Ω :=∫_(−∞) ^( ∞) ((log(2+x^( 2) ))/(4+x^( 2) ))dx=?


Question and Answers Forum

All Questions Topic List
Differentiation Questions
Previous in All Question Next in All Question
Previous in Differentiation Next in Differentiation
Question Number 143783 by mnjuly1970 last updated on 18/Jun/21

$Ω := \int_{- \infty}^{\infty} \frac{l o g (2 + x^{2})}{4 + x^{2}} d x = ?$
	Answered by mindispower last updated on 18/Jun/21

	$f (a) = \int_{- \infty}^{\infty} \frac{l n (2 + 2 {a x}^{2})}{4 + x^{2}}, a ⩾ 0$ $f^{'} (a) = \int_{- \infty}^{\infty} \frac{2 x^{2}}{(4 + x^{2}) (2 + 2 {a x}^{2})}$ $b y r e s i d u e t h e o r e m$ $f^{'} (a) = 2 i π . (\frac{2 {(2 i)}^{2}}{2.2 i (2 + 2 a . {(2 i)}^{2})} + \frac{2 . (\frac{- 1}{a})}{(4 - \frac{1}{a}) . 4 a . \frac{i}{\sqrt{a}}}$ $= 2 i π (. \frac{- 8}{4 i (2 - 8 a)} - \frac{2}{(4 a - 1) . 4 i \sqrt{a}})$ $= \frac{2 π}{4 a - 1} - \frac{π}{(4 a - 1) \sqrt{a}}$ $f (a) = \frac{π}{2} l n ∣ 4 a - 1 ∣ + π \int (\frac{d x}{2 x + 1} - \frac{1}{2 x - 1})$ $= \frac{π}{2} l n ∣ 4 a - 1 ∣ + \frac{π}{2} l n ∣ \frac{2 \sqrt{a} + 1}{2 \sqrt{a} - 1} ∣ + c$ $Missing \left or extra \right$ $\frac{π}{2} l n ∣ 4 a + 4 \sqrt{a} + 1 ∣ + \frac{l n (2) π}{2} = f (a)$ $Ω = f (\frac{1}{2}) = \frac{π}{2} l n (6 + 4 \sqrt{2}) ≃ 3, 85$
	Commented by mnjuly1970 last updated on 18/Jun/21

