nice-calculus-prove-that-0-ln-1-x-2-1-x-2-2-dx-pi-2-ln-2-pi-4-

Question Number 136555 by mnjuly1970 last updated on 23/Mar/21

\dots . n i c e \dots \dots c a l c u l u s \dots .

p r o v e t h a t

ϕ = \int_{0}^{\infty} \frac{l n (1 + x^{2})}{{(1 + x^{2})}^{2}} d x = \frac{π}{2} l n (2) - \frac{π}{4}

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Answered by mnjuly1970 last updated on 23/Mar/21

f (a) = \int_{0}^{\infty} \frac{d x}{{(1 + x^{2})}^{a}} \dots (a > \frac{1}{2})

f (a) \overset{x^{2} = y}{=} \frac{1}{2} \int_{0}^{\infty} \frac{y^{\frac{- 1}{2}}}{{(1 + y)}^{a}} d y = β (\frac{1}{2}, a - \frac{1}{2})

f (a) = \frac{1}{2} (\frac{\sqrt{π} Γ (a - \frac{1}{2})}{Γ (a)})

g o a l :: ϕ = - f^{'} (2)

f^{'} (a) = \frac{\sqrt{π}}{2} (\frac{Γ^{'} (a - \frac{1}{2}) . Γ (a) - Γ^{'} (a) . Γ (a - \frac{1}{2})}{Γ^{2} (a)})

f^{'} (2) = \frac{\sqrt{π}}{2} (\frac{ψ (\frac{3}{2}) Γ (\frac{3}{2}) - ψ (2) Γ (\frac{3}{2})}{1^{2}})

= \frac{\sqrt{π}}{2} (\frac{\sqrt{π}}{2} (2 - γ - 2 l n (2) - (1 - γ)))

∴ - ϕ = \frac{π}{4} (2 - γ - 2 l n (2) - 1 + γ)

= \frac{π}{4} (1 - 2 l n (2)) = \frac{π}{4} - \frac{π}{2} l n (2) . . ✓

∴ ϕ = \frac{π}{2} l n (2) - \frac{π}{4} \dots .

m . n

Answered by mindispower last updated on 23/Mar/21

= \int_{0}^{\frac{π}{2}} \frac{l n (\frac{1}{{c o s}^{2} (x)})}{{(1 + {t g}^{2} (x))}^{2}} d t g (x)

= \int_{0}^{\frac{π}{2}} - 2 {c o s}^{2} (x) l n (c o s (x)) d x . . A

β (a, b) = 2 \int_{0}^{\frac{π}{2}} {c o s}^{2 a - 1} (x) . {s i n}^{2 b - 1} (x) d x

A = - \frac{1}{2} \partial^{a} β (a, b) ∣ (a, b) = (\frac{3}{2}, \frac{1}{2})

= - \frac{1}{2} β (a, b) (Ψ (a) - Ψ (a + b))

= - \frac{1}{2} . \frac{Γ (\frac{1}{2}) Γ (\frac{3}{2})}{Γ (2)} (Ψ (\frac{3}{2}) - Ψ (2))

= - \frac{1}{2} . \frac{1}{2} . π (2 + Ψ (\frac{1}{2}) - (1 - γ))

= - \frac{π}{4} (1 - 2 l n (2)) = \frac{π l n (2)}{4} - \frac{π}{4}

Commented by mnjuly1970 last updated on 23/Mar/21

t h a n k s a l o t \dots .

Answered by Dwaipayan Shikari last updated on 23/Mar/21

ϑ (α) = \int_{0}^{\infty} \frac{{(1 + x^{2})}^{α}}{{(1 + x^{2})}^{2}} d x = \int_{0}^{\infty} \frac{1}{{(1 + x^{2})}^{2 - α}} d x

= \frac{1}{2} \int_{0}^{\infty} \frac{u^{\frac{1}{2} - 1}}{{(1 + u)}^{\frac{1}{2} + \frac{3}{2} - α}} d u = \frac{1}{2} . \frac{Γ (\frac{1}{2}) Γ (\frac{3}{2} - α)}{Γ (2 - α)}

ϑ^{'} (α) = \frac{- \sqrt{π} Γ (2 - α) Γ^{'} (\frac{3}{2} - α) + \sqrt{π} Γ (2 - α) ψ (2 - α) Γ (\frac{3}{2} - α)}{2 Γ^{2} (2 - α)}

ϑ^{'} (0) = \frac{- \frac{π}{2} ψ (\frac{3}{2}) + \frac{π}{2} ψ (2)}{2} = \frac{π}{4} (- γ + 1 + γ - 2 l o g (2))

= \frac{π}{4} - \frac{π}{2} l o g (2)

- ϑ^{'} (0) = \int_{0}^{\infty} \frac{l o g (x^{2} + 1)}{{(x^{2} + 1)}^{2}} d x = \frac{π}{2} l o g (2) - \frac{π}{4}

Commented by mnjuly1970 last updated on 23/Mar/21

t h a n k i n g \dots