log-2-x-2-4-x-2-dx-

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 \dots

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 = \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})