	$t h a n k s a l o t . . .$
	Commented by mindispower last updated on 18/Jun/21

	$w i t h e p l e a s u r$
	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
	$Φ=∫_(−∞) ^(+∞) ((log(2+x^2 ))/(4+x^2 ))dx ⇒Φ=_(x=2t) ∫_(−∞) ^(+∞) ((log(2+4t^2 ))/(4(1+t^2 )))(2dt) =(1/2) ∫_(−∞) ^(+∞) ((log2 +log(1+2t^2 ))/(1+t^2 ))dt =(1/2)log2 [arctant]_(−∞) ^(+∞) +(1/2)∫_(−∞) ^(+∞) ((log(2t^2 +1))/(t^2 +1))dt =πlog2 +(1/2)∫_(−∞) ^(+∞) ((log(2t^2 +1))/(t^2 +1))dt ϕ(a)=∫_(−∞) ^(+∞) ((log(2t^2 +a))/(t^2 +1))dt (a>0) ⇒ ϕ^′ (a)=∫_(−∞) ^(+∞) (1/((t^2 +1)(2t^2 +a)))dt =(1/2)∫_(−∞) ^(+∞) (dt/((t^2 +1)(t^2 +((√(a/2)))^2 ))) =(1/2)2iπ{Res(ϕ^∼ ,i) +Res(ϕ^∼ ,i(√(a/2)))} ϕ^∼ (z)=(1/((z−i)(z+i)(z−i(√(a/2)))(z+i(√(a/2))))) Res(ϕ,i)=(1/(2i(−1+(a/2)))) Res(ϕ^∼ ,i(√(a/2)))=(1/(((a/2)+1)2i(√(a/2)))) ⇒ ϕ^′ (a)=iπ{(1/(2i((a/2)−1)))+(1/(2i((a/2)+1)(√(a/2))))} =(π/(a−2)) +(π/((a+2)(√(a/2)))) ⇒ ϕ(a)=πlog∣a−2∣ +π∫ (da/((a+2)(√(a/2)))) +K ∫ (da/((a+2)(√(a/2)))) =_((√(a/2))=t) ∫ ((4tdt)/((2t^2 +2)t)) (a=2t^2 ) =4∫ (dt/(2t^2 +2))=2 ∫ (dt/(t^2 +1))=2arctan((√(a/2))) ⇒ ϕ(a)=πlog∣a−2∣+2π arctan((√(a/2))) +K ϕ(o)=∫_(−∞) ^(+∞) ((log(2t^2 ))/(t^2 +1))dt =2∫_0 ^∞ ((log2+2logt)/(t^2 +1))dt =2log2 ×(π/2)+4∫_0 ^∞ ((logt)/(1+t^2 ))dt=πlog2 +0=πlog2+K ⇒K=0 ϕ(a)=πlog∣a−2∣+2πarctan((√(a/2))) ϕ(1)=2πarctan((1/( (√2)))) ⇒ Φ=πlog2 +π arctan((1/( (√2)))) =πlog2+π((π/2)−arctan(√2)) =πlog2+(π^2 /2)−π arctan((√2))$
	$Φ = \int_{- \infty}^{+ \infty} \frac{\log (2 + x^{2})}{4 + x^{2}} dx \Rightarrow Φ =_{x = 2 t} \int_{- \infty}^{+ \infty} \frac{\log (2 + {4 t}^{2})}{4 (1 + t^{2})} (2 dt)$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\log 2 + \log (1 + {2 t}^{2})}{1 + t^{2}} dt$ $= \frac{1}{2} \log 2 {[arctant]}_{- \infty}^{+ \infty} + \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\log ({2 t}^{2} + 1)}{t^{2} + 1} dt$ $= π \log 2 + \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{\log ({2 t}^{2} + 1)}{t^{2} + 1} dt$ $φ (a) = \int_{- \infty}^{+ \infty} \frac{\log ({2 t}^{2} + a)}{t^{2} + 1} dt (a > 0) \Rightarrow$ $φ^{^{'}} (a) = \int_{- \infty}^{+ \infty} \frac{1}{(t^{2} + 1) ({2 t}^{2} + a)} dt$ $= \frac{1}{2} \int_{- \infty}^{+ \infty} \frac{dt}{(t^{2} + 1) (t^{2} + {(\sqrt{\frac{a}{2}})}^{2})}$ $= \frac{1}{2} 2 i π {Res (\tilde{φ}, i) + Res (\tilde{φ}, i \sqrt{\frac{a}{2}})}$ $\tilde{φ} (z) = \frac{1}{(z - i) (z + i) (z - i \sqrt{\frac{a}{2}}) (z + i \sqrt{\frac{a}{2}})}$ $Res (φ, i) = \frac{1}{2 i (- 1 + \frac{a}{2})}$ $Res (\tilde{φ}, i \sqrt{\frac{a}{2}}) = \frac{1}{(\frac{a}{2} + 1) 2 i \sqrt{\frac{a}{2}}} \Rightarrow$ $φ^{^{'}} (a) = i π {\frac{1}{2 i (\frac{a}{2} - 1)} + \frac{1}{2 i (\frac{a}{2} + 1) \sqrt{\frac{a}{2}}}}$ $= \frac{π}{a - 2} + \frac{π}{(a + 2) \sqrt{\frac{a}{2}}} \Rightarrow$ $φ (a) = π \log ∣ a - 2 ∣ + π \int \frac{da}{(a + 2) \sqrt{\frac{a}{2}}} + K$ $\int \frac{da}{(a + 2) \sqrt{\frac{a}{2}}} =_{\sqrt{\frac{a}{2}} = t} \int \frac{4 tdt}{({2 t}^{2} + 2) t} (a = {2 t}^{2})$ $= 4 \int \frac{dt}{{2 t}^{2} + 2} = 2 \int \frac{dt}{t^{2} + 1} = 2 \arctan (\sqrt{\frac{a}{2}}) \Rightarrow$ $φ (a) = π \log ∣ a - 2 ∣ + 2 π \arctan (\sqrt{\frac{a}{2}}) + K$ $φ (o) = \int_{- \infty}^{+ \infty} \frac{\log ({2 t}^{2})}{t^{2} + 1} dt = 2 \int_{0}^{\infty} \frac{\log 2 + 2 logt}{t^{2} + 1} dt$ $= 2 \log 2 \times \frac{π}{2} + 4 \int_{0}^{\infty} \frac{logt}{1 + t^{2}} dt = π \log 2 + 0 = π \log 2 + K \Rightarrow K = 0$ $φ (a) = π \log ∣ a - 2 ∣ + 2 π \arctan (\sqrt{\frac{a}{2}})$ $φ (1) = 2 π \arctan (\frac{1}{\sqrt{2}}) \Rightarrow$ $Φ = π \log 2 + π \arctan (\frac{1}{\sqrt{2}}) = π \log 2 + π (\frac{π}{2} - \arctan \sqrt{2})$ $= π \log 2 + \frac{π^{2}}{2} - π \arctan (\sqrt{2})$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum