm{1}+\mathrm{2t}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log2}\:\left[\mathrm{arctant}\right]_{−\infty} ^{+\infty} \:+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2t}^{\mathrm{2}} \:+\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt} \\ $$$$=\pi\mathrm{log2}\:+\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt} \\ $$$$\varphi\left(\mathrm{a}\right)=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2t}^{\mathrm{2}} \:+\mathrm{a}\right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:\:\:\:\:\:\left(\mathrm{a}>\mathrm{0}\right)\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{1}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{2t}^{\mathrm{2}} \:+\mathrm{a}\right)}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{−\infty} ^{+\infty} \:\frac{\mathrm{dt}}{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)\left(\mathrm{t}^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{2i}\pi\left\{\mathrm{Res}\left(\overset{\sim} {\varphi}\:\:,\mathrm{i}\right)\:+\mathrm{Res}\left(\overset{\sim} {\varphi}\:,\mathrm{i}\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)\right\} \\ $$$$\overset{\sim} {\varphi}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\left(\mathrm{z}−\mathrm{i}\right)\left(\mathrm{z}+\mathrm{i}\right)\left(\mathrm{z}−\mathrm{i}\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)\left(\mathrm{z}+\mathrm{i}\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)} \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)=\frac{\mathrm{1}}{\mathrm{2i}\left(−\mathrm{1}+\frac{\mathrm{a}}{\mathrm{2}}\right)} \\ $$$$\mathrm{Res}\left(\overset{\sim} {\varphi},\mathrm{i}\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\left(\frac{\mathrm{a}}{\mathrm{2}}+\mathrm{1}\right)\mathrm{2i}\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}}\:\Rightarrow \\ $$$$\varphi^{'} \left(\mathrm{a}\right)=\mathrm{i}\pi\left\{\frac{\mathrm{1}}{\mathrm{2i}\left(\frac{\mathrm{a}}{\mathrm{2}}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2i}\left(\frac{\mathrm{a}}{\mathrm{2}}+\mathrm{1}\right)\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}}\right\} \\ $$$$=\frac{\pi}{\mathrm{a}−\mathrm{2}}\:+\frac{\pi}{\left(\mathrm{a}+\mathrm{2}\right)\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}}\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)=\pi\mathrm{log}\mid\mathrm{a}−\mathrm{2}\mid\:+\pi\int\:\:\frac{\mathrm{da}}{\left(\mathrm{a}+\mathrm{2}\right)\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}}\:\:+\mathrm{K} \\ $$$$\int\:\:\:\frac{\mathrm{da}}{\left(\mathrm{a}+\mathrm{2}\right)\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}}\:=_{\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}=\mathrm{t}} \:\:\int\:\:\:\:\:\frac{\mathrm{4tdt}}{\left(\mathrm{2t}^{\mathrm{2}} +\mathrm{2}\right)\mathrm{t}}\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{a}=\mathrm{2t}^{\mathrm{2}} \right) \\ $$$$=\mathrm{4}\int\:\frac{\mathrm{dt}}{\mathrm{2t}^{\mathrm{2}} \:+\mathrm{2}}=\mathrm{2}\:\int\:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}=\mathrm{2arctan}\left(\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\varphi\left(\mathrm{a}\right)=\pi\mathrm{log}\mid\mathrm{a}−\mathrm{2}\mid+\mathrm{2}\pi\:\mathrm{arctan}\left(\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right)\:+\mathrm{K} \\ $$$$\varphi\left(\mathrm{o}\right)=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{log}\left(\mathrm{2t}^{\mathrm{2}} \right)}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt}\:=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log2}+\mathrm{2logt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\mathrm{dt} \\ $$$$=\mathrm{2log2}\:×\frac{\pi}{\mathrm{2}}+\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{logt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\pi\mathrm{log2}\:+\mathrm{0}=\pi\mathrm{log2}+\mathrm{K}\:\Rightarrow\mathrm{K}=\mathrm{0} \\ $$$$\varphi\left(\mathrm{a}\right)=\pi\mathrm{log}\mid\mathrm{a}−\mathrm{2}\mid+\mathrm{2}\pi\mathrm{arctan}\left(\sqrt{\frac{\mathrm{a}}{\mathrm{2}}}\right) \\ $$$$\varphi\left(\mathrm{1}\right)=\mathrm{2}\pi\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\Phi=\pi\mathrm{log2}\:+\pi\:\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\pi\mathrm{log2}+\pi\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\sqrt{\mathrm{2}}\right) \\ $$$$=\pi\mathrm{log2}+\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\pi\:\mathrm{arctan}\left(\sqrt{\mathrm{2}}\right) \\ $